Chapter 16 Probability

Given:

Two coins are tossed once. The coins are distinguishable: one is a one rupee coin and the other is a two rupee coin.

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To Find:

The sample space of the experiment.

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Solution:

Let H represent the outcome 'Heads' and T represent the outcome 'Tails' for a single coin toss.

When the one rupee coin is tossed, the possible outcomes are H or T.

When the two rupee coin is tossed, the possible outcomes are H or T.

Since the two coins are distinct, we consider the outcome for each coin separately. We can represent the outcome as an ordered pair where the first element is the outcome of the one rupee coin and the second element is the outcome of the two rupee coin.

Possible outcomes for the one rupee coin: {H, T}

Possible outcomes for the two rupee coin: {H, T}

The sample space S is the set of all possible combinations of outcomes from the two coins.

Using a table to list the possibilities:

One Rupee Coin	Two Rupee Coin	Combined Outcome (Sample Point)
H	H	(H, H)
H	T	(H, T)
T	H	(T, H)
T	T	(T, T)

The sample space S is the set containing all these combined outcomes.

$S = \{ (H, H), (H, T), (T, H), (T, T) \}$

The number of elements in the sample space is $n(S) = 4$.

Given:

A pair of dice (one blue and one red) is rolled once. The dice are distinguishable by color.

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To Find:

The sample space associated with this experiment and the number of elements in the sample space.

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Solution:

When a single die is rolled, the possible outcomes are the numbers 1, 2, 3, 4, 5, or 6. So, there are 6 possible outcomes for a single die.

In this experiment, we are rolling two distinguishable dice (blue and red).

Let the outcome of the blue die be the first component of an ordered pair, and the outcome of the red die be the second component.

The possible outcomes for the blue die are $\{1, 2, 3, 4, 5, 6\}$.

The possible outcomes for the red die are $\{1, 2, 3, 4, 5, 6\}$.

The sample space $S$ is the set of all possible ordered pairs $(b, r)$, where $b$ is the outcome of the blue die and $r$ is the outcome of the red die.

We can list the sample points systematically:

Red Die $\to$	1	2	3	4	5	6
Blue Die $\downarrow$
1	(1, 1)	(1, 2)	(1, 3)	(1, 4)	(1, 5)	(1, 6)
2	(2, 1)	(2, 2)	(2, 3)	(2, 4)	(2, 5)	(2, 6)
3	(3, 1)	(3, 2)	(3, 3)	(3, 4)	(3, 5)	(3, 6)
4	(4, 1)	(4, 2)	(4, 3)	(4, 4)	(4, 5)	(4, 6)
5	(5, 1)	(5, 2)	(5, 3)	(5, 4)	(5, 5)	(5, 6)
6	(6, 1)	(6, 2)	(6, 3)	(6, 4)	(6, 5)	(6, 6)

The sample space $S$ is the set of all these 36 ordered pairs:

$S = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),$

$(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),$

$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),$

$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),$

$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),$

$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}$

The number of elements in the sample space, $n(S)$, is the total count of these ordered pairs.

Since there are 6 possible outcomes for the blue die and 6 possible outcomes for the red die, the total number of outcomes when rolling both is the product of the individual outcomes.

$n(S) = (\text{Number of outcomes for blue die}) \times (\text{Number of outcomes for red die})$

$n(S) = 6 \times 6 = 36$

Thus, the sample space consists of 36 elements.

Part (i):

Given:

A boy has three distinguishable coins: a $\textsf{₹}1$ coin, a $\textsf{₹}2$ coin, and a $\textsf{₹}5$ coin. He takes out two coins one after the other (order matters, and sampling is without replacement).

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To Find:

The sample space for the experiment.

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Solution:

Let the coins be denoted by their values: $C_1$ (for $\textsf{₹}1$), $C_2$ (for $\textsf{₹}2$), and $C_5$ (for $\textsf{₹}5$).

The boy takes out one coin first, then another from the remaining two coins.

The outcomes are ordered pairs representing the value of the first coin drawn and the value of the second coin drawn.

Possible outcomes for the first draw: $\textsf{₹}1, \textsf{₹}2, \textsf{₹}5$.

Possible outcomes for the second draw (given the first draw):

• If the first coin is $\textsf{₹}1$, the remaining coins are $\textsf{₹}2$ and $\textsf{₹}5$. Possible second draws: $\textsf{₹}2, \textsf{₹}5$. Outcomes: $(\textsf{₹}1, \textsf{₹}2), (\textsf{₹}1, \textsf{₹}5)$.
• If the first coin is $\textsf{₹}2$, the remaining coins are $\textsf{₹}1$ and $\textsf{₹}5$. Possible second draws: $\textsf{₹}1, \textsf{₹}5$. Outcomes: $(\textsf{₹}2, \textsf{₹}1), (\textsf{₹}2, \textsf{₹}5)$.
• If the first coin is $\textsf{₹}5$, the remaining coins are $\textsf{₹}1$ and $\textsf{₹}2$. Possible second draws: $\textsf{₹}1, \textsf{₹}2$. Outcomes: $(\textsf{₹}5, \textsf{₹}1), (\textsf{₹}5, \textsf{₹}2)$.

The sample space S is the set of all these possible ordered pairs.

$S = \{ (\textsf{₹}1, \textsf{₹}2), (\textsf{₹}1, \textsf{₹}5), (\textsf{₹}2, \textsf{₹}1), (\textsf{₹}2, \textsf{₹}5), (\textsf{₹}5, \textsf{₹}1), (\textsf{₹}5, \textsf{₹}2) \}$

The number of elements in the sample space is $n(S) = 6$. This is a permutation problem: $P(3, 2) = \frac{3!}{(3-2)!} = \frac{6}{1} = 6$.

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Part (ii):

Given:

A person is noting down the number of accidents along a busy highway during a year.

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To Find:

An appropriate sample space for this experiment.

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Solution:

The number of accidents along a highway in a year must be a non-negative integer. It can be 0 (no accidents), 1, 2, 3, and so on. Theoretically, there is no definite upper limit to the number of accidents that can occur, although in practice it would be finite.

Therefore, an appropriate sample space for this experiment is the set of all non-negative integers, representing the possible count of accidents.

The sample space S can be represented as:

$S = \{0, 1, 2, 3, 4, ... \}$

This is an infinite sample space, but it appropriately represents all possible outcomes for the number of accidents.

Given:

An experiment is conducted in two stages:

1. A coin is tossed. The possible outcomes are Head (H) or Tail (T).

2. The second stage depends on the outcome of the first:

• If the outcome is a Head (H), a ball is drawn from a bag containing 3 blue and 4 white balls.
• If the outcome is a Tail (T), a standard six-sided die is thrown.

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To Find:

The sample space of this entire experiment.

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Solution:

The sample space is the set of all possible outcomes of the experiment. We can find all outcomes by considering the two possible results of the coin toss.

Let's denote the 3 blue balls as $B_1, B_2, B_3$ and the 4 white balls as $W_1, W_2, W_3, W_4$.

Case 1: The coin toss results in a Head (H).

Following this outcome, a ball is drawn from the bag. The possible balls that can be drawn are any of the 7 balls. The outcomes for this branch of the experiment are represented as ordered pairs (Coin Toss, Ball Drawn):

$(H, B_1), (H, B_2), (H, B_3), (H, W_1), (H, W_2), (H, W_3), (H, W_4)$

Case 2: The coin toss results in a Tail (T).

Following this outcome, a die is thrown. The possible results from rolling a die are the numbers 1, 2, 3, 4, 5, and 6. The outcomes for this branch of the experiment are represented as ordered pairs (Coin Toss, Die Roll):

$(T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)$

The complete sample space (S) for the entire experiment is the collection of all possible outcomes from both cases.

Therefore, the sample space S is:

$S = \{ (H, B_1), (H, B_2), (H, B_3), (H, W_1), (H, W_2), (H, W_3), (H, W_4), \ $$ (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6) \}$

Given:

A coin is tossed repeatedly until a head comes up.

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To Find:

The sample space of this experiment.

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Solution:

In this experiment, we continue tossing the coin until we observe the first Head (H). Let T represent the outcome Tail.

The possible outcomes are:

• The first toss is a Head. The experiment stops. The outcome is H.
• The first toss is a Tail, and the second toss is a Head. The experiment stops. The outcome is TH.
• The first two tosses are Tails, and the third toss is a Head. The experiment stops. The outcome is TTH.
• The first three tosses are Tails, and the fourth toss is a Head. The experiment stops. The outcome is TTTH.

This process can continue indefinitely, with any number of consecutive Tails followed by a single Head.

The sample space S is the set of all possible sequences of outcomes until the first Head appears.

$S = \{ H, TH, TTH, TTTH, TTTTH, ... \}$

Each element in the sample space is a sequence of zero or more Tails followed by exactly one Head.

This is an example of an infinite sample space.

Given:

A coin is tossed three times.

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To Find:

The sample space for this experiment.

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Solution:

When a single coin is tossed, there are two possible outcomes: Head (H) and Tail (T).

Since the coin is tossed three times, the outcome of the experiment is an ordered sequence of three results, where each result is either H or T.

We can determine the sample space by listing all possible combinations:

• First toss: H or T
• Second toss: H or T
• Third toss: H or T

Let's systematically list all the possible ordered triplets:

If the first toss is H:

• Second toss is H:
• Third toss is H: HHH
• Third toss is T: HHT
• Second toss is T:
• Third toss is H: HTH
• Third toss is T: HTT

If the first toss is T:

• Second toss is H:
• Third toss is H: THH
• Third toss is T: THT
• Second toss is T:
• Third toss is H: TTH
• Third toss is T: TTT

The sample space $S$ is the set containing all these possible outcomes:

$S = \{ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT \}$

The number of elements in the sample space is $n(S) = 2^3 = 8$, as there are 2 outcomes for each of the 3 independent tosses.

Given:

A die is thrown two times.

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To Find:

The sample space for this experiment.

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Solution:

When a standard six-sided die is thrown once, the set of possible outcomes is $\{1, 2, 3, 4, 5, 6\}$.

Since the die is thrown two times, the outcome of the experiment is an ordered pair where the first element is the result of the first throw and the second element is the result of the second throw.

Let $(i, j)$ represent an outcome where $i$ is the result of the first throw and $j$ is the result of the second throw, with $i, j \in \{1, 2, 3, 4, 5, 6\}$.

The sample space $S$ is the set of all such ordered pairs:

Second Throw $\to$	1	2	3	4	5	6
First Throw $\downarrow$
1	(1, 1)	(1, 2)	(1, 3)	(1, 4)	(1, 5)	(1, 6)
2	(2, 1)	(2, 2)	(2, 3)	(2, 4)	(2, 5)	(2, 6)
3	(3, 1)	(3, 2)	(3, 3)	(3, 4)	(3, 5)	(3, 6)
4	(4, 1)	(4, 2)	(4, 3)	(4, 4)	(4, 5)	(4, 6)
5	(5, 1)	(5, 2)	(5, 3)	(5, 4)	(5, 5)	(5, 6)
6	(6, 1)	(6, 2)	(6, 3)	(6, 4)	(6, 5)	(6, 6)

So, the sample space $S$ is:

$S = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),$

$(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),$

$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),$

$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),$

$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),$

$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}$

The number of elements in the sample space is $n(S) = 6 \times 6 = 36$, since there are 6 possible outcomes for the first throw and 6 possible outcomes for the second throw, and the throws are independent.

Given:

A coin is tossed four times.

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To Find:

The sample space for this experiment.

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Solution:

When a coin is tossed once, there are two possible outcomes: Head (H) and Tail (T).

Since the coin is tossed four times, the outcome of the experiment is an ordered sequence of four results, where each result is either H or T.

The total number of possible outcomes is $2^4 = 16$.

The sample space $S$ is the set of all such ordered sequences of length four.

Let's list all the possible outcomes:

• All Heads: HHHH
• Three Heads and one Tail: HHHT, HHTH, HTHH, THHH
• Two Heads and two Tails: HHTT, HTHT, HTTH, THHT, THTH, TTHH
• One Head and three Tails: HTTT, THTT, TTHT, TTTH
• All Tails: TTTT

Combining these, the sample space S is:

$S = \{ HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, \ $$ HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, \ $$ TTTH, TTTT \}$

The number of elements in the sample space is $n(S) = 16$.

Given:

An experiment where a coin is tossed and a die is thrown.

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To Find:

The sample space for this experiment.

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Solution:

This experiment consists of two independent parts: tossing a coin and throwing a die.

The possible outcomes when tossing a coin are Head (H) and Tail (T).

The possible outcomes when throwing a standard six-sided die are the numbers 1, 2, 3, 4, 5, and 6.

The outcome of the combined experiment is an ordered pair, where the first element is the result of the coin toss and the second element is the result of the die throw.

Let's list all possible combinations:

• If the coin toss is Head (H), the die throw can be any number from 1 to 6. This gives the outcomes: (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6).
• If the coin toss is Tail (T), the die throw can be any number from 1 to 6. This gives the outcomes: (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6).

The sample space $S$ is the set of all these possible ordered pairs:

$S = \{ (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6),$

$(T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6) \}$

The number of elements in the sample space is $n(S) = (\text{Number of outcomes for coin}) \times (\text{Number of outcomes for die})$

$n(S) = 2 \times 6 = 12$

Given:

An experiment where a coin is tossed, and if a head is obtained, a die is rolled. If a tail is obtained, the experiment stops.

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To Find:

The sample space for this experiment.

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Solution:

Let H represent the outcome 'Head' and T represent the outcome 'Tail' for the coin toss.

Let the outcomes of the die roll be the numbers 1, 2, 3, 4, 5, 6.

The experiment proceeds as follows:

• The first step is tossing the coin. The outcomes are H or T.
• If the coin shows Head (H), then a die is rolled. The possible outcomes for the die are 1, 2, 3, 4, 5, or 6. The combined outcome is represented as a pair (Coin outcome, Die outcome). The possible outcomes are (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6).
• If the coin shows Tail (T), the experiment ends. The outcome is just T.

The sample space S is the set of all distinct possible outcomes of the entire experiment.

Combining the outcomes from both cases, the sample space S is:

$S = \{ (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), T \}$

Note that the outcome 'T' is a single outcome, while the outcomes (H, i) are pairs, representing the sequence of events.

The number of elements in the sample space is $n(S) = 6 + 1 = 7$.

Given:

Room X contains 2 boys and 2 girls.

Room Y contains 1 boy and 3 girls.

An experiment is conducted by first selecting a room and then selecting a person from that room.

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To Find:

The sample space for this experiment.

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Solution:

Let's denote the rooms as X and Y.

Let the boys in Room X be $B_{X1}$ and $B_{X2}$.

Let the girls in Room X be $G_{X1}$ and $G_{X2}$.

Let the boy in Room Y be $B_{Y1}$.

Let the girls in Room Y be $G_{Y1}$, $G_{Y2}$, and $G_{Y3}$.

The experiment consists of two stages:

Stage 1: Select a room. The possible outcomes are selecting Room X or selecting Room Y.

Stage 2: Select a person from the chosen room.

If Room X is selected in Stage 1, the possible persons in Room X are $B_{X1}, B_{X2}, G_{X1}, G_{X2}$.

The outcomes in this case are represented as an ordered pair (Room, Person).

• Select Room X, then select $B_{X1}$: $(X, B_{X1})$
• Select Room X, then select $B_{X2}$: $(X, B_{X2})$
• Select Room X, then select $G_{X1}$: $(X, G_{X1})$
• Select Room X, then select $G_{X2}$: $(X, G_{X2})$

If Room Y is selected in Stage 1, the possible persons in Room Y are $B_{Y1}, G_{Y1}, G_{Y2}, G_{Y3}$.

The outcomes in this case are represented as an ordered pair (Room, Person).

• Select Room Y, then select $B_{Y1}$: $(Y, B_{Y1})$
• Select Room Y, then select $G_{Y1}$: $(Y, G_{Y1})$
• Select Room Y, then select $G_{Y2}$: $(Y, G_{Y2})$
• Select Room Y, then select $G_{Y3}$: $(Y, G_{Y3})$

The sample space S is the set of all these distinct possible outcomes.

$S = \{ (X, B_{X1}), (X, B_{X2}), (X, G_{X1}), (X, G_{X2}), (Y, B_{Y1}), (Y, G_{Y1}), \ $$ (Y, G_{Y2}), (Y, G_{Y3}) \}$

The number of elements in the sample space is $n(S) = 4 + 4 = 8$.

Given:

A bag contains three distinguishable dice: one red, one white, and one blue.

An experiment is conducted in two stages: first, one die is selected at random, and then the selected die is rolled. The colour of the die and the number on its uppermost face are noted.

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To Find:

The sample space for this experiment.

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Solution:

The experiment involves selecting one of the three coloured dice and then rolling it.

Let the colours of the dice be represented by R (Red), W (White), and B (Blue).

When a standard die is rolled, the possible numbers on the uppermost face are $\{1, 2, 3, 4, 5, 6\}$.

The outcome of the experiment is an ordered pair representing the colour of the selected die and the number obtained on rolling it.

Let's list the possible outcomes based on which die is selected:

Case 1: The Red die is selected.

The possible outcomes when rolling the red die are 1, 2, 3, 4, 5, or 6.

The combined outcomes for this case are: (R, 1), (R, 2), (R, 3), (R, 4), (R, 5), (R, 6).

Case 2: The White die is selected.

The possible outcomes when rolling the white die are 1, 2, 3, 4, 5, or 6.

The combined outcomes for this case are: (W, 1), (W, 2), (W, 3), (W, 4), (W, 5), (W, 6).

Case 3: The Blue die is selected.

The possible outcomes when rolling the blue die are 1, 2, 3, 4, 5, or 6.

The combined outcomes for this case are: (B, 1), (B, 2), (B, 3), (B, 4), (B, 5), (B, 6).

The sample space S is the set of all these possible outcomes from the three cases.

$S = \{ (R, 1), (R, 2), (R, 3), (R, 4), (R, 5), (R, 6),$

$(W, 1), (W, 2), (W, 3), (W, 4), (W, 5), (W, 6),$

$(B, 1), (B, 2), (B, 3), (B, 4), (B, 5), (B, 6) \}$

The number of elements in the sample space is $n(S) = 3 \times 6 = 18$, as there are 3 choices for the die colour and 6 possible outcomes for the roll.

Given:

An experiment recording the boy-girl composition of families with 2 children.

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To Find:

(i) The sample space when the order of births (boy or girl) is considered.

(ii) The sample space when the number of girls in the family is considered.

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Solution:

Part (i): Sample space when the order of births is considered.

Let B represent a Boy and G represent a Girl.

Since there are 2 children and we are interested in the order of their births, we list the possible sequences of genders for the two children.

The first child can be a Boy or a Girl.

The second child can be a Boy or a Girl.

The possible ordered outcomes are:

• First child is Boy, Second child is Boy: BB
• First child is Boy, Second child is Girl: BG
• First child is Girl, Second child is Boy: GB
• First child is Girl, Second child is Girl: GG

The sample space $S_1$ for this part of the experiment is the set of these ordered pairs of genders:

$S_1 = \{ BB, BG, GB, GG \}$

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Part (ii): Sample space when the number of girls in the family is considered.

In this case, we are only interested in the count of girls among the two children, regardless of the order of birth.

Referring to the outcomes from part (i):

• Outcome BB: Number of girls = 0
• Outcome BG: Number of girls = 1
• Outcome GB: Number of girls = 1
• Outcome GG: Number of girls = 2

The possible number of girls in a family with 2 children are 0, 1, or 2.

The sample space $S_2$ for this part of the experiment is the set of these possible numbers of girls:

$S_2 = \{ 0, 1, 2 \}$

Given:

A box contains 1 red ball (R) and 3 identical white balls (W).

Two balls are drawn at random in succession without replacement.

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To Find:

The sample space for this experiment.

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Solution:

Let R represent the red ball and W represent a white ball. Since the white balls are identical, we cannot distinguish between them when we draw a white ball; the outcome is simply 'White'.

We are drawing two balls in succession without replacement, which means the order of the draws matters, and the first ball drawn is not returned to the box before the second draw.

Let's consider the possible outcomes for the first and second draw:

Case 1: The first ball drawn is Red (R).

If the first ball is Red, only the single red ball can be drawn. After drawing the red ball, there are 3 identical white balls remaining in the box.

The second ball drawn must be White (W).

The outcome for this case is the ordered pair (Red, White).

Case 2: The first ball drawn is White (W).

If the first ball is White, one of the three white balls is drawn. After drawing one white ball, there is 1 red ball and 2 identical white balls remaining in the box.

For the second draw, we can draw either the Red ball (R) or one of the remaining White balls (W).

• If the second ball is Red (R): The outcome is the ordered pair (White, Red).
• If the second ball is White (W): The outcome is the ordered pair (White, White).

The sample space S is the set of all these distinct possible ordered outcomes from the two cases:

$S = \{ (R, W), (W, R), (W, W) \}$

The number of elements in the sample space is $n(S) = 3$.

Given:

An experiment with the following steps:

1. Toss a coin.

2. If the result is Head (H), toss the coin a second time.

3. If the result is Tail (T), roll a die once.

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To Find:

The sample space for this experiment.

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Solution:

Let H represent the outcome 'Head' and T represent the outcome 'Tail' for a coin toss.

Let the outcomes of rolling a die be the numbers 1, 2, 3, 4, 5, 6.

We need to list all possible outcomes based on the sequence of events in the experiment.

Case 1: The first toss is Head (H).

According to the experiment description, if the first toss is H, the coin is tossed a second time.

The possible outcomes for the second toss are H or T.

The combined outcomes for this case are represented as ordered pairs (first toss, second toss):

• First toss H, Second toss H: (H, H)
• First toss H, Second toss T: (H, T)

Case 2: The first toss is Tail (T).

According to the experiment description, if the first toss is T, a die is rolled once.

The possible outcomes for the die roll are 1, 2, 3, 4, 5, or 6.

The combined outcomes for this case are represented as ordered pairs (first toss, die result):

• First toss T, Die result 1: (T, 1)
• First toss T, Die result 2: (T, 2)
• First toss T, Die result 3: (T, 3)
• First toss T, Die result 4: (T, 4)
• First toss T, Die result 5: (T, 5)
• First toss T, Die result 6: (T, 6)

The sample space S is the set of all these distinct possible outcomes from both cases.

$S = \{ (H, H), (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6) \}$

The number of elements in the sample space is $n(S) = 2 + 6 = 8$.

Given:

Three bulbs are selected at random from a lot.

Each selected bulb is classified as Defective (D) or Non-defective (N).

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To Find:

The sample space of this experiment.

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Solution:

The experiment consists of selecting three bulbs and classifying each one.

For each bulb selected, there are two possible outcomes: it is either Defective (D) or Non-defective (N).

Since there are 3 bulbs selected, and the classification of each bulb is independent of the others, the outcome of the experiment is an ordered sequence of classifications for the three bulbs.

Let's list all possible sequences of outcomes for the three bulbs:

• First bulb: D or N
• Second bulb: D or N
• Third bulb: D or N

We can systematically list all the possible combinations of D and N for the three positions:

• All three are Defective: DDD
• Two Defective, one Non-defective: DDN, DND, NDD
• One Defective, two Non-defective: DNN, NDN, NND
• All three are Non-defective: NNN

The sample space $S$ is the set of all these possible ordered outcomes of length three.

$S = \{ DDD, DDN, DND, DNN, NDD, NDN, NND, NNN \}$

The number of elements in the sample space is $n(S) = 2^3 = 8$, as there are 2 possible outcomes for each of the 3 bulbs.

Given:

An experiment involving a coin toss and potential die rolls based on the coin outcome.

The sequence of events is:

1. Toss a coin.

2. If Head (H) occurs, throw a die.

3. If the die shows an even number after getting a Head on the coin, throw the die again.

4. If Tail (T) occurs on the first toss, the experiment stops.

5. If the die shows an odd number after getting a Head on the coin, the experiment stops.

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To Find:

The sample space for this experiment.

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Solution:

Let H denote Head and T denote Tail for the coin toss.

Let the possible outcomes for a die roll be $D = \{1, 2, 3, 4, 5, 6\}$.

We examine the outcomes based on the result of the first coin toss.

Case 1: The first toss is Tail (T).

If the first toss is T, the experiment stops. The outcome is simply T.

Case 2: The first toss is Head (H).

If the first toss is H, a die is thrown. Let the result of this first die throw be $d_1 \in D$.

• If $d_1$ is an odd number ($d_1 \in \{1, 3, 5\}$), the experiment stops. The outcome is the ordered pair $(H, d_1)$. The possible outcomes here are (H, 1), (H, 3), (H, 5).
• If $d_1$ is an even number ($d_1 \in \{2, 4, 6\}$), the die is thrown again. Let the result of this second die throw be $d_2 \in D$. The outcome is the ordered triplet $(H, d_1, d_2)$.

Listing the outcomes when the first toss is H and the first die roll is even ($d_1 \in \{2, 4, 6\}$):

• If $d_1 = 2$, the second die can be any value from 1 to 6. Outcomes: (H, 2, 1), (H, 2, 2), (H, 2, 3), (H, 2, 4), (H, 2, 5), (H, 2, 6).
• If $d_1 = 4$, the second die can be any value from 1 to 6. Outcomes: (H, 4, 1), (H, 4, 2), (H, 4, 3), (H, 4, 4), (H, 4, 5), (H, 4, 6).
• If $d_1 = 6$, the second die can be any value from 1 to 6. Outcomes: (H, 6, 1), (H, 6, 2), (H, 6, 3), (H, 6, 4), (H, 6, 5), (H, 6, 6).

The sample space S is the set of all distinct possible outcomes from both cases.

$S = \{ T, (H, 1), (H, 3), (H, 5), (H, 2, 1), (H, 2, 2), (H, 2, 3), (H, 2, 4), \ $$ (H, 2, 5), (H, 2, 6), (H, 4, 1), (H, 4, 2), (H, 4, 3), (H, 4, 4), (H, 4, 5), \ $$ (H, 4, 6), (H, 6, 1), (H, 6, 2), (H, 6, 3), (H, 6, 4), (H, 6, 5), (H, 6, 6) \}$

The number of elements in the sample space is $n(S) = 1 (\text{for T}) + 3 (\text{for H, odd}) + 3 \times 6 (\text{for H, even, any}) \ $$ = 1 + 3 + 18 = 22$.

Given:

Four slips of paper with numbers 1, 2, 3, and 4 written on them.

Two slips are drawn from the box one after the other without replacement.

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To Find:

The sample space for this experiment.

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Solution:

Since the slips are drawn one after the other and without replacement, the order of the drawn slips matters, and the same slip cannot be drawn twice.

The outcome of the experiment is an ordered pair $(a, b)$, where $a$ is the number on the first slip drawn and $b$ is the number on the second slip drawn, with $a \neq b$ and $a, b \in \{1, 2, 3, 4\}$.

Let's list all possible ordered pairs:

• If the first slip drawn is 1, the second slip can be 2, 3, or 4. The outcomes are (1, 2), (1, 3), (1, 4).
• If the first slip drawn is 2, the second slip can be 1, 3, or 4. The outcomes are (2, 1), (2, 3), (2, 4).
• If the first slip drawn is 3, the second slip can be 1, 2, or 4. The outcomes are (3, 1), (3, 2), (3, 4).
• If the first slip drawn is 4, the second slip can be 1, 2, or 3. The outcomes are (4, 1), (4, 2), (4, 3).

The sample space S is the set of all these possible ordered pairs:

$S = \{ (1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), \ $$ (4, 1), (4, 2), (4, 3) \}$

The number of elements in the sample space is the number of permutations of 4 items taken 2 at a time, which is $P(4, 2) = \frac{4!}{(4-2)!} = \frac{24}{2} = 12$.

Given:

An experiment involving rolling a die and then tossing a coin based on the die result.

The steps are:

1. Roll a die.

2. If the die result is an even number, toss a coin once.

3. If the die result is an odd number, toss a coin twice.

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To Find:

The sample space for this experiment.

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Solution:

Let the outcome of the die roll be $d \in \{1, 2, 3, 4, 5, 6\}$.

Let H represent Head and T represent Tail for the coin toss.

We analyze the outcomes based on whether the die roll is even or odd.

Case 1: The die shows an even number.

The possible even numbers are $\{2, 4, 6\}$.

If the die result is 2, 4, or 6, a coin is tossed once. The possible coin outcomes are H or T.

The combined outcomes in this case are ordered pairs (die result, coin toss):

• If die is 2: (2, H), (2, T)
• If die is 4: (4, H), (4, T)
• If die is 6: (6, H), (6, T)

Case 2: The die shows an odd number.

The possible odd numbers are $\{1, 3, 5\}$.

If the die result is 1, 3, or 5, a coin is tossed twice. The possible outcomes for tossing a coin twice are HH, HT, TH, TT.

The combined outcomes in this case are represented as pairs (die result, sequence of two coin tosses):

• If die is 1: (1, HH), (1, HT), (1, TH), (1, TT)
• If die is 3: (3, HH), (3, HT), (3, TH), (3, TT)
• If die is 5: (5, HH), (5, HT), (5, TH), (5, TT)

The sample space S is the set of all these distinct possible outcomes from both cases.

$S = \{ (2, H), (2, T), (4, H), (4, T), (6, H), (6, T),$

$(1, HH), (1, HT), (1, TH), (1, TT),$

$(3, HH), (3, HT), (3, TH), (3, TT),$

$(5, HH), (5, HT), (5, TH), (5, TT) \}$

The number of elements in the sample space is $n(S) = (3 \times 2) + (3 \times 4) = 6 + 12 = 18$.

Given:

An experiment involving a coin toss and a subsequent action based on the outcome:

• Toss a coin.
• If Tail (T), draw a ball from a box containing 2 red (R) and 3 black (B) balls.
• If Head (H), throw a standard six-sided die (outcomes 1 to 6).

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To Find:

The sample space for this experiment.

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Solution:

The sample space is the set of all possible outcomes of the entire experiment.

The experiment starts with a coin toss, which has two possible outcomes: Head (H) or Tail (T).

Case 1: The coin shows Tail (T).

If the coin outcome is T, a ball is drawn from a box with 2 red balls (R) and 3 black balls (B). The possible outcomes for the ball colour are Red (R) or Black (B).

The combined outcomes in this case are represented as ordered pairs (Coin outcome, Ball colour):

• First outcome T, second outcome R: (T, R)
• First outcome T, second outcome B: (T, B)

Case 2: The coin shows Head (H).

If the coin outcome is H, a standard die is thrown. The possible outcomes for the die throw are the numbers 1, 2, 3, 4, 5, or 6.

The combined outcomes in this case are represented as ordered pairs (Coin outcome, Die result):

• First outcome H, second outcome 1: (H, 1)
• First outcome H, second outcome 2: (H, 2)
• First outcome H, second outcome 3: (H, 3)
• First outcome H, second outcome 4: (H, 4)
• First outcome H, second outcome 5: (H, 5)
• First outcome H, second outcome 6: (H, 6)

The sample space S is the set of all these distinct possible outcomes from both cases.

$S = \{ (T, R), (T, B), (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6) \}$

The number of elements in the sample space is $n(S) = 2 + 6 = 8$.

Given:

An experiment consists of repeatedly throwing a standard six-sided die. The experiment stops as soon as the outcome '6' is observed.

---

To Find:

The sample space for this experiment.

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Solution:

The sample space is the set of all possible outcomes. An outcome is a sequence of die rolls. Let's analyze the possible sequences based on when the first '6' appears.

Let S denote the outcome of getting a '6'. Let F denote the outcome of not getting a '6', which means getting any number from the set {1, 2, 3, 4, 5}.

Case 1: The experiment ends on the first throw.

This happens if a '6' comes up on the very first throw. The outcome is simply:

6

Case 2: The experiment ends on the second throw.

This happens if the first throw is not a '6' (an F), and the second throw is a '6' (an S). The outcomes are sequences of the form (F, 6).

(1, 6), (2, 6), (3, 6), (4, 6), (5, 6)

Case 3: The experiment ends on the third throw.

This happens if the first two throws are not '6's (F, F), and the third throw is a '6' (S). The outcomes are sequences of the form (F, F, 6).

For example: (1, 1, 6), (1, 2, 6), (2, 4, 6), etc.

Case 4: The experiment ends on the fourth throw.

This happens if the first three throws are not '6's (F, F, F), and the fourth throw is a '6' (S). The outcomes are sequences of the form (F, F, F, 6).

For example: (1, 3, 5, 6), (4, 2, 1, 6), etc.

This process can continue indefinitely, as there is no theoretical limit to how many times one might roll the die without getting a '6'. Therefore, the sample space is an infinite set.

The sample space (S) is the collection of all such possible sequences.

S = { 6,

(1, 6), (2, 6), (3, 6), (4, 6), (5, 6),

(1, 1, 6), (1, 2, 6), ..., (5, 5, 6),

(1, 1, 1, 6), (1, 1, 2, 6), ..., (5, 5, 5, 6),

... }

More formally, the sample space S can be described as the set of all finite sequences of numbers $(x_1, x_2, \dots, x_k)$ where $x_k = 6$ and $x_i \in \{1, 2, 3, 4, 5\}$ for all $i < k$.

Given:

The experiment is rolling a die.

Event A: 'getting a prime number'.

Event B: 'getting an odd number'.

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To Find:

The sets representing the events (i) A or B, (ii) A and B, (iii) A but not B, (iv) ‘not A’.

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Solution:

The sample space S for rolling a die is the set of all possible outcomes:

$S = \{1, 2, 3, 4, 5, 6\}$

Event A is 'getting a prime number'. The prime numbers in S are 2, 3, and 5.

$A = \{2, 3, 5\}$

Event B is 'getting an odd number'. The odd numbers in S are 1, 3, and 5.

$B = \{1, 3, 5\}$

Now, we can find the sets for the required events:

(i) A or B: This event occurs if the outcome is in set A or in set B (or both). This is the union of sets A and B, denoted by $A \cup B$.

$A \cup B = \{2, 3, 5\} \cup \{1, 3, 5\}$

$A \cup B = \{1, 2, 3, 5\}$

(ii) A and B: This event occurs if the outcome is in both set A and set B. This is the intersection of sets A and B, denoted by $A \cap B$.

$A \cap B = \{2, 3, 5\} \cap \{1, 3, 5\}$

$A \cap B = \{3, 5\}$

(iii) A but not B: This event occurs if the outcome is in set A but not in set B. This is the set difference $A - B$. Alternatively, it can be represented as the intersection of A and the complement of B ($A \cap B'$).

$A - B = \{2, 3, 5\} - \{1, 3, 5\}$

$A - B = \{2\}$

(iv) ‘not A’: This event occurs if the outcome is not in set A. This is the complement of set A with respect to the sample space S, denoted by $A'$ or $S - A$.

$A' = S - A = \{1, 2, 3, 4, 5, 6\} - \{2, 3, 5\}$

$A' = \{1, 4, 6\}$

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Summary of the events as sets:

(i) A or B: $\{1, 2, 3, 5\}$

(ii) A and B: $\{3, 5\}$

(iii) A but not B: $\{2\}$

(iv) ‘not A’: $\{1, 4, 6\}$

Given:

The experiment is throwing two dice and noting the sum of the numbers. The events are:

A: ‘the sum is even’.

B: ‘the sum is a multiple of 3’.

C: ‘the sum is less than 4’.

D: ‘the sum is greater than 11’.

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To Find:

Which pairs of these events are mutually exclusive.

Two events are mutually exclusive if they cannot happen at the same time. In terms of sets, this means their intersection is the empty set ($\emptyset$).

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Solution:

First, let's list the outcomes (as ordered pairs) for each event. The sample space consists of 36 outcomes, from (1, 1) to (6, 6).

Event A: ‘the sum is even’

The possible sums are 2, 4, 6, 8, 10, 12.

$A = \{(1,1), (1,3), (2,2), (3,1), (1,5), (2,4), (3,3), (4,2), (5,1), (2,6), \ $$ (3,5), (4,4), (5,3), (6,2), (4,6), (5,5), (6,4), (6,6)\}$

Event B: ‘the sum is a multiple of 3’

The possible sums are 3, 6, 9, 12.

$B = \{(1,2), (2,1), (1,5), (2,4), (3,3), (4,2), (5,1), (3,6), (4,5), (5,4), \ $$ (6,3), (6,6)\}$

Event C: ‘the sum is less than 4’

The possible sums are 2, 3.

$C = \{(1,1), (1,2), (2,1)\}$

Event D: ‘the sum is greater than 11’

The possible sum is 12.

$D = \{(6,6)\}$

Now, we check each pair of events for common outcomes (intersection).

1. Pair (A, B):

$A \cap B = \{(1,5), (2,4), (3,3), (4,2), (5,1), (6,6)\}$. The intersection is not empty (e.g., a sum of 6 is both even and a multiple of 3). Therefore, A and B are not mutually exclusive.

2. Pair (A, C):

$A \cap C = \{(1,1)\}$. The intersection is not empty (a sum of 2 is both even and less than 4). Therefore, A and C are not mutually exclusive.

3. Pair (A, D):

$A \cap D = \{(6,6)\}$. The intersection is not empty (a sum of 12 is both even and greater than 11). Therefore, A and D are not mutually exclusive.

4. Pair (B, C):

$B \cap C = \{(1,2), (2,1)\}$. The intersection is not empty (a sum of 3 is both a multiple of 3 and less than 4). Therefore, B and C are not mutually exclusive.

5. Pair (B, D):

$B \cap D = \{(6,6)\}$. The intersection is not empty (a sum of 12 is both a multiple of 3 and greater than 11). Therefore, B and D are not mutually exclusive.

6. Pair (C, D):

Event C requires the sum to be less than 4, while event D requires the sum to be greater than 11. It is impossible for a single outcome to satisfy both conditions. The sets have no common elements.

$C \cap D = \emptyset$. Therefore, C and D are mutually exclusive.

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Answer:

The only pair of events that are mutually exclusive is C and D.

Given:

The experiment is tossing a coin three times.

The events are:

A: ‘No head appears’

B: ‘Exactly one head appears’

C: ‘At least two heads appear’

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To Find:

Whether events A, B, and C form a set of mutually exclusive and exhaustive events.

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Solution:

First, let's determine the sample space S for tossing a coin three times. The possible outcomes are sequences of Heads (H) and Tails (T).

$S = \{ HHH, HHT, HTH, THH, HTT, THT, TTH, TTT \}$

Now, let's write the given events as subsets of the sample space:

Event A: ‘No head appears’

This means all three tosses are Tails.

$A = \{ TTT \}$

Event B: ‘Exactly one head appears’

This means one Head and two Tails in any order.

$B = \{ HTT, THT, TTH \}$

Event C: ‘At least two heads appear’

This means two Heads and one Tail, or three Heads.

$C = \{ HHT, HTH, THH, HHH \}$

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Check for Mutually Exclusive Events:

Events are mutually exclusive if no two events can occur at the same time, i.e., their intersections are empty.

Check the intersections:

$A \cap B = \{ TTT \} \cap \{ HTT, THT, TTH \} = \emptyset$

$A \cap C = \{ TTT \} \cap \{ HHT, HTH, THH, HHH \} = \emptyset$

$B \cap C = \{ HTT, THT, TTH \} \cap \{ HHT, HTH, THH, HHH \} = \emptyset$

Since the intersection of every pair of events is the empty set, events A, B, and C are mutually exclusive.

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Check for Exhaustive Events:

Events are exhaustive if their union is equal to the sample space S, meaning they cover all possible outcomes of the experiment.

Check the union of A, B, and C:

$A \cup B \cup C = \{ TTT \} \cup \{ HTT, THT, TTH \} \cup \{ HHT, HTH, \ $$ THH, HHH \}$

$A \cup B \cup C = \{ TTT, HTT, THT, TTH, HHT, HTH, THH, HHH \}$

Comparing this with the sample space S:

$S = \{ HHH, HHT, HTH, THH, HTT, THT, TTH, TTT \}$

We see that $A \cup B \cup C = S$. Thus, the events A, B, and C are exhaustive.

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Conclusion:

Since the events A, B, and C are both mutually exclusive and exhaustive, they form a set of mutually exclusive and exhaustive events.

Given:

An experiment where a standard six-sided die is rolled.

Event E: “die shows 4”

Event F: “die shows even number”

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To Find:

Whether events E and F are mutually exclusive.

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Solution:

The sample space S for rolling a standard six-sided die is the set of all possible outcomes:

$S = \{1, 2, 3, 4, 5, 6\}$

Now, let's represent the given events E and F as subsets of the sample space:

Event E is "die shows 4". The outcome where the die shows 4 is the number 4 itself.

$E = \{4\}$

Event F is "die shows even number". The even numbers in the sample space S are 2, 4, and 6.

$F = \{2, 4, 6\}$

Two events are considered mutually exclusive if their intersection is the empty set ($\emptyset$). This means they cannot both occur at the same time in a single trial of the experiment.

Let's find the intersection of events E and F, denoted by $E \cap F$:

$E \cap F = \{4\} \cap \{2, 4, 6\}$

The elements common to both sets E and F are the outcomes that satisfy both conditions (showing a 4 and showing an even number). The number 4 is present in both sets.

$E \cap F = \{4\}$

Since the intersection $E \cap F = \{4\}$, which is not the empty set, there is an outcome (rolling a 4) that belongs to both event E and event F. This means that events E and F can occur simultaneously.

Therefore, events E and F are not mutually exclusive.

Given:

An experiment: A standard six-sided die is thrown.

Events A, B, C, D, E, and F are defined based on the outcome of the die roll.

---

To Find:

Describe the sets for events A, B, C, D, E, F. Also find the results of the following set operations: A ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, A – C, D – E, E ∩ F′, F′.

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Solution:

The sample space S for rolling a standard six-sided die is the set of all possible outcomes:

$S = \{1, 2, 3, 4, 5, 6\}$

Now, we describe each given event as a subset of the sample space S:

(i) A: a number less than 7

The numbers in S that are less than 7 are 1, 2, 3, 4, 5, and 6.

$A = \{1, 2, 3, 4, 5, 6\} = S$

(ii) B: a number greater than 7

There are no numbers in S that are greater than 7.

$B = \emptyset$ (the empty set)

(iii) C: a multiple of 3

The multiples of 3 in S are 3 and 6.

$C = \{3, 6\}$

(iv) D: a number less than 4

The numbers in S that are less than 4 are 1, 2, and 3.

$D = \{1, 2, 3\}$

(v) E: an even number greater than 4

The even numbers in S are 2, 4, 6. The numbers in S greater than 4 are 5, 6. The numbers that are both even and greater than 4 is 6.

$E = \{6\}$

(vi) F: a number not less than 3

A number not less than 3 means a number greater than or equal to 3 ($\geq 3$). The numbers in S that are not less than 3 are 3, 4, 5, and 6.

$F = \{3, 4, 5, 6\}$

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Now, we perform the required set operations:

A ∪ B: The union of A and B contains all elements that are in A or B (or both).

$A \cup B = \{1, 2, 3, 4, 5, 6\} \cup \emptyset = \{1, 2, 3, 4, 5, 6\} = S$

A ∩ B: The intersection of A and B contains all elements that are in both A and B.

$A \cap B = \{1, 2, 3, 4, 5, 6\} \cap \emptyset = \emptyset$

B ∪ C: The union of B and C contains all elements that are in B or C (or both).

$B \cup C = \emptyset \cup \{3, 6\} = \{3, 6\} = C$

E ∩ F: The intersection of E and F contains all elements that are in both E and F.

$E \cap F = \{6\} \cap \{3, 4, 5, 6\} = \{6\}$

D ∩ E: The intersection of D and E contains all elements that are in both D and E.

$D \cap E = \{1, 2, 3\} \cap \{6\} = \emptyset$

A – C: The set difference A minus C contains all elements that are in A but not in C.

$A – C = \{1, 2, 3, 4, 5, 6\} – \{3, 6\} = \{1, 2, 4, 5\}$

D – E: The set difference D minus E contains all elements that are in D but not in E.

$D – E = \{1, 2, 3\} – \{6\} = \{1, 2, 3\} = D$

F′: The complement of F is the set of all elements in the sample space S that are not in F. $F' = S - F$.

$F' = \{1, 2, 3, 4, 5, 6\} – \{3, 4, 5, 6\} = \{1, 2\}$

E ∩ F′: The intersection of E and F′ contains all elements that are in both E and F′.

$E \cap F' = \{6\} \cap \{1, 2\} = \emptyset$

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Summary of the results:

$A = \{1, 2, 3, 4, 5, 6\}$

$B = \emptyset$

$C = \{3, 6\}$

$D = \{1, 2, 3\}$

$E = \{6\}$

$F = \{3, 4, 5, 6\}$

$A \cup B = \{1, 2, 3, 4, 5, 6\}$

$A \cap B = \emptyset$

$B \cup C = \{3, 6\}$

$E \cap F = \{6\}$

$D \cap E = \emptyset$

$A – C = \{1, 2, 4, 5\}$

$D – E = \{1, 2, 3\}$

$F' = \{1, 2\}$

$E \cap F' = \emptyset$

Given:

The experiment is rolling a pair of dice. The sample space consists of 36 ordered pairs, from (1, 1) to (6, 6).

The events are defined as:

• A: the sum is greater than 8.
• B: 2 occurs on either die.
• C: the sum is at least 7 and a multiple of 3.

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To Find:

First, describe the set of outcomes for each event. Then, identify which pairs of these events are mutually exclusive.

Two events are mutually exclusive if they have no outcomes in common, meaning their intersection is the empty set ($\emptyset$).

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Solution:

Step 1: Describe the events by listing their outcomes

Event A: The sum is greater than 8

This means the sum can be 9, 10, 11, or 12.

$A = \{(3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)\}$

Event B: 2 occurs on either die

This means at least one of the dice shows the number 2.

$B = \{(1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (2, 4), (4, 2), (2, 5), (5, 2), (2, 6), \ $$ (6, 2)\}$

Event C: The sum is at least 7 and a multiple of 3

"At least 7" means the sum is $\ge 7$. "A multiple of 3" means the sum could be 3, 6, 9, 12. The conditions that satisfy both are sums of 9 or 12.

$C = \{(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)\}$

Step 2: Check which pairs are mutually exclusive

Pair (A, B):

We check if there are any common outcomes between A and B. Event A contains outcomes with sums greater than 8. The largest possible sum with a '2' on one die is with a '6' on the other, which is $2+6=8$. Since 8 is not greater than 8, no outcome in B can have a sum greater than 8.

$A \cap B = \emptyset$.

Therefore, A and B are mutually exclusive.

Pair (A, C):

We check for common outcomes between A and C.

A = $\{(3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)\}$

C = $\{(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)\}$

We can see that all the outcomes in C are also present in A.

$A \cap C = \{(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)\}$.

Since the intersection is not empty, A and C are not mutually exclusive.

Pair (B, C):

We check for common outcomes between B and C. Event B requires a '2' on at least one die. Event C contains the outcomes $\{(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)\}$. None of these outcomes have a '2'.

$B \cap C = \emptyset$.

Therefore, B and C are mutually exclusive.

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Answer:

The pairs of events that are mutually exclusive are (A, B) and (B, C).

First, let's define the sample space and list the outcomes for each event.

Sample Space (S)

When three coins are tossed, the set of all possible outcomes is:

$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$

Describing the Events

• A: ‘three heads show’

  $A = \{HHH\}$

• B: ‘two heads and one tail show’

  $B = \{HHT, HTH, THH\}$

• C: ‘three tails show’

  $C = \{TTT\}$

• D: ‘a head shows on the first coin’

  $D = \{HHH, HHT, HTH, HTT\}$

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(i) Mutually Exclusive Events

Two events are mutually exclusive if they have no outcomes in common (their intersection is the empty set, $\emptyset$). Let's check the pairs:

• A and B: $A \cap B = \{HHH\} \cap \{HHT, HTH, THH\} = \emptyset$.

  Yes, A and B are mutually exclusive.

• A and C: $A \cap C = \{HHH\} \cap \{TTT\} = \emptyset$.

  Yes, A and C are mutually exclusive.

• A and D: $A \cap D = \{HHH\} \cap \{HHH, HHT, HTH, HTT\} \ $$ = \{HHH\}$.

  No, A and D are not mutually exclusive as they share the outcome {HHH}.

• B and C: $B \cap C = \{HHT, HTH, THH\} \cap \{TTT\} = \emptyset$.

  Yes, B and C are mutually exclusive.

• B and D: $B \cap D = \{HHT, HTH, THH\} \cap \{HHH, HHT, \ $$ HTH, HTT\} \ $$ = \{HHT, HTH\}$.

  No, B and D are not mutually exclusive.

• C and D: $C \cap D = \{TTT\} \cap \{HHH, HHT, HTH, HTT\} = \emptyset$.

  Yes, C and D are mutually exclusive.

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(ii) Simple Events

A simple event (or elementary event) is an event with only one outcome.

• Event A = {HHH}. It has one outcome. So, A is a simple event.
• Event B = {HHT, HTH, THH}. It has three outcomes. So, B is not a simple event.
• Event C = {TTT}. It has one outcome. So, C is a simple event.
• Event D = {HHH, HHT, HTH, HTT}. It has four outcomes. So, D is not a simple event.

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(iii) Compound Events

A compound event is an event with more than one outcome.

• Event A has only one outcome. So, A is not a compound event.
• Event B has three outcomes. So, B is a compound event.
• Event C has only one outcome. So, C is not a compound event.
• Event D has four outcomes. So, D is a compound event.

First, let's establish the sample space (S) for the experiment of tossing three coins. Let H denote a head and T denote a tail.

$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$

Now we will describe a set of events for each scenario as requested.

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(i) Two events which are mutually exclusive.

Two events are mutually exclusive if they cannot happen at the same time (i.e., they have no common outcomes).

• Event A: “Getting at least two heads”

  $A = \{HHH, HHT, HTH, THH\}$

• Event B: “Getting at least two tails”

  $B = \{HTT, THT, TTH, TTT\}$

The intersection of these two events is $A \cap B = \emptyset$. Therefore, events A and B are mutually exclusive.

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(ii) Three events which are mutually exclusive and exhaustive.

Events are mutually exclusive if no two have a common outcome. They are exhaustive if their union makes up the entire sample space.

• Event A: “Getting no heads”

  $A = \{TTT\}$

• Event B: “Getting exactly one head”

  $B = \{HTT, THT, TTH\}$

• Event C: “Getting at least two heads”

  $C = \{HHH, HHT, HTH, THH\}$

Here, $A \cap B = \emptyset$, $A \cap C = \emptyset$, and $B \cap C = \emptyset$, so they are mutually exclusive. Also, $A \cup B \cup C = S$, so they are exhaustive.

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(iii) Two events, which are not mutually exclusive.

Two events are not mutually exclusive if they have at least one common outcome.

• Event A: “Getting at most two tails” (i.e., 0, 1, or 2 tails)

  $A = \{HHH, HHT, HTH, THH, HTT, THT, TTH\}$

• Event B: “Getting exactly two tails”

  $B = \{HTT, THT, TTH\}$

The intersection is $A \cap B = \{HTT, THT, TTH\}$, which is not empty. Therefore, events A and B are not mutually exclusive.

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(iv) Two events which are mutually exclusive but not exhaustive.

The events must have no common outcomes, but their union must not be the entire sample space.

• Event A: “Getting exactly one head”

  $A = \{HTT, THT, TTH\}$

• Event B: “Getting exactly two heads”

  $B = \{HHT, HTH, THH\}$

Here, $A \cap B = \emptyset$, so they are mutually exclusive. However, $A \cup B$ does not include $\{HHH\}$ or $\{TTT\}$, so the events are not exhaustive.

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(v) Three events which are mutually exclusive but not exhaustive.

The events must be pairwise disjoint, but their union must not be the entire sample space.

• Event A: “Getting exactly one tail”

  $A = \{HHT, HTH, THH\}$

• Event B: “Getting exactly two tails”

  $B = \{HTT, THT, TTH\}$

• Event C: “Getting exactly three tails”

  $C = \{TTT\}$

Here, $A \cap B = \emptyset$, $A \cap C = \emptyset$, and $B \cap C = \emptyset$, so they are mutually exclusive. However, their union, $A \cup B \cup C$, does not include the outcome $\{HHH\}$, so the events are not exhaustive.

First, let's define the sample space (S) for rolling two dice. It consists of 36 ordered pairs:

$S = \{(x, y) : x, y \in \{1, 2, 3, 4, 5, 6\}\}$

Now, let's list the outcomes for the given events A, B, and C.

• A: getting an even number on the first die.

  $A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), \ $$ (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$

• B: getting an odd number on the first die.

  $B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), \ $$ (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$

• C: getting the sum of the numbers on the dice ≤ 5.

  $C = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), \ $$ (3,2), (4,1)\}$

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(i) A′

A′ (complement of A) means 'not A', which is the event of not getting an even number on the first die. This is the same as getting an odd number on the first die, which is event B.

$A' = B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), \ $$ (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$

(ii) not B

'not B' is the complement of B, denoted as B′. It's the event of not getting an odd number on the first die, which is the same as getting an even number on the first die. This is event A.

$B' = A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), \ $$ (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$

(iii) A or B

This corresponds to the union of A and B ($A \cup B$). Since the number on the first die must be either even (event A) or odd (event B), this covers all possible outcomes.

$A \cup B = S$ (the entire sample space).

(iv) A and B

This corresponds to the intersection of A and B ($A \cap B$). An outcome in this event must have an even number and an odd number on the first die simultaneously, which is impossible.

$A \cap B = \emptyset$ (the empty set).

(v) A but not C

This corresponds to the set difference $A - C$. It includes all outcomes in A that are not in C. These are outcomes where the first die is even and the sum is greater than 5.

$A - C = \{(2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), \ $$ (6,2), (6,3), (6,4), (6,5), (6,6)\}$

(vi) B or C

This corresponds to the union $B \cup C$. It includes all outcomes that are in B, or in C, or in both.

$B \cup C = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), \ $$ (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (2,1), (2,2), \ $$ (2,3), (4,1)\}$

(vii) B and C

This corresponds to the intersection $B \cap C$. It includes outcomes where the first die is odd and the sum is less than or equal to 5.

$B \cap C = \{(1,1), (1,2), (1,3), (1,4), (3,1), (3,2)\}$

(viii) A ∩ B′ ∩ C′

Let's break this down. We already know that B′ is the same as event A. So the expression becomes $A \cap A \cap C'$, which simplifies to $A \cap C'$. This is the same as "A but not C" from part (v).

$A \cap B' \cap C' = A - C = \{(2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), \ $$ (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$

Let's first recall the events from question 6:

S = Sample space for rolling two dice.

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice ≤ 5.

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(i) A and B are mutually exclusive

True.

Reason: Two events are mutually exclusive if they cannot occur at the same time, meaning their intersection is the empty set ($\emptyset$). Event A requires the first die to be even {2, 4, 6}, while event B requires the first die to be odd {1, 3, 5}. It is impossible for the first die to be both even and odd simultaneously. Therefore, $A \cap B = \emptyset$.

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(ii) A and B are mutually exclusive and exhaustive

True.

Reason: We already established they are mutually exclusive. Two events are exhaustive if their union equals the entire sample space ($A \cup B = S$). Since the first die must show either an even number (event A) or an odd number (event B), there are no other possibilities. Every outcome in the sample space belongs to either A or B. Thus, $A \cup B = S$.

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(iii) A = B′

True.

Reason: B′ represents the complement of B, which means "not B". The event "not getting an odd number on the first die" is exactly the same as the event "getting an even number on the first die", which is event A. Therefore, the set of outcomes for A is identical to the set of outcomes for B′.

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(iv) A and C are mutually exclusive

False.

Reason: For A and C to be mutually exclusive, they must have no outcomes in common. However, there are outcomes that belong to both events. For example, the outcome (2, 1) is in A (first die is even) and also in C (sum is 3, which is ≤ 5). Since $A \cap C = \{(2,1), (2,2), (2,3), (4,1)\} \neq \emptyset$, they are not mutually exclusive.

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(v) A and B′ are mutually exclusive.

False.

Reason: As established in part (iii), B′ is the same event as A. The intersection of an event with itself is the event itself ($A \cap A = A$). Since A is not an empty set, $A \cap B'$ is not empty. Therefore, they are not mutually exclusive.

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(vi) A′, B′, C are mutually exclusive and exhaustive.

False.

Reason: Let's check the conditions. First, we identify the sets: A′ = B and B′ = A. So the events are B, A, and C.

Are they mutually exclusive? No. We already know that $A \cap C \neq \emptyset$ and $B \cap C \neq \emptyset$. For a set of three events to be mutually exclusive, every pair must be mutually exclusive. Since this condition fails, the statement is false.

Are they exhaustive? No. Their union $A \cup B \cup C$ is not the entire sample space. For example, the outcome (6, 6) is in S, but it is not in A, B, or C. Actually, since $A \cup B = S$, it follows that $A \cup B \cup C = S$. So the events are exhaustive. However, since they are not mutually exclusive, the overall statement is false.

For an assignment of probabilities to be valid for a sample space with $n$ outcomes {$ω_1, ω_2, ..., ω_n$}, two conditions must be met:

1. The probability of each individual outcome must be non-negative and less than or equal to 1. That is, $0 \le P(ω_i) \le 1$ for all $i = 1, 2, \dots, n$.

2. The sum of the probabilities of all possible outcomes must be equal to 1. That is, $\sum\limits_{i=1}^{n} P(ω_i) = 1$.

We will check each assignment against these two conditions.

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(a) $P(\omega_1) = \frac{1}{6}, P(\omega_2) = \frac{1}{6}, \dots, P(\omega_6) = \frac{1}{6}$

Condition 1: Each probability is $\frac{1}{6}$, which is between 0 and 1. This condition is met.

Condition 2: Sum of probabilities = $\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = 6 \times \frac{1}{6} = 1$. This condition is met.

Result: This assignment is Valid.

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(b) $P(\omega_1) = 1, P(\omega_2) = 0, \dots, P(\omega_6) = 0$

Condition 1: All probabilities (1 and 0) are between 0 and 1. This condition is met.

Condition 2: Sum of probabilities = $1 + 0 + 0 + 0 + 0 + 0 = 1$. This condition is met.

Result: This assignment is Valid.

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(c) $P(\omega_5) = -\frac{1}{4}, P(\omega_6) = -\frac{1}{3}$, etc.

Condition 1: The probabilities for $\omega_5$ and $\omega_6$ are negative. Probability cannot be negative. This condition is not met.

Result: This assignment is Not Valid.

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(d) $P(\omega_6) = \frac{3}{2}$, etc.

Condition 1: The probability for $\omega_6$ is $\frac{3}{2} = 1.5$, which is greater than 1. Probability cannot be greater than 1. This condition is not met.

Result: This assignment is Not Valid.

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(e) $P(\omega_1) = 0.1, P(\omega_2) = 0.2, \dots, P(\omega_6) = 0.6$

Condition 1: All individual probabilities are between 0 and 1. This condition is met.

Condition 2: Sum of probabilities = $0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 = 2.1$. The sum is not equal to 1. This condition is not met.

Result: This assignment is Not Valid.

The experiment is drawing one card from a standard deck of 52 cards. The total number of possible outcomes in the sample space is $n(S) = 52$. Since each outcome is equally likely, the probability of any event E is given by the formula:

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(i) a diamond

A standard deck has 13 diamond cards. So, the number of favourable outcomes is 13.

$P(\text{a diamond}) = \frac{13}{52} = \frac{1}{4}$

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(ii) not an ace

There are 4 aces in a deck. The number of cards that are not aces is $52 - 4 = 48$. So, the number of favourable outcomes is 48.

$P(\text{not an ace}) = \frac{48}{52} = \frac{12}{13}$

Alternate Method: Let A be the event that the card is an ace. $P(A) = \frac{4}{52} = \frac{1}{13}$. The probability of 'not an ace' is $P(A') = 1 - P(A) = 1 - \frac{1}{13} = \frac{12}{13}$.

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(iii) a black card (i.e., a club or, a spade)

There are 13 clubs and 13 spades, both of which are black suits. The total number of black cards is $13 + 13 = 26$. So, the number of favourable outcomes is 26.

$P(\text{a black card}) = \frac{26}{52} = \frac{1}{2}$

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(iv) not a diamond

There are 13 diamonds. The number of cards that are not diamonds is $52 - 13 = 39$. So, the number of favourable outcomes is 39.

$P(\text{not a diamond}) = \frac{39}{52} = \frac{3}{4}$

Alternate Method: From part (i), we know $P(\text{a diamond}) = \frac{1}{4}$. The probability of 'not a diamond' is $1 - P(\text{a diamond}) = 1 - \frac{1}{4} = \frac{3}{4}$.

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(v) not a black card.

A card that is not black must be red. There are 13 hearts and 13 diamonds, which are red suits. The total number of red cards is $13 + 13 = 26$. So, the number of favourable outcomes is 26.

$P(\text{not a black card}) = \frac{26}{52} = \frac{1}{2}$

Alternate Method: From part (iii), we know $P(\text{a black card}) = \frac{1}{2}$. The probability of 'not a black card' is $1 - P(\text{a black card}) = 1 - \frac{1}{2} = \frac{1}{2}$.

Given Information:

Total number of discs in the bag = 9.

• Number of red discs = 4
• Number of blue discs = 3
• Number of yellow discs = 2

The total number of possible outcomes is the total number of discs, so $n(S) = 9$.

The probability of an event E is calculated as $P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$.

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(i) Probability that the disc will be red.

The number of red discs (favourable outcomes) is 4.

$P(\text{red}) = \frac{\text{Number of red discs}}{\text{Total number of discs}} = \frac{4}{9}$

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(ii) Probability that the disc will be yellow.

The number of yellow discs (favourable outcomes) is 2.

$P(\text{yellow}) = \frac{\text{Number of yellow discs}}{\text{Total number of discs}} = \frac{2}{9}$

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(iii) Probability that the disc will be blue.

The number of blue discs (favourable outcomes) is 3.

$P(\text{blue}) = \frac{\text{Number of blue discs}}{\text{Total number of discs}} = \frac{3}{9} = \frac{1}{3}$

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(iv) Probability that the disc will not be blue.

This can be solved in two ways:

Method 1: The discs that are 'not blue' are the red and yellow ones. Number of non-blue discs = $4 (\text{red}) + 2 (\text{yellow}) = 6$.

$P(\text{not blue}) = \frac{\text{Number of non-blue discs}}{\text{Total number of discs}} = \frac{6}{9} = \frac{2}{3}$

Method 2: Using the complement rule. The event 'not blue' is the complement of the event 'blue'.

$P(\text{not blue}) = 1 - P(\text{blue}) = 1 - \frac{1}{3} = \frac{2}{3}$

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(v) Probability that the disc will be either red or blue.

This means we want the probability of drawing a red disc OR a blue disc. Since these are mutually exclusive events (a disc cannot be both red and blue), we can add their probabilities.

$P(\text{red or blue}) = P(\text{red}) + P(\text{blue})$

$P(\text{red or blue}) = \frac{4}{9} + \frac{3}{9} = \frac{7}{9}$

Alternatively, the number of favourable outcomes is the number of red discs plus the number of blue discs = $4 + 3 = 7$.

$P(\text{red or blue}) = \frac{7}{9}$

Given Information:

Let A be the event that Anil qualifies.

Let B be the event that Ashima qualifies.

We are given the following probabilities:

• $P(A) = 0.05$
• $P(B) = 0.10$
• $P(A \cap B) = P(\text{both qualify}) = 0.02$

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(a) Both Anil and Ashima will not qualify the examination.

This is the event 'not A' AND 'not B', which can be written as $P(A' \cap B')$.

Using De Morgan's Law, we know that $P(A' \cap B') = P((A \cup B)')$.

And $P((A \cup B)') = 1 - P(A \cup B)$.

First, we find $P(A \cup B)$ using the addition rule:

Now, we can find the probability that both will not qualify:

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(b) At least one of them will not qualify the examination.

This is the event 'not A' OR 'not B', which can be written as $P(A' \cup B')$.

Using De Morgan's Law, we know that $P(A' \cup B') = P((A \cap B)')$.

This is the complement of the event that 'both qualify'.

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(c) Only one of them will qualify the examination.

This means either (Anil qualifies AND Ashima does not) OR (Anil does not qualify AND Ashima does). This can be written as $P((A \cap B') \cup (A' \cap B))$.

A simpler way to think about this is: it's the probability that "at least one qualifies" minus the probability that "both qualify".

Given Information:

A committee of 2 persons is to be selected from a group of 2 men and 2 women.

Total number of people = 4.

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Step 1: Find the total number of possible committees

The total number of ways to select 2 people from 4 is a combination problem, calculated as $^4C_2$.

So, there are 6 possible committees that can be formed. This is our total number of outcomes.

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(a) Probability that the committee will have no man.

If there are no men, the committee must consist of 2 women. We need to choose 2 women from the 2 available women.

Number of ways to choose 2 women from 2 = $\binom{2}{2} = 1$.

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(b) Probability that the committee will have one man.

If there is one man, the committee must also have one woman. We need to choose 1 man from 2 men AND 1 woman from 2 women.

Number of ways to choose 1 man from 2 = $\binom{2}{1} = 2$.

Number of ways to choose 1 woman from 2 = $\binom{2}{1} = 2$.

Total number of ways to form this committee = $2 \times 2 = 4$.

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(c) Probability that the committee will have two men.

If there are two men, the committee must consist of 2 men. We need to choose 2 men from the 2 available men.

Number of ways to choose 2 men from 2 = $\binom{2}{2} = 1$.

For an assignment of probabilities to be valid, two fundamental conditions must be met:

1. The probability of each individual outcome must be a number between 0 and 1 (inclusive). That is, $0 \le P(\omega_i) \le 1$.

2. The sum of the probabilities of all the outcomes in the sample space must equal 1. That is, $\sum\limits_{i=1}^{7} P(\omega_i) = 1$.

Let's check each assignment.

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(a)

Condition 1: All probabilities (0.1, 0.01, 0.05, 0.03, 0.01, 0.2, 0.6) are positive and less than 1. This condition is met.

Condition 2: Sum = $0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1.00$. This condition is met.

Conclusion: This is a valid assignment.

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(b)

Condition 1: Each probability is $\frac{1}{7}$, which is between 0 and 1. This condition is met.

Condition 2: Sum = $\frac{1}{7} \times 7 = 1$. This condition is met.

Conclusion: This is a valid assignment.

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(c)

Condition 1: All probabilities are positive and less than or equal to 1. This condition is met.

Condition 2: Sum = $0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8$. The sum is not equal to 1. This condition is not met.

Conclusion: This is not a valid assignment.

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(d)

Condition 1: The probabilities for $\omega_1$ (–0.1) and $\omega_5$ (–0.2) are negative. Probability cannot be negative. This condition is not met.

Conclusion: This is not a valid assignment.

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(e)

Condition 1: The probability for $\omega_7$ is $\frac{15}{14}$, which is greater than 1. Probability cannot be greater than 1. This condition is not met.

Conclusion: This is not a valid assignment.

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Answer:

The assignments that cannot be valid are (c), (d), and (e).

Given:

A coin is tossed twice.

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To Find:

The probability that at least one tail occurs.

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Solution:

The sample space S for tossing a coin twice is the set of all possible ordered outcomes:

$S = \{ HH, HT, TH, TT \}$

The total number of outcomes in the sample space is $n(S) = 4$.

Since the coin is assumed to be fair and the tosses are independent, each outcome in the sample space is equally likely. The probability of each outcome is $\frac{1}{n(S)} = \frac{1}{4}$.

Let E be the event that at least one tail occurs.

'At least one tail' means the outcome has one tail or two tails.

Looking at the sample space S, the outcomes with at least one tail are HT, TH, and TT.

$E = \{ HT, TH, TT \}$

The number of outcomes favourable to event E is $n(E) = 3$.

The probability of event E is $P(E) = \frac{n(E)}{n(S)} = \frac{3}{4}$.

Alternatively, consider the complementary event, E′. E′ is the event that at least one tail does not occur, which means no tails occur. The only outcome with no tails is HH.

$E' = \{ HH \}$

The number of outcomes favourable to E′ is $n(E') = 1$.

The probability of event E′ is $P(E') = \frac{n(E')}{n(S)} = \frac{1}{4}$.

The probability of event E is the complement of the probability of E′:

$P(E) = 1 - P(E')$

$P(E) = 1 - \frac{1}{4} = \frac{4-1}{4} = \frac{3}{4}$.

The probability that at least one tail occurs is $\frac{3}{4}$.

Given:

A standard six-sided die is thrown.

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To Find:

The probability of the given events.

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Solution:

The sample space S for rolling a standard six-sided die is:

$S = \{1, 2, 3, 4, 5, 6\}$

The total number of outcomes in the sample space is $n(S) = 6$.

Assuming the die is fair, each outcome is equally likely, with a probability of $\frac{1}{6}$.

For any event E, the probability is $P(E) = \frac{\text{Number of outcomes favourable to E}}{\text{Total number of outcomes}} = \frac{n(E)}{n(S)}$.

Let's find the probability for each event:

(i) A prime number will appear.

Let $E_1$ be the event that a prime number appears.

The prime numbers in the sample space S are 2, 3, and 5.

$E_1 = \{2, 3, 5\}$

Number of outcomes favourable to $E_1$, $n(E_1) = 3$.

Probability $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{3}{6} = \frac{1}{2}$.

(ii) A number greater than or equal to 3 will appear.

Let $E_2$ be the event that a number greater than or equal to 3 appears.

The numbers in S that are $\geq 3$ are 3, 4, 5, and 6.

$E_2 = \{3, 4, 5, 6\}$

Number of outcomes favourable to $E_2$, $n(E_2) = 4$.

Probability $P(E_2) = \frac{n(E_2)}{n(S)} = \frac{4}{6} = \frac{2}{3}$.

(iii) A number less than or equal to one will appear.

Let $E_3$ be the event that a number less than or equal to one appears.

The numbers in S that are $\leq 1$ is only 1.

$E_3 = \{1\}$

Number of outcomes favourable to $E_3$, $n(E_3) = 1$.

Probability $P(E_3) = \frac{n(E_3)}{n(S)} = \frac{1}{6}$.

(iv) A number more than 6 will appear.

Let $E_4$ be the event that a number more than 6 appears.

There are no numbers in the sample space S that are greater than 6.

$E_4 = \emptyset$ (the empty set)

Number of outcomes favourable to $E_4$, $n(E_4) = 0$.

Probability $P(E_4) = \frac{n(E_4)}{n(S)} = \frac{0}{6} = 0$.

This is an impossible event.

(v) A number less than 6 will appear.

Let $E_5$ be the event that a number less than 6 appears.

The numbers in S that are less than 6 are 1, 2, 3, 4, and 5.

$E_5 = \{1, 2, 3, 4, 5\}$

Number of outcomes favourable to $E_5$, $n(E_5) = 5$.

Probability $P(E_5) = \frac{n(E_5)}{n(S)} = \frac{5}{6}$.

Alternatively, the complementary event is getting a number that is not less than 6, which means getting 6. Let $E'_5$ be the event of getting 6. $E'_5 = \{6\}$, $P(E'_5) = \frac{1}{6}$. $P(E_5) = 1 - P(E'_5) = 1 - \frac{1}{6} = \frac{5}{6}$.

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Summary of probabilities:

(i) $P(\text{prime number}) = \frac{1}{2}$

(ii) $P(\text{number} \geq 3) = \frac{2}{3}$

(iii) $P(\text{number} \leq 1) = \frac{1}{6}$

(iv) $P(\text{number} > 6) = 0$

(v) $P(\text{number} < 6) = \frac{5}{6}$

Given:

A standard pack of 52 cards.

One card is selected from the pack.

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To Find:

(a) The number of points in the sample space.

(b) The probability that the card is an ace of spades.

(c) The probability that the card is (i) an ace, (ii) black card.

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Solution:

The experiment is selecting one card from a well-shuffled deck of 52 cards. Assuming each card has an equal chance of being selected, each outcome is equally likely.

(a) How many points are there in the sample space?

The sample space S consists of all possible outcomes when drawing one card. Since there are 52 distinct cards in the deck, each card represents a unique outcome (a point in the sample space).

The total number of points in the sample space is the total number of cards in the deck.

$n(S) = 52$

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(b) Calculate the probability that the card is an ace of spades.

Let $E_b$ be the event that the card drawn is the ace of spades.

In a standard deck of 52 cards, there is exactly one ace of spades.

The number of outcomes favourable to $E_b$ is $n(E_b) = 1$.

The probability of event $E_b$ is $P(E_b) = \frac{n(E_b)}{n(S)}$.

$P(\text{ace of spades}) = \frac{1}{52}$.

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(c) Calculate the probability that the card is:

(i) an ace

Let $E_{ci}$ be the event that the card drawn is an ace.

In a standard deck of 52 cards, there are 4 aces (one in each suit: Hearts, Diamonds, Clubs, Spades).

The number of outcomes favourable to $E_{ci}$ is $n(E_{ci}) = 4$.

The probability of event $E_{ci}$ is $P(E_{ci}) = \frac{n(E_{ci})}{n(S)}$.

$P(\text{an ace}) = \frac{4}{52} = \frac{\cancel{4}^{1}}{\cancel{52}_{13}} = \frac{1}{13}$.

(ii) black card

Let $E_{cii}$ be the event that the card drawn is a black card.

The black suits are Clubs and Spades. There are 13 cards in the Clubs suit and 13 cards in the Spades suit.

The total number of black cards is $13 + 13 = 26$.

The number of outcomes favourable to $E_{cii}$ is $n(E_{cii}) = 26$.

The probability of event $E_{cii}$ is $P(E_{cii}) = \frac{n(E_{cii})}{n(S)}$.

$P(\text{a black card}) = \frac{26}{52} = \frac{\cancel{26}^{1}}{\cancel{52}_{2}} = \frac{1}{2}$.

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Summary of results:

(a) Number of points in the sample space = 52

(b) $P(\text{ace of spades}) = \frac{1}{52}$

(c) (i) $P(\text{an ace}) = \frac{1}{13}$

(c) (ii) $P(\text{a black card}) = \frac{1}{2}$

Given:

A fair coin with faces marked 1 and 6.

A fair standard six-sided die.

Both are tossed.

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To Find:

The probability that the sum of the numbers that turn up is (i) 3, (ii) 12.

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Solution:

The experiment consists of two independent events: tossing the modified coin and throwing the die.

Possible outcomes for the coin: {1, 6}

Possible outcomes for the die: {1, 2, 3, 4, 5, 6}

The sample space S for the combined experiment consists of ordered pairs (coin outcome, die outcome). The total number of outcomes is the product of the number of outcomes for each device.

Total number of outcomes, $n(S) = (\text{Outcomes for coin}) \times (\text{Outcomes for die}) = 2 \times 6 = 12$.

The sample space S is:

$S = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),$

$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}$

Since the coin and the die are fair, each of these 12 outcomes is equally likely, with a probability of $\frac{1}{12}$.

We are interested in the sum of the numbers that turn up.

(i) Probability that the sum of numbers is 3.

Let $E_1$ be the event that the sum is 3. We look for pairs $(c, d)$ in the sample space such that $c + d = 3$, where $c \in \{1, 6\}$ and $d \in \{1, 2, 3, 4, 5, 6\}$.

If $c = 1$, then $1 + d = 3 \implies d = 2$. The outcome is (1, 2).

If $c = 6$, then $6 + d = 3 \implies d = -3$. This is not a possible outcome for the die.

The only outcome where the sum is 3 is (1, 2).

$E_1 = \{ (1, 2) \}$

Number of outcomes favourable to $E_1$, $n(E_1) = 1$.

Probability $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{1}{12}$.

(ii) Probability that the sum of numbers is 12.

Let $E_2$ be the event that the sum is 12. We look for pairs $(c, d)$ in the sample space such that $c + d = 12$, where $c \in \{1, 6\}$ and $d \in \{1, 2, 3, 4, 5, 6\}$.

If $c = 1$, then $1 + d = 12 \implies d = 11$. This is not a possible outcome for the die.

If $c = 6$, then $6 + d = 12 \implies d = 6$. The outcome is (6, 6).

The only outcome where the sum is 12 is (6, 6).

$E_2 = \{ (6, 6) \}$

Number of outcomes favourable to $E_2$, $n(E_2) = 1$.

Probability $P(E_2) = \frac{n(E_2)}{n(S)} = \frac{1}{12}$.

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Summary of probabilities:

(i) $P(\text{sum is 3}) = \frac{1}{12}$

(ii) $P(\text{sum is 12}) = \frac{1}{12}$

Given:

City council consists of 4 men and 6 women.

Total number of council members = $4 + 6 = 10$.

One council member is selected at random for a committee.

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To Find:

The probability that the selected council member is a woman.

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Solution:

The experiment is selecting one council member at random from the 10 members.

The sample space S consists of the 10 individual council members.

Total number of outcomes in the sample space, $n(S) = 10$.

Since the selection is random, each council member is equally likely to be selected.

The probability of selecting any specific member is $\frac{1}{10}$.

Let W be the event that the selected council member is a woman.

The number of women on the council is 6.

The number of outcomes favourable to event W (selecting a woman) is the number of women available.

Number of outcomes favourable to W, $n(W) = 6$.

The probability of event W is $P(W) = \frac{n(W)}{n(S)}$.

$P(\text{selecting a woman}) = \frac{6}{10} = \frac{\cancel{6}^{3}}{\cancel{10}_{5}} = \frac{3}{5}$.

The likelihood that the selected council member is a woman is $\frac{3}{5}$. This can also be expressed as a decimal (0.6) or a percentage (60%).

Step 1: Define the Sample Space

A fair coin is tossed four times. The total number of possible outcomes in the sample space is $2^4 = 16$. Each outcome is equally likely, so the probability of any single outcome is $\frac{1}{16}$.

The sample space S consists of all possible sequences of Heads (H) and Tails (T):

$S = \{HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, \ $$ HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, \ $$ TTTH, TTTT\}$

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Step 2: Calculate the Different Amounts of Money

The person wins ₹ 1 for each head and loses ₹ 1.50 for each tail. Let $h$ be the number of heads and $t$ be the number of tails. The net amount is calculated as:

Amount = $(h \times \textsf{₹ } 1) - (t \times \textsf{₹ } 1.50)$

We can have the following combinations of heads and tails:

• Case 1: 4 Heads, 0 Tails (HHHH)

  Amount = $(4 \times 1) - (0 \times 1.50) = \textsf{₹ } 4.00$

• Case 2: 3 Heads, 1 Tail (e.g., HHHT)

  Amount = $(3 \times 1) - (1 \times 1.50) = \textsf{₹ } 1.50$

• Case 3: 2 Heads, 2 Tails (e.g., HHTT)

  Amount = $(2 \times 1) - (2 \times 1.50) = 2 - 3 = -\textsf{₹ } 1.00$ (a loss of ₹ 1)

• Case 4: 1 Head, 3 Tails (e.g., HTTT)

  Amount = $(1 \times 1) - (3 \times 1.50) = 1 - 4.50 = -\textsf{₹ } 3.50$ (a loss of ₹ 3.50)

• Case 5: 0 Heads, 4 Tails (TTTT)

  Amount = $(0 \times 1) - (4 \times 1.50) = -\textsf{₹ } 6.00$ (a loss of ₹ 6)

So, there are 5 different amounts of money you can have after four tosses: ₹ 4.00, ₹ 1.50, -₹ 1.00, -₹ 3.50, and -₹ 6.00.

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Step 3: Calculate the Probability of Each Amount

To find the probability of each amount, we need to count how many outcomes in the sample space lead to that amount.

Event (Heads, Tails)	Net Amount	Number of Favourable Outcomes	Probability
4 Heads, 0 Tails	Win ₹ 4.00	1 (HHHH)	$\frac{1}{16}$
3 Heads, 1 Tail	Win ₹ 1.50	4 (HHHT, HHTH, HTHH, THHH)	$\frac{4}{16} = \frac{1}{4}$
2 Heads, 2 Tails	Lose ₹ 1.00	6 (HHTT, HTHT, HTTH, THHT, THTH, TTHH)	$\frac{6}{16} = \frac{3}{8}$
1 Head, 3 Tails	Lose ₹ 3.50	4 (HTTT, THTT, TTHT, TTTH)	$\frac{4}{16} = \frac{1}{4}$
0 Heads, 4 Tails	Lose ₹ 6.00	1 (TTTT)	$\frac{1}{16}$

Sample Space and Total Outcomes

When three fair coins are tossed, the total number of possible outcomes is $2 \times 2 \times 2 = 8$. The sample space (S) is:

$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$

The probability of any event E is calculated as $P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$.

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(i) 3 heads

Favourable outcome: {HHH}. Number of favourable outcomes = 1.

$P(3 \text{ heads}) = \frac{1}{8}$.

(ii) 2 heads

Favourable outcomes: {HHT, HTH, THH}. Number of favourable outcomes = 3.

$P(2 \text{ heads}) = \frac{3}{8}$.

(iii) atleast 2 heads

This means 2 heads OR 3 heads. Favourable outcomes: {HHT, HTH, THH, HHH}. Number of favourable outcomes = 4.

$P(\text{atleast 2 heads}) = \frac{4}{8} = \frac{1}{2}$.

(iv) atmost 2 heads

This means 0, 1, or 2 heads. It's easier to find the complement: the event 'more than 2 heads', which is '3 heads'.

$P(\text{atmost 2 heads}) = 1 - P(3 \text{ heads}) = 1 - \frac{1}{8} = \frac{7}{8}$.

(v) no head

This means all tails. Favourable outcome: {TTT}. Number of favourable outcomes = 1.

$P(\text{no head}) = \frac{1}{8}$.

(vi) 3 tails

Favourable outcome: {TTT}. Number of favourable outcomes = 1.

$P(3 \text{ tails}) = \frac{1}{8}$.

(vii) exactly two tails

Favourable outcomes: {HTT, THT, TTH}. Number of favourable outcomes = 3.

$P(\text{exactly two tails}) = \frac{3}{8}$.

(viii) no tail

This means all heads. Favourable outcome: {HHH}. Number of favourable outcomes = 1.

$P(\text{no tail}) = \frac{1}{8}$.

(ix) atmost two tails

This means 0, 1, or 2 tails. The complement is the event 'more than 2 tails', which is '3 tails'.

$P(\text{atmost two tails}) = 1 - P(3 \text{ tails}) = 1 - \frac{1}{8} = \frac{7}{8}$.

Given:

The probability of an event A, $P(A) = \frac{2}{11}$.

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To Find:

The probability of the event ‘not A’, which is denoted as $P(A')$.

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Solution:

The probability of an event and its complement always add up to 1. This is a fundamental rule of probability:

To find the probability of 'not A', we rearrange the formula:

Substitute the given value of $P(A)$:

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Answer:

The probability of the event ‘not A’ is $\frac{9}{11}$.

Step 1: Analyze the Word

The word is ‘ASSASSINATION’.

Total number of letters = 13. This is our total number of possible outcomes, $n(S) = 13$.

Let's count the number of vowels and consonants.

The vowels are A, E, I, O, U.

• Number of 'A's = 3
• Number of 'I's = 2
• Number of 'O's = 1

Total number of vowels = $3 + 2 + 1 = 6$.

The consonants are the remaining letters.

• Number of 'S's = 4
• Number of 'N's = 2
• Number of 'T's = 1

Total number of consonants = $4 + 2 + 1 = 7$.

(Check: $6 \text{ vowels} + 7 \text{ consonants} = 13 \text{ total letters}$. The count is correct.)

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(i) Probability that the letter is a vowel

The number of favourable outcomes (choosing a vowel) is 6.

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(ii) Probability that the letter is a consonant

The number of favourable outcomes (choosing a consonant) is 7.

Alternatively, since a letter is either a vowel or a consonant, these are complementary events. So, $P(\text{consonant}) = 1 - P(\text{vowel}) = 1 - \frac{6}{13} = \frac{7}{13}$.

Given:

A person chooses 6 different natural numbers from 1 to 20. The order of selection does not matter.

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To Find:

The probability of winning the prize.

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Solution:

This is a problem of combinations, since the order of the numbers chosen does not matter.

Step 1: Calculate the total number of possible outcomes.

The total number of outcomes is the number of ways a person can choose 6 different numbers from 20. This is calculated using the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Here, $n=20$ and $k=6$.

Let's simplify the calculation:

• $6 \times 3 = 18$ (cancels with the 18 in the numerator)
• $4 \times 5 = 20$ (cancels with the 20 in the numerator)
• $\frac{16}{2} = 8$

$ = 323 \times 120$

$ = 38760$

So, there are 38,760 possible combinations of 6 numbers.

Step 2: Determine the number of favourable outcomes.

To win the prize, the person must choose the single, specific combination of 6 numbers that has been fixed by the lottery committee. Therefore, there is only 1 favourable (winning) outcome.

Number of favourable outcomes = 1.

Step 3: Calculate the probability.

The probability of an event is the ratio of favourable outcomes to the total number of outcomes.

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Answer:

The probability of winning the prize in the game is $\frac{1}{38760}$.

To check if probabilities are consistently defined, we must verify that they do not violate any fundamental rules of probability. Two key rules are:

1. The probability of an intersection of two events cannot be greater than the probability of either individual event. That is, $P(A \cap B) \le P(A)$ and $P(A \cap B) \le P(B)$.

2. The addition rule must hold: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

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(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6

Let's check Rule 1.

Is $P(A \cap B) \le P(A)$?

Is $0.6 \le 0.5$? This is false.

The probability of both A and B occurring (0.6) cannot be greater than the probability of A occurring (0.5). Therefore, the probabilities are not consistently defined.

Conclusion: Not consistently defined.

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(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8

Here we are given the union. Let's use the addition rule (Rule 2) to find the implied probability of the intersection, $P(A \cap B)$, and then check if it is valid.

Now, we must check if this implied probability is consistent. The value $P(A \cap B) = 0.1$ must be between 0 and 1 (which it is) and must satisfy Rule 1.

Is $P(A \cap B) \le P(A)$? Is $0.1 \le 0.5$? Yes, this is true.

Is $P(A \cap B) \le P(B)$? Is $0.1 \le 0.4$? Yes, this is true.

Since all rules of probability are satisfied, the probabilities are consistently defined.

Conclusion: Consistently defined.

To fill in the blanks, we will use the Addition Rule of Probability, which relates the probabilities of two events with their intersection and union:

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(i) Find P(A ∪ B)

Given: $P(A) = \frac{1}{3}$, $P(B) = \frac{1}{5}$, $P(A \cap B) = \frac{1}{15}$.

Using the formula:

To add and subtract the fractions, we find a common denominator, which is 15.

The missing value is $\frac{7}{15}$.

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(ii) Find P(B)

Given: $P(A) = 0.35$, $P(A \cap B) = 0.25$, $P(A \cup B) = 0.6$.

We rearrange the addition rule formula to solve for P(B):

We must check if this is a valid probability. Since $P(A \cap B) = 0.25 \le P(B) = 0.5$, the value is consistent.

The missing value is 0.5.

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(iii) Find P(A ∩ B)

Given: $P(A) = 0.5$, $P(B) = 0.35$, $P(A \cup B) = 0.7$.

We rearrange the addition rule formula to solve for $P(A \cap B)$:

We must check if this is a valid probability. The value 0.15 is between 0 and 1, and it is less than both P(A) and P(B). The value is consistent.

The missing value is 0.15.

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Completed Table:

	P(A)	P(B)	P(A ∩ B)	P(A ∪ B)
(i)	$\frac{1}{3}$	$\frac{1}{5}$	$\frac{1}{15}$	$\frac{7}{15}$
(ii)	0.35	0.5	0.25	0.6
(iii)	0.5	0.35	0.15	0.7

Given:

Probability of event A, $P(A) = \frac{3}{5}$.

Probability of event B, $P(B) = \frac{1}{5}$.

Events A and B are mutually exclusive.

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To Find:

The probability of 'A or B', which is $P(A \cup B)$.

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Solution:

For any two events A and B, the probability of their union is given by the Addition Rule:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

We are given that events A and B are mutually exclusive.

Mutually exclusive events are events that cannot occur at the same time. This means their intersection is the empty set ($\emptyset$).

The probability of the intersection of mutually exclusive events is 0.

$P(A \cap B) = P(\emptyset) = 0$.

Substitute the given probabilities and $P(A \cap B) = 0$ into the Addition Rule:

$P(A \cup B) = P(A) + P(B) - 0$

$P(A \cup B) = P(A) + P(B)$

$P(A \cup B) = \frac{3}{5} + \frac{1}{5}$

$P(A \cup B) = \frac{3 + 1}{5} = \frac{4}{5}$.

The probability of 'A or B' is $\frac{4}{5}$.

Given:

Probability of event E, $P(E) = \frac{1}{4}$.

Probability of event F, $P(F) = \frac{1}{2}$.

Probability of event 'E and F', $P(E \cap F) = \frac{1}{8}$.

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To Find:

(i) $P(E \text{ or } F)$, which is $P(E \cup F)$.

(ii) $P(\text{not } E \text{ and not } F)$, which is $P(E' \cap F')$.

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Solution:

(i) P(E or F):

We use the Addition Rule for probability:

$P(E \cup F) = P(E) + P(F) - P(E \cap F)$

Substitute the given values:

$P(E \cup F) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8}$

Find a common denominator for the fractions (the least common multiple of 4, 2, and 8 is 8):

$\frac{1}{4} = \frac{1 \times 2}{4 \times 2} = \frac{2}{8}$

$\frac{1}{2} = \frac{1 \times 4}{2 \times 4} = \frac{4}{8}$

$P(E \cup F) = \frac{2}{8} + \frac{4}{8} - \frac{1}{8}$

$P(E \cup F) = \frac{2 + 4 - 1}{8} = \frac{5}{8}$.

(ii) P(not E and not F):

'not E' is the complement of E, denoted by E′.

'not F' is the complement of F, denoted by F′.

'not E and not F' is the intersection of E′ and F′, denoted by $E' \cap F'$.

By De Morgan's Law, $(E \cup F)' = E' \cap F'$.

So, $P(E' \cap F') = P((E \cup F)')$.

The probability of the complement of an event is 1 minus the probability of the event:

$P((E \cup F)') = 1 - P(E \cup F)$.

From part (i), we found $P(E \cup F) = \frac{5}{8}$.

$P(E' \cap F') = 1 - \frac{5}{8}$

$P(E' \cap F') = \frac{8}{8} - \frac{5}{8} = \frac{8 - 5}{8} = \frac{3}{8}$.

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Summary of results:

(i) $P(E \text{ or } F) = \frac{5}{8}$

(ii) $P(\text{not } E \text{ and not } F) = \frac{3}{8}$

Given:

Events E and F.

$P(\text{not E or not F}) = 0.25$.

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To State:

Whether E and F are mutually exclusive, with reason.

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Solution:

The event 'not E or not F' can be written using set notation as $E' \cup F'$.

We are given $P(E' \cup F') = 0.25$.

By De Morgan's Law, the union of the complements of two events is equal to the complement of their intersection:

$E' \cup F' = (E \cap F)'$

So, $P(E' \cup F') = P((E \cap F)')$.

We have $P((E \cap F)') = 0.25$.

The probability of the complement of an event is 1 minus the probability of the event:

$P((E \cap F)') = 1 - P(E \cap F)$.

Substituting the given probability:

$0.25 = 1 - P(E \cap F)$.

Now, we can solve for $P(E \cap F)$:

$P(E \cap F) = 1 - 0.25$

$P(E \cap F) = 0.75$.

Two events E and F are mutually exclusive if their intersection is the empty set, which means the probability of their intersection is 0 ($P(E \cap F) = 0$).

In this case, we found $P(E \cap F) = 0.75$.

Since $P(E \cap F) = 0.75 \neq 0$, the events E and F have common outcomes.

Therefore, events E and F are not mutually exclusive.

Conclusion: E and F are not mutually exclusive.

Reason: We found that $P(E \cap F) = 0.75$. Since the probability of their intersection is not 0, the events can occur simultaneously, and thus they are not mutually exclusive.

Given:

Probability of event A, $P(A) = 0.42$.

Probability of event B, $P(B) = 0.48$.

Probability of event 'A and B' (intersection of A and B), $P(A \cap B) = 0.16$.

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To Determine:

(i) Probability of 'not A', $P(A')$.

(ii) Probability of 'not B', $P(B')$.

(iii) Probability of 'A or B', $P(A \cup B)$.

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Solution:

(i) P(not A):

The event 'not A' is the complement of event A, denoted by $A'$. The probability of the complement of an event is given by $P(A') = 1 - P(A)$.

Substituting the given value for $P(A)$:

$P(A') = 1 - 0.42$

$P(A') = 0.58$

(ii) P(not B):

The event 'not B' is the complement of event B, denoted by $B'$. The probability of the complement of an event is given by $P(B') = 1 - P(B)$.

Substituting the given value for $P(B)$:

$P(B') = 1 - 0.48$

$P(B') = 0.52$

(iii) P(A or B):

The event 'A or B' is the union of events A and B, denoted by $A \cup B$. We use the Addition Rule for probability:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Substitute the given values for $P(A)$, $P(B)$, and $P(A \cap B)$:

$P(A \cup B) = 0.42 + 0.48 - 0.16$

$P(A \cup B) = 0.90 - 0.16$

$P(A \cup B) = 0.74$

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Summary of results:

(i) $P(\text{not A}) = 0.58$

(ii) $P(\text{not B}) = 0.52$

(iii) $P(A \text{ or } B) = 0.74$

Given:

Let M be the event that a randomly selected student studies Mathematics.

Let B be the event that a randomly selected student studies Biology.

We are given the following probabilities:

$P(M) = 40\% = \frac{40}{100} = 0.40$

$P(B) = 30\% = \frac{30}{100} = 0.30$

The probability that a student studies both Mathematics and Biology is the probability of the intersection of M and B.

$P(M \cap B) = 10\% = \frac{10}{100} = 0.10$

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To Find:

The probability that a randomly selected student studies Mathematics or Biology, which is $P(M \cup B)$.

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Solution:

The event 'studying Mathematics or Biology' is the union of the events M and B. We use the Addition Rule for probability:

$P(M \cup B) = P(M) + P(B) - P(M \cap B)$

Substitute the given probabilities:

$P(M \cup B) = 0.40 + 0.30 - 0.10$

$P(M \cup B) = 0.70 - 0.10$

$P(M \cup B) = 0.60$

The probability that a randomly selected student studies Mathematics or Biology is 0.60, or 60%.

Given:

Let $E_1$ be the event that a student passes the first examination.

Let $E_2$ be the event that a student passes the second examination.

We are given the following probabilities:

Probability of passing the first examination, $P(E_1) = 0.8$.

Probability of passing the second examination, $P(E_2) = 0.7$.

The probability of passing at least one of them is the probability of the union of $E_1$ and $E_2$.

$P(E_1 \cup E_2) = 0.95$.

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To Find:

The probability of passing both examinations, which is $P(E_1 \cap E_2)$.

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Solution:

We use the Addition Rule for probability:

$P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$

We want to find $P(E_1 \cap E_2)$. We can rearrange the formula:

$P(E_1 \cap E_2) = P(E_1) + P(E_2) - P(E_1 \cup E_2)$

Substitute the given probabilities:

$P(E_1 \cap E_2) = 0.8 + 0.7 - 0.95$

$P(E_1 \cap E_2) = 1.5 - 0.95$

$P(E_1 \cap E_2) = 0.55$

The probability of a randomly chosen student passing both examinations is 0.55.

Given:

Let E be the event that a student passes the English examination.

Let H be the event that a student passes the Hindi examination.

We are given the following probabilities:

Probability of passing both English and Hindi, $P(E \cap H) = 0.5$.

Probability of passing neither English nor Hindi, $P(\text{neither E nor H}) = 0.1$.

Probability of passing the English examination, $P(E) = 0.75$.

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To Find:

The probability of passing the Hindi examination, $P(H)$.

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Solution:

The event 'passing neither English nor Hindi' is the complement of the event 'passing at least one of them'.

The event 'passing at least one of them' is the union of events E and H, $E \cup H$.

The event 'passing neither E nor H' is the complement of $E \cup H$, which is $(E \cup H)'$.

So, $P((E \cup H)') = 0.1$.

Using the property of complementary events, $P(E \cup H) + P((E \cup H)') = 1$.

$P(E \cup H) + 0.1 = 1$

$P(E \cup H) = 1 - 0.1 = 0.9$.

Now we know the probability of the union of E and H, $P(E \cup H) = 0.9$.

We also know the Addition Rule for probability:

$P(E \cup H) = P(E) + P(H) - P(E \cap H)$

We have $P(E \cup H) = 0.9$, $P(E) = 0.75$, and $P(E \cap H) = 0.5$. We want to find $P(H)$.

Substitute the known values into the Addition Rule:

$0.9 = 0.75 + P(H) - 0.5$

$0.9 = (0.75 - 0.5) + P(H)$

$0.9 = 0.25 + P(H)$

Solve for $P(H)$:

$P(H) = 0.9 - 0.25$

$P(H) = 0.65$

The probability of a randomly chosen student passing the Hindi examination is 0.65.

Given:

Total number of students in the class, $n(S) = 60$.

Let C be the event that a student opted for NCC.

Let S be the event that a student opted for NSS.

Number of students who opted for NCC, $n(C) = 30$.

Number of students who opted for NSS, $n(N) = 32$.

Number of students who opted for both NCC and NSS, $n(C \cap N) = 24$.

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(i) The student opted for NCC or NSS.

We need to find the probability of the event 'C or N', which is $P(C \cup N)$.

First, we find the number of students who opted for NCC or NSS, which is $n(C \cup N)$.

Using the formula for the union of two sets:

So, 38 students opted for at least one of the two.

The probability is:

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(ii) The student has opted neither NCC nor NSS.

This is the event 'not C and not N', which is the complement of the event 'C or N'. We need to find $P((C \cup N)')$.

Using the rule for complementary events, $P((C \cup N)') = 1 - P(C \cup N)$.

Alternatively (using counts):

Number of students who opted for neither = Total students - Number of students who opted for at least one.

Number = $60 - 38 = 22$.

Probability = $\frac{22}{60} = \frac{11}{30}$.

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(iii) The student has opted NSS but not NCC.

This is the event 'N and not C'. We need to find the number of students who are in NSS only.

This can be found by subtracting the number of students in both from the total number of students in NSS.

The probability is:

Sample Space and Total Outcomes

The experiment is arranging the four cities (A, B, C, D) in a random order. The total number of possible arrangements (permutations) is $4! = 4 \times 3 \times 2 \times 1 = 24$.

The entire sample space (S) with all 24 possible outcomes is:

ABCD	BACD	CABD	DABC
ABDC	BADC	CADB	DACB
ACBD	BCAD	CBAD	DBAC
ACDB	BCDA	CBDA	DBCA
ADBC	BDAC	CDAB	DCAB
ADCB	BDCA	CDBA	DCBA

The total number of outcomes is $n(S) = 24$.

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(i) Probability that she visits A before B

We need to count all the arrangements in the sample space where A appears before B.

Favourable outcomes are:

{ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, CABD, CADB, CDAB, DABC, DACB, DCAB}

Number of favourable outcomes = 12.

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(ii) Probability that she visits A before B and B before C

We need to count arrangements where the relative order is A...B...C.

Favourable outcomes are:

{ABCD, ABDC, ADBC, DABC}

Number of favourable outcomes = 4.

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(iii) Probability that she visits A first and B last

We need to count arrangements that start with A and end with B (A _ _ B).

Favourable outcomes are:

{ACDB, ADCB}

Number of favourable outcomes = 2.

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(iv) Probability that she visits A either first or second

We count the arrangements where A is in the first position or the second position.

Favourable outcomes where A is first: {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB} (6 outcomes)

Favourable outcomes where A is second: {BACD, BADC, CABD, CADB, DABC, DACB} (6 outcomes)

Total number of favourable outcomes = $6 + 6 = 12$.

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(v) Probability that she visits A just before B

We need to count arrangements where the sequence 'AB' appears together.

Favourable outcomes are:

{ABCD, ABDC, CABD, DABC, CDAB, DCAB}

Number of favourable outcomes = 6.

Step 1: Find the total number of possible hands

The total number of ways to draw 7 cards from a deck of 52 is given by the combination formula $\binom{52}{7}$.

This is the total number of outcomes in our sample space.

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(i) Probability that the hand contains all Kings

This means the hand must contain all 4 Kings and 3 other cards (which are not Kings).

Number of ways to choose 4 Kings from 4 Kings = $\binom{4}{4} = 1$.

Number of ways to choose the remaining 3 cards from the 48 non-King cards = $\binom{48}{3}$.

$\binom{48}{3} = \frac{48 \times 47 \times 46}{3 \times 2 \times 1} = 16 \times 47 \times 23 = 17296$.

Total favourable outcomes = $1 \times 17296 = 17296$.

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(ii) Probability that the hand contains 3 Kings

This means the hand must contain exactly 3 Kings and 4 other cards (which are not Kings).

Number of ways to choose 3 Kings from 4 Kings = $\binom{4}{3} = 4$.

Number of ways to choose the remaining 4 cards from the 48 non-King cards = $\binom{48}{4}$.

$\binom{48}{4} = \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1} = 194580$.

Total favourable outcomes = $4 \times 194580 = 778320$.

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(iii) Probability that the hand contains atleast 3 Kings.

"At least 3 Kings" means the hand contains either exactly 3 Kings OR exactly 4 Kings. Since these two events are mutually exclusive, we can add their probabilities.

$P(\text{atleast 3 Kings}) = P(\text{3 Kings}) + P(\text{4 Kings})$

We already have the number of favourable outcomes for each case:

• Number of hands with 3 Kings = 778,320
• Number of hands with 4 Kings = 17,296

Total favourable outcomes = $778,320 + 17,296 = 795,616$.

To Prove:

$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) \ $$ – P ( B \cap C) + P ( A \cap B \cap C)$

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Proof:

We will start with the known addition rule for two events: $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$.

Let's treat the event $(A \cup B)$ as a single event. We can write $P(A \cup B \cup C)$ as $P((A \cup B) \cup C)$.

Applying the two-event rule, let $X = (A \cup B)$ and $Y = C$:

Now, let's expand the terms in equation (1).

1. Expand $P(A \cup B)$ using the two-event rule:

2. Expand the intersection term $P((A \cup B) \cap C)$. Using the distributive law of sets, $(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$. So, we need to find $P((A \cap C) \cup (B \cap C))$.

Again, apply the two-event rule, this time with $X = (A \cap C)$ and $Y = (B \cap C)$:

The intersection of $(A \cap C)$ and $(B \cap C)$ is simply $(A \cap B \cap C)$. So, the expression becomes:

Now, substitute equations (2) and (3) back into equation (1):

$P(A \cup B \cup C) = \underbrace{[P(A) + P(B) - P(A \cap B)]}_{P(A \cup B)} + P(C) \ $$ - \underbrace{[P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)]}_{P((A \cup B) \cap C)}$

Remove the brackets and rearrange the terms:

$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) \ $$ - P(B \cap C) + P(A \cap B \cap C)$

Hence Proved.

Total Possible Outcomes

There are 5 teams (A, B, C, D, E). The total number of different finishing orders (permutations) is $5!$.

There are 120 possible outcomes in the sample space.

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(a) What is the probability that A, B and C finish first, second and third, respectively.

We need to find the number of outcomes where the finishing order starts with A, B, C.

The first 3 positions are fixed: 1st - A, 2nd - B, 3rd - C.

The remaining 2 teams (D and E) can finish in the 4th and 5th positions in $2!$ ways (DE or ED).

Number of favourable outcomes = $1 \times 1 \times 1 \times 2! = 2$.

The probability is:

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(b) What is the probability that A, B and C are first three to finish (in any order).

We need to find the number of outcomes where the top 3 finishers are the teams A, B, and C.

Ways to arrange the top 3: The 3 teams (A, B, C) can be arranged in the first 3 positions in $3!$ ways.

$3! = 3 \times 2 \times 1 = 6$.

Ways to arrange the bottom 2: The remaining 2 teams (D, E) can be arranged in the last 2 positions in $2!$ ways.

$2! = 2 \times 1 = 2$.

The total number of favourable outcomes is the product of these two numbers.

Number of favourable outcomes = $6 \times 2 = 12$.

The probability is:

Given:

A box contains:

• 10 red marbles
• 20 blue marbles
• 30 green marbles

Total number of marbles = $10 + 20 + 30 = 60$.

Number of marbles to be drawn = 5.

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Total Number of Possible Outcomes

The total number of ways to draw 5 marbles from 60 is given by the combination formula $\binom{60}{5}$.

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(i) Probability that all will be blue

We need to find the number of ways to draw 5 blue marbles from the 20 available blue marbles.

Number of favourable outcomes = $\binom{20}{5}$.

The probability is the ratio of favourable outcomes to total outcomes:

This fraction simplifies to $\frac{34}{11977}$.

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(ii) Probability that atleast one will be green

It is easier to calculate the probability of the complementary event: "none of the marbles are green", and then subtract this from 1.

The event "none are green" means all 5 marbles are drawn from the non-green marbles (10 red + 20 blue = 30 marbles).

Number of ways to draw 5 non-green marbles from 30 = $\binom{30}{5}$.

The probability of drawing no green marbles is:

The probability of drawing "at least one green" is $1 - P(\text{no green})$.

This fraction simplifies to $\frac{131}{134}$.

Given:

A hand of 4 cards is drawn from a standard deck of 52 cards.

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To Find:

The probability of obtaining 3 diamonds and one spade.

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Solution:

Step 1: Find the total number of possible hands.

The total number of ways to draw 4 cards from 52 is given by the combination $\binom{52}{4}$.

This is the total number of outcomes in the sample space.

Step 2: Find the number of favourable outcomes.

We want to obtain a hand with 3 diamonds and 1 spade.

There are 13 diamond cards and 13 spade cards in the deck.

• The number of ways to choose 3 diamonds from 13 is $\binom{13}{3}$.
• The number of ways to choose 1 spade from 13 is $\binom{13}{1}$.

Calculating these combinations:

By the multiplication principle, the total number of ways to get 3 diamonds and 1 spade is:

Number of favourable outcomes = $\binom{13}{3} \times \binom{13}{1} = 286 \times 13 = 3718$.

Step 3: Calculate the probability.

Probability = $\frac{\text{Favourable outcomes}}{\text{Total outcomes}}$.

Given:

A die has 6 faces with the following numbers:

• Number ‘1’ appears on 2 faces.
• Number ‘2’ appears on 3 faces.
• Number ‘3’ appears on 1 face.

Total number of outcomes = 6.

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(i) P(2)

The event is rolling a '2'. The number of faces with '2' is 3.

Number of favourable outcomes = 3.

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(ii) P(1 or 3)

The event is rolling a '1' or a '3'. These are mutually exclusive events.

Number of faces with '1' = 2.

Number of faces with '3' = 1.

Total favourable outcomes = $2 + 1 = 3$.

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(iii) P(not 3)

This is the complement of the event 'getting a 3'.

First, find the probability of getting a '3'. There is 1 face with '3'.

$P(3) = \frac{1}{6}$.

Now, use the complement rule: $P(\text{not } 3) = 1 - P(3)$.

Given:

Total tickets = 10,000.

Number of prize tickets = 10.

Number of non-prize (losing) tickets = $10,000 - 10 = 9990$.

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(a) one ticket

We want the probability that the one ticket you buy is a non-prize ticket.

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(b) two tickets

We want the probability that both of the two tickets you buy are non-prize tickets. This involves combinations.

Total ways to choose 2 tickets from 10,000 = $\binom{10000}{2}$.

Ways to choose 2 non-prize tickets from 9,990 = $\binom{9990}{2}$.

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(c) 10 tickets

We want the probability that all 10 tickets you buy are non-prize tickets.

Total ways to choose 10 tickets from 10,000 = $\binom{10000}{10}$.

Ways to choose 10 non-prize tickets from 9,990 = $\binom{9990}{10}$.

Given:

Total students = 100.

Section 1 has 40 students.

Section 2 has 60 students.

We are interested in the placement of two specific students: you and your friend.

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(a) Probability that you both enter the same section.

This can happen in two mutually exclusive ways: either both of you are in Section 1, OR both of you are in Section 2. We can find the probability of each case and add them.

Case 1: Both in Section 1 (40 students)

The probability that you are in Section 1 is $\frac{40}{100}$.

Given that you are in Section 1, there are now 39 spots left in that section and 99 students remaining. So, the probability that your friend is also in Section 1 is $\frac{39}{99}$.

$P(\text{both in Section 1}) = \frac{40}{100} \times \frac{39}{99}$

Case 2: Both in Section 2 (60 students)

The probability that you are in Section 2 is $\frac{60}{100}$.

Given that you are in Section 2, there are now 59 spots left and 99 students remaining. The probability that your friend is also in Section 2 is $\frac{59}{99}$.

$P(\text{both in Section 2}) = \frac{60}{100} \times \frac{59}{99}$

Total Probability:

$P(\text{same section}) = P(\text{both in Section 1}) + P(\text{both in Section 2})$

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(b) Probability that you both enter the different sections.

The event "entering different sections" is the complement of the event "entering the same section".

Therefore, we can calculate this probability as:

Alternate Method for (b):

This can happen in two ways: (You in Section 1 and friend in Section 2) OR (You in Section 2 and friend in Section 1).

$P(\text{You in S1, friend in S2}) = \frac{40}{100} \times \frac{60}{99}$

$P(\text{You in S2, friend in S1}) = \frac{60}{100} \times \frac{40}{99}$

$P(\text{different sections}) = \frac{40 \times 60}{100 \times 99} + \frac{60 \times 40}{100 \times 99} = 2 \times \frac{2400}{9900} = \frac{4800}{9900} = \frac{48}{99} \ $$ = \frac{16}{33}$.

Total Number of Outcomes

Let the three letters be $L_1, L_2, L_3$ and their corresponding correct envelopes be $E_1, E_2, E_3$.

The total number of ways to place the 3 letters into the 3 envelopes is the number of permutations of 3 items, which is $3!$.

Favourable Outcomes vs. Complementary Event

The event we are interested in is "at least one letter is in its proper envelope".

It is often easier to calculate the probability of the complementary event, which is "no letter is in its proper envelope", and then subtract this from 1.

Finding the Number of Unfavourable Outcomes

Let's find the number of ways to place the letters so that none of them are in the correct envelope. This is known as a derangement.

The possible arrangements (Letter in Envelope) are:

1. $L_1 \to E_1, L_2 \to E_2, L_3 \to E_3$ (All correct)
2. $L_1 \to E_1, L_2 \to E_3, L_3 \to E_2$ (One correct)
3. $L_1 \to E_2, L_2 \to E_1, L_3 \to E_3$ (One correct)
4. $L_1 \to E_3, L_2 \to E_2, L_3 \to E_1$ (One correct)
5. $L_1 \to E_2, L_2 \to E_3, L_3 \to E_1$ (None correct)
6. $L_1 \to E_3, L_2 \to E_1, L_3 \to E_2$ (None correct)

There are 2 outcomes where no letter is in its proper envelope.

Calculating the Probabilities

The probability of the event "no letter is in its proper envelope" is:

The probability of "at least one letter is in its proper envelope" is the complement of this.

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Answer:

The probability that at least one letter is in its proper envelope is $\frac{2}{3}$.

Given:

$P(A) = 0.54$

$P(B) = 0.69$

$P(A \cap B) = 0.35$

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(i) P(A ∪ B)

We use the Addition Rule of Probability:

Answer: 0.88

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(ii) P(A´ ∩ B´)

This is the probability that neither A nor B occurs. By De Morgan's Law, $A' \cap B' = (A \cup B)'$.

So, we need to find the probability of the complement of $(A \cup B)$.

Using the result from part (i):

Answer: 0.12

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(iii) P(A ∩ B´)

This is the probability that event A occurs but event B does not. It can be found by subtracting the probability of the intersection from the probability of A.

Answer: 0.19

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(iv) P(B ∩ A´)

This is the probability that event B occurs but event A does not. It can be found by subtracting the probability of the intersection from the probability of B.

Answer: 0.34

Given:

There is a group of 5 persons. A spokesperson is selected at random from this group.

Total number of outcomes in the sample space, $n(S) = 5$.

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To Find:

The probability that the spokesperson will be either male OR over 35 years old.

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Solution:

Let's define the two events:

• Let M be the event that the spokesperson is male.
• Let O be the event that the spokesperson is over 35 years old.

We need to find the probability of the union of these two events, $P(M \cup O)$.

We will use the Addition Rule of Probability: $P(M \cup O) = P(M) + P(O) - P(M \cap O)$.

Step 1: Find the number of outcomes for each event.

From the table, we list the people who satisfy each condition:

• Event M (Male): Harish, Rohan, Salim.

  Number of favourable outcomes, $n(M) = 3$.

• Event O (Over 35 years): Sheetal (46), Salim (41).

  Number of favourable outcomes, $n(O) = 2$.

• Event M ∩ O (Male AND Over 35 years): Salim.

  Number of favourable outcomes, $n(M \cap O) = 1$.

Step 2: Calculate the probabilities for each event.

Step 3: Apply the Addition Rule.

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Alternate Method (Direct Counting):

We can directly count the number of people who are either male or over 35 years old. We must be careful not to double-count anyone who satisfies both conditions.

The people are: Harish (male), Rohan (male), Sheetal (over 35), Salim (male AND over 35).

The unique individuals are: Harish, Rohan, Sheetal, Salim.

Number of favourable outcomes = 4.

Probability = $\frac{\text{Favourable outcomes}}{\text{Total outcomes}} = \frac{4}{5}$.

Given Digits: {0, 1, 3, 5, 7}

Condition 1: The number must be a 4-digit number greater than 5,000. This means the first digit (thousands place) must be 5 or 7. Critically, the number 5,000 itself is excluded.

Condition 2: The number must be divisible by 5. This means the last digit (units place) must be either 0 or 5.

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(i) The digits are repeated

Total Possible Numbers (Sample Space):

• Thousands place: 2 choices (5 or 7)
• Hundreds place: 5 choices (0, 1, 3, 5, 7)
• Tens place: 5 choices (0, 1, 3, 5, 7)
• Units place: 5 choices (0, 1, 3, 5, 7)

Total combinations = $2 \times 5 \times 5 \times 5 = 250$.

However, this count includes the number 5000, which is not "greater than 5,000". So, we must subtract it.

Total valid numbers = $250 - 1 = 249$. So, $n(S) = 249$.

Favourable Numbers (Divisible by 5):

• Thousands place: 2 choices (5 or 7)
• Hundreds place: 5 choices
• Tens place: 5 choices
• Units place: 2 choices (0 or 5)

Favourable combinations = $2 \times 5 \times 5 \times 2 = 100$.

This count also includes the number 5000. Since we must exclude it, the number of favourable outcomes is:

Favourable numbers = $100 - 1 = 99$. So, $n(E) = 99$.

Probability:

Simplifying by dividing the numerator and denominator by 3:

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(ii) The repetition of digits is not allowed

Total Possible Numbers (Sample Space):

• Thousands place: 2 choices (5 or 7)
• Hundreds place: 4 choices (any of the 4 remaining digits)
• Tens place: 3 choices (any of the 3 remaining digits)
• Units place: 2 choices (any of the 2 remaining digits)

Total numbers = $2 \times 4 \times 3 \times 2 = 48$. (Note: 5000 cannot be formed here, so no subtraction is needed).

Favourable Numbers (Divisible by 5):

We must consider two separate cases based on the last digit, as the choice of '5' for the last digit restricts the choice for the first digit.

Case 1: The number ends in 0.

• Units place: 1 choice (0)
• Thousands place: 2 choices (5 or 7)
• Hundreds place: 3 choices (remaining)
• Tens place: 2 choices (remaining)

Number of favourable outcomes = $2 \times 3 \times 2 \times 1 = 12$.

Case 2: The number ends in 5.

• Units place: 1 choice (5)
• Thousands place: 1 choice (must be 7, as it must be > 5 and '5' is already used)
• Hundreds place: 3 choices (remaining)
• Tens place: 2 choices (remaining)

Number of favourable outcomes = $1 \times 3 \times 2 \times 1 = 6$.

Total favourable numbers = $12 (\text{from Case 1}) + 6 (\text{from Case 2}) = 18$.

Probability:

Given:

A number lock with 4 wheels, using digits 0 to 9. The opening sequence consists of 4 unique digits (no repeats).

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To Find:

The probability of guessing the right sequence.

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Solution:

Step 1: Find the total number of possible sequences.

We are forming a 4-digit sequence from 10 distinct digits (0-9) where the order matters and digits cannot be repeated. This is a permutation problem.

The number of ways to arrange 4 items chosen from 10 is given by the permutation formula $^{n}P_k = \frac{n!}{(n-k)!}$.

Here, $n=10$ and $k=4$.

So, there are 5,040 possible sequences to open the lock.

Step 2: Determine the number of favourable outcomes.

There is only one single, correct sequence that will open the suitcase.

Number of favourable outcomes = 1.

Step 3: Calculate the probability.

The probability is the ratio of favourable outcomes to the total number of outcomes.