Chapter 15 Probability (Additional Questions)

The correct answer is (C) $0 \leq P(E) \leq 1$.

Explanation:

The probability of any event E, denoted by P(E), is a measure of the likelihood that the event will occur.

By definition, the probability of an impossible event is 0, and the probability of a sure event is 1.

For any event E, its probability P(E) must lie between 0 and 1, inclusive.

Mathematically, this is expressed as:

$0 \leq P(E) \leq 1$

Options (A), (B), and (D) are incorrect because probability cannot be negative, cannot be greater than 1, and is specifically restricted to the range [0, 1], not any real number.

The correct answer is (C) 0.25.

Explanation:

Let E be the event that the student passes the exam.

The probability of the student passing is given as P(E) = 0.75.

Let E' be the event that the student fails the exam (which is the complement of event E).

The sum of the probability of an event and its complement is always 1.

So, P(E) + P(E') = 1.

We want to find P(E').

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

Substitute the given value of P(E):

$P(E') = 1 - 0.75$

$P(E') = 0.25$

Thus, the probability of the student failing the exam is 0.25.

The correct answer is (B) $\frac{3}{6}$.

Solution:

When a fair die is thrown once, the total number of possible outcomes is the number of faces on the die, which is 6.

The sample space is $S = \{1, 2, 3, 4, 5, 6\}$.

Total number of outcomes = 6.

Let E be the event of getting an even number.

The even numbers in the sample space are 2, 4, and 6.

The favorable outcomes for event E are $\{2, 4, 6\}$.

Number of favorable outcomes = 3.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

$P(\text{getting an even number}) = \frac{3}{6}$

The probability can be simplified as:

$P(\text{getting an even number}) = \frac{\cancel{3}^{1}}{\cancel{6}_{2}} = \frac{1}{2}$

Comparing this with the given options, option (B) is $\frac{3}{6}$, which is equivalent to $\frac{1}{2}$.

The correct answer is (C) $\frac{1}{2}$.

Explanation:

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

The total number of possible outcomes is 2.

The event of getting a head has one favorable outcome (H).

The theoretical probability of an event is given by the formula:

$P(\text{event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

For the event of getting a head:

$P(\text{Head}) = \frac{1}{2}$

Therefore, the theoretical probability of getting a head when tossing a fair coin is $\frac{1}{2}$.

The correct answer is (C) $\frac{5}{8}$.

Explanation:

The total number of balls in the bag is the sum of red balls and blue balls.

Total number of balls = Number of red balls + Number of blue balls

Total number of balls = 5 + 3 = 8

The number of favorable outcomes for drawing a red ball is the number of red balls, which is 5.

The probability of an event is calculated as:

$P(\text{event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

In this case, the event is drawing a red ball.

$P(\text{drawing a red ball}) = \frac{\text{Number of red balls}}{\text{Total number of balls}}$

$P(\text{drawing a red ball}) = \frac{5}{8}$

Therefore, the probability of drawing a red ball is $\frac{5}{8}$.

The correct answer is (C) 0.

Explanation:

When a single die is thrown, the possible outcomes are the numbers on its faces: 1, 2, 3, 4, 5, and 6.

The sample space is $S = \{1, 2, 3, 4, 5, 6\}$.

The total number of possible outcomes is 6.

The event is getting a number greater than 6.

Let E be this event.

We look for numbers in the sample space that are greater than 6. There are no such numbers in the set $\{1, 2, 3, 4, 5, 6\}$.

Therefore, the number of favorable outcomes for event E is 0.

The probability of an event E is given by:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{getting a number greater than 6}) = \frac{0}{6}$

$P(\text{getting a number greater than 6}) = 0$

Since it is impossible to get a number greater than 6 when throwing a standard die, the probability is 0. This is an example of an impossible event.

The correct answer is (C) 4.

Explanation:

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

When a coin is tossed two times, we list all possible sequences of outcomes for the two tosses.

Possible outcomes for the first toss: H, T

Possible outcomes for the second toss: H, T

The possible combinations for two tosses are:

• Head on the first toss and Head on the second toss (HH)
• Head on the first toss and Tail on the second toss (HT)
• Tail on the first toss and Head on the second toss (TH)
• Tail on the first toss and Tail on the second toss (TT)

The sample space for tossing a coin two times is $S = \{HH, HT, TH, TT\}$.

The total number of possible outcomes is the number of elements in the sample space.

Total number of outcomes = 4.

Alternatively, if an event has $n_1$ outcomes and a second independent event has $n_2$ outcomes, the total number of outcomes for both events occurring is $n_1 \times n_2$.

For the first toss, there are 2 outcomes (H, T).

For the second toss, there are 2 outcomes (H, T).

Total outcomes = $2 \times 2 = 4$.

The correct answer is (D) $\frac{1}{2}$.

Solution:

When two coins are tossed simultaneously, the total possible outcomes are:

• Head, Head (HH)
• Head, Tail (HT)
• Tail, Head (TH)
• Tail, Tail (TT)

The sample space is $S = \{HH, HT, TH, TT\}$.

The total number of possible outcomes is 4.

We are interested in the event of getting exactly one head.

The outcomes with exactly one head are HT and TH.

The favorable outcomes are $\{HT, TH\}$.

The number of favorable outcomes is 2.

The probability of an event E is given by:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

The probability of getting exactly one head is:

$P(\text{exactly one head}) = \frac{2}{4}$

This fraction can be simplified:

$\frac{\cancel{2}^{1}}{\cancel{4}_{2}} = \frac{1}{2}$

So, the probability of getting exactly one head is $\frac{1}{2}$, which corresponds to option (D).

Note that option (B) $\frac{2}{4}$ is also mathematically equivalent to $\frac{1}{2}$. However, the simplest form is usually preferred, and $\frac{1}{2}$ is provided as option (D).

The correct answer is (B) $\frac{4}{52}$.

Explanation:

A standard deck of playing cards contains 52 cards in total.

Total number of possible outcomes = 52.

We are interested in the event of drawing a King.

In a standard deck, there are 4 Kings (King of Spades, King of Hearts, King of Diamonds, King of Clubs).

Number of favorable outcomes (getting a King) = 4.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{getting a King}) = \frac{\text{Number of Kings}}{\text{Total number of cards}}$

$P(\text{getting a King}) = \frac{4}{52}$

This probability can also be simplified to $\frac{\cancel{4}^{1}}{\cancel{52}_{13}} = \frac{1}{13}$. However, option (B) is given as $\frac{4}{52}$, which is the direct result before simplification.

The correct answer is (A) $\frac{6}{52}$.

Explanation:

A standard deck of playing cards has 52 cards.

Total number of possible outcomes = 52.

We need to find the number of red face cards.

Face cards are King, Queen, and Jack.

There are two red suits: Hearts and Diamonds.

In each red suit, there are 3 face cards (King, Queen, Jack).

Number of red face cards = (Number of face cards per red suit) $\times$ (Number of red suits)

Number of red face cards = $3 \times 2 = 6$.

The favorable outcomes are the 6 red face cards.

The probability of an event E is given by:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{getting a red face card}) = \frac{\text{Number of red face cards}}{\text{Total number of cards}}$

$P(\text{getting a red face card}) = \frac{6}{52}$

This fraction can be simplified to $\frac{\cancel{6}^{3}}{\cancel{52}_{26}} = \frac{3}{26}$, but the option (A) is provided in the unsimplified form $\frac{6}{52}$.

The correct answer is (A) Both A and R are true and R is the correct explanation of A.

Explanation:

Assertion (A) states that the probability of a sure event is 1. This statement is true. By definition, a sure event (or certain event) is an event that contains all the sample points of the sample space and is certain to occur. The probability of such an event is always the maximum possible probability, which is 1.

Reason (R) states that a sure event is an event that is certain to happen. This statement is also true. This is the definition of a sure event.

Furthermore, the reason R directly explains why the probability of a sure event (as stated in A) is 1. The certainty of the event's occurrence is what gives it a probability of 1.

Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).

The correct answer is (A) Both A and R are true and R is the correct explanation of A.

Explanation:

Assertion (A) states that if a coin is tossed 100 times and heads appear 60 times, the experimental probability of getting a head is 0.6. The experimental probability is calculated based on actual performed experiments.

Experimental Probability = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

In this case, the event is getting a head. It occurred 60 times in 100 trials.

Experimental Probability (Heads) = $\frac{60}{100} = 0.6$.

So, Assertion (A) is true.

Reason (R) states that experimental probability is calculated as (Number of favorable outcomes / Total number of trials).

This is the standard definition of experimental probability, also known as empirical probability.

So, Reason (R) is also true.

Furthermore, the calculation performed to verify Assertion (A) directly uses the formula given in Reason (R). The number of favorable outcomes is the number of times heads appeared (60), and the total number of trials is 100. The formula in R explains exactly how the value 0.6 in A was obtained.

Therefore, both A and R are true, and R is the correct explanation of A.

The correct answer is (A) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b).

Explanation:

When a standard die is rolled, the possible outcomes are the set $S = \{1, 2, 3, 4, 5, 6\}$. The total number of possible outcomes is 6.

We calculate the probability for each event in Column A:

**(i) Getting a number less than 1:** The numbers in the sample space that are less than 1 are none. This is an impossible event.

Number of favorable outcomes = 0.

$P(\text{number less than 1}) = \frac{0}{6} = 0$. This matches with (c).

**(ii) Getting an even prime number:** The prime numbers in the sample space are 2, 3, 5. The even numbers are 2, 4, 6. The only number that is both even and prime is 2.

Number of favorable outcomes = 1.

$P(\text{even prime number}) = \frac{1}{6}$. This matches with (d).

**(iii) Getting an odd number:** The odd numbers in the sample space are 1, 3, 5.

Number of favorable outcomes = 3.

$P(\text{odd number}) = \frac{3}{6}$. This matches with (a).

**(iv) Getting a number less than 7:** The numbers in the sample space that are less than 7 are 1, 2, 3, 4, 5, 6. All outcomes are less than 7. This is a sure event.

Number of favorable outcomes = 6.

$P(\text{number less than 7}) = \frac{6}{6} = 1$. This matches with (b).

Thus, the correct matching is (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b).

The correct answer is (A) (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a).

Explanation:

A standard deck of playing cards contains 52 cards.

Total number of possible outcomes = 52.

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

Let's calculate the probability for each event in Column A:

**(i) Getting a red card:** There are two red suits (Hearts and Diamonds), each with 13 cards. So, the total number of red cards is $13 + 13 = 26$.

Number of favorable outcomes = 26.

$P(\text{red card}) = \frac{26}{52}$. This matches with (c).

**(ii) Getting a Spade:** There is one suit of Spades, which contains 13 cards.

Number of favorable outcomes = 13.

$P(\text{Spade}) = \frac{13}{52}$. This matches with (d).

**(iii) Getting a Queen of Diamonds:** There is only one Queen of Diamonds in a deck of 52 cards.

Number of favorable outcomes = 1.

$P(\text{Queen of Diamonds}) = \frac{1}{52}$. This matches with (b).

**(iv) Getting a face card:** Face cards are Jack, Queen, and King. There are 3 face cards in each of the 4 suits (Spades, Hearts, Diamonds, Clubs). So, the total number of face cards is $3 \times 4 = 12$.

Number of favorable outcomes = 12.

$P(\text{face card}) = \frac{12}{52}$. This matches with (a).

Combining the matches, we get: (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a), which corresponds to option (A).

The correct answer is (B) $\frac{50}{100}$.

Solution:

The experimental probability of an event is based on the results of an experiment or survey.

Experimental Probability = $\frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

In this case study:

The total number of trials is the total number of families surveyed, which is 100.

The event is that a randomly chosen family has exactly 2 children.

The number of times this event occurred is the number of families with exactly 2 children, which is 50.

Using the formula:

$P(\text{family has 2 children}) = \frac{\text{Number of families with 2 children}}{\text{Total number of families surveyed}}$

$P(\text{family has 2 children}) = \frac{50}{100}$

This probability can be simplified to $\frac{1}{2}$ or 0.5, but option (B) is given in the form $\frac{50}{100}$.

The correct answer is (D) $\frac{50+30}{100} = \frac{80}{100}$.

Solution:

Based on the survey results from Question 15:

Total number of families surveyed = 100.

Number of families with 1 child = 20.

Number of families with 2 children = 50.

Number of families with 3 children = 30.

The event we are interested in is a randomly chosen family having more than 1 child.

Families with more than 1 child include those with 2 children and those with 3 children.

Number of families with more than 1 child = (Number of families with 2 children) + (Number of families with 3 children)

Number of families with more than 1 child = $50 + 30 = 80$.

The experimental probability is calculated as:

$P(\text{event}) = \frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

$P(\text{more than 1 child}) = \frac{\text{Number of families with more than 1 child}}{\text{Total number of families surveyed}}$

$P(\text{more than 1 child}) = \frac{80}{100}$

This corresponds to option (D).

The correct answer is (B) $-1.5$.

Explanation:

The probability of any event E, denoted by P(E), must satisfy a fundamental property:

The probability must be a number greater than or equal to 0 and less than or equal to 1.

Mathematically, this is written as:

$0 \leq P(E) \leq 1$

Let's examine the given options:

(A) 0.07: This value is between 0 and 1 ($0 \leq 0.07 \leq 1$). This can be a probability.

(B) $-1.5$: This value is negative ($-1.5 < 0$). Probability cannot be negative.

(C) $\frac{15}{100}$: This fraction is equal to 0.15. This value is between 0 and 1 ($0 \leq 0.15 \leq 1$). This can be a probability.

(D) $\frac{2}{3}$: This fraction is approximately 0.666... . This value is between 0 and 1 ($0 \leq \frac{2}{3} \leq 1$). This can be a probability.

Therefore, the value that cannot be the probability of an event is $-1.5$ because it is less than 0.

The correct answer is (C) 1.

Explanation:

Two events E and F are called complementary events if the occurrence of F is equivalent to the non-occurrence of E. In other words, F is the complement of E (often denoted as E' or $\bar{E}$), and vice versa.

For any event E and its complementary event E' (or F), the sum of their probabilities is always equal to 1.

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

Since E and F are complementary events, we have:

$P(E) + P(F) = 1$

This is a fundamental property of probability. The probability of an event happening plus the probability of the event not happening is always 1.

The correct answer is (B) $\frac{40}{52}$.

Explanation:

A standard deck of 52 playing cards contains 12 face cards (3 face cards in each of the 4 suits: Jack, Queen, King).

Number of face cards = 12.

Total number of cards = 52.

The number of cards that are not face cards is the total number of cards minus the number of face cards.

Number of non-face cards = Total cards - Number of face cards

Number of non-face cards = $52 - 12 = 40$.

The probability of an event is given by the formula:

$P(\text{event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

In this case, the event is getting a card that is not a face card.

$P(\text{not a face card}) = \frac{\text{Number of non-face cards}}{\text{Total number of cards}}$

$P(\text{not a face card}) = \frac{40}{52}$

This fraction can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 4:

$\frac{\cancel{40}^{10}}{\cancel{52}_{13}} = \frac{10}{13}$

However, option (B) provides the probability before simplification as $\frac{40}{52}$.

Alternatively, we can first find the probability of getting a face card and then use the concept of complementary events.

$P(\text{face card}) = \frac{\text{Number of face cards}}{\text{Total number of cards}} = \frac{12}{52}$.

The event of not getting a face card is the complement of the event of getting a face card.

$P(\text{not a face card}) = 1 - P(\text{face card})$

$P(\text{not a face card}) = 1 - \frac{12}{52} = \frac{52}{52} - \frac{12}{52} = \frac{52-12}{52} = \frac{40}{52}$.

The correct answer is (B) 36.

Explanation:

When a single standard die is thrown, there are 6 possible outcomes: $\{1, 2, 3, 4, 5, 6\}$.

When two dice are thrown simultaneously, the outcome of each die is independent of the other.

To find the total number of possible outcomes when two independent events occur, we multiply the number of outcomes for each event.

Number of outcomes for the first die = 6

Number of outcomes for the second die = 6

Total number of possible outcomes = (Outcomes for die 1) $\times$ (Outcomes for die 2)

Total number of possible outcomes = $6 \times 6 = 36$.

The sample space for throwing two dice consists of ordered pairs $(d_1, d_2)$, where $d_1$ is the result of the first die and $d_2$ is the result of the second die. There are 6 possibilities for $d_1$ and 6 possibilities for $d_2$, resulting in $6 \times 6 = 36$ unique pairs.

For example, some outcomes are (1,1), (1,2), ..., (1,6), (2,1), (2,2), ..., (6,6).

The correct answer is (D) $\frac{1}{6}$.

Solution:

When two standard dice are thrown, the total number of possible outcomes is $6 \times 6 = 36$. Each outcome is an ordered pair $(d_1, d_2)$, where $d_1$ is the result of the first die and $d_2$ is the result of the second die.

Total number of outcomes = 36.

We are interested in the event that the sum of the numbers on the faces is 7.

Let's list the pairs of outcomes $(d_1, d_2)$ where $d_1 + d_2 = 7$:

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

The favorable outcomes are $\{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\}$.

The number of favorable outcomes is 6.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{sum is 7}) = \frac{6}{36}$

This fraction can be simplified:

$\frac{\cancel{6}^{1}}{\cancel{36}_{6}} = \frac{1}{6}$

Thus, the probability that the sum of the numbers on the faces is 7 is $\frac{1}{6}$.

The correct answer is (C) 0.

Explanation:

The bag contains only lemon flavoured candies.

This means that there are no orange flavoured candies in the bag.

Let the event be E: getting an orange flavoured candy.

The number of favorable outcomes for event E is the number of orange flavoured candies in the bag, which is 0.

The total number of possible outcomes is the total number of candies in the bag. Although we don't know the exact number, let's say there are $N$ candies, where $N > 0$ since Malini takes out one candy.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{getting an orange flavoured candy}) = \frac{\text{Number of orange flavoured candies}}{\text{Total number of candies}}$

$P(\text{getting an orange flavoured candy}) = \frac{0}{N}$

Since the numerator is 0, the probability is 0.

$P(\text{getting an orange flavoured candy}) = 0$

This is an example of an impossible event, which has a probability of 0.

The correct answer is (C) $\frac{28}{52}$.

Solution:

A standard deck of 52 playing cards has:

Total number of cards = 52.

Number of red cards = 26 (13 Hearts + 13 Diamonds).

Number of Kings = 4 (King of Spades, King of Hearts, King of Diamonds, King of Clubs).

We are looking for the probability of drawing a card that is a red card or a King.

Let R be the event of drawing a red card.

$P(R) = \frac{\text{Number of red cards}}{\text{Total number of cards}} = \frac{26}{52}$

Let K be the event of drawing a King.

$P(K) = \frac{\text{Number of Kings}}{\text{Total number of cards}} = \frac{4}{52}$

The event "red card or a King" means the card is either red, or a King, or both. When events are not mutually exclusive (they can happen at the same time), the probability of their union (A or B) is given by:

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

Here, the event $(R \cap K)$ is drawing a card that is both red and a King. These are the King of Hearts and the King of Diamonds.

Number of red Kings = 2.

$P(R \cap K) = \frac{\text{Number of red Kings}}{\text{Total number of cards}} = \frac{2}{52}$

Now, we can calculate the probability of drawing a red card or a King:

$P(R \cup K) = P(R) + P(K) - P(R \cap K)$

$P(R \cup K) = \frac{26}{52} + \frac{4}{52} - \frac{2}{52}$

$P(R \cup K) = \frac{26 + 4 - 2}{52}$

$P(R \cup K) = \frac{30 - 2}{52}$

$P(R \cup K) = \frac{28}{52}$

This fraction can be simplified by dividing the numerator and denominator by 4:

$\frac{\cancel{28}^{7}}{\cancel{52}_{13}} = \frac{7}{13}$

Looking at the options, option (C) matches the unsimplified probability $\frac{28}{52}$.

The correct answer is (C) 8.

Explanation:

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

Total number of outcomes for one coin toss = 2.

When three coins are tossed simultaneously, the outcome of each coin is independent of the others.

To find the total number of possible outcomes for multiple independent events, we multiply the number of outcomes for each individual event.

Total number of possible outcomes = (Outcomes for Coin 1) $\times$ (Outcomes for Coin 2) $\times$ (Outcomes for Coin 3)

Total number of possible outcomes = $2 \times 2 \times 2$

Total number of possible outcomes = 8.

The sample space (list of all possible outcomes) when tossing three coins is:

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

Counting these outcomes, we find there are 8 distinct possibilities.

The correct answer is (C) 8.

Explanation:

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

Total number of outcomes for one coin toss = 2.

When three coins are tossed simultaneously, the outcome of each coin is independent of the others.

To find the total number of possible outcomes for multiple independent events, we multiply the number of outcomes for each individual event.

Total number of possible outcomes = (Outcomes for Coin 1) $\times$ (Outcomes for Coin 2) $\times$ (Outcomes for Coin 3)

Total number of possible outcomes = $2 \times 2 \times 2$

Total number of possible outcomes = 8.

The sample space (list of all possible outcomes) when tossing three coins is:

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

Counting these outcomes, we find there are 8 distinct possibilities.

The correct answer is (B) 8.

Solution:

Given:

Total number of marbles in the jar = 24.

The marbles are either green or blue.

Probability of drawing a green marble, $P(\text{Green}) = \frac{2}{3}$.

The probability of an event is defined as:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

For the event of drawing a green marble:

$P(\text{Green}) = \frac{\text{Number of green marbles}}{\text{Total number of marbles}}$

We are given $P(\text{Green}) = \frac{2}{3}$ and Total number of marbles = 24.

So, we can write the equation:

$\frac{2}{3} = \frac{\text{Number of green marbles}}{24}$

To find the number of green marbles, we can multiply both sides by 24:

Number of green marbles = $\frac{2}{3} \times 24$

Number of green marbles = $\frac{2 \times \cancel{24}^{8}}{\cancel{3}_{1}}$

Number of green marbles = $2 \times 8 = 16$

So, there are 16 green marbles in the jar.

The jar contains only green and blue marbles.

Total number of marbles = Number of green marbles + Number of blue marbles

$24 = 16 + \text{Number of blue marbles}$

To find the number of blue marbles, subtract the number of green marbles from the total:

Number of blue marbles = $24 - 16$

Number of blue marbles = 8

Thus, there are 8 blue marbles in the jar.

The correct answer is (A) $\frac{25}{100}$.

Solution:

The numbers from 1 to 100 are the possible outcomes when a number is selected.

Total number of possible outcomes = 100.

We need to find the number of prime numbers between 1 and 100.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

The prime numbers between 1 and 100 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Counting these numbers, we find that there are 25 prime numbers between 1 and 100.

Number of favorable outcomes (getting a prime number) = 25.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{getting a prime number}) = \frac{\text{Number of prime numbers}}{\text{Total number of numbers}}$

$P(\text{getting a prime number}) = \frac{25}{100}$

This probability can be simplified to $\frac{\cancel{25}^{1}}{\cancel{100}_{4}} = \frac{1}{4}$ or 0.25. However, option (A) provides the probability in the unsimplified form $\frac{25}{100}$.

The correct answer is (C) Getting a specific pair like (3, 4).

Explanation:

When rolling two standard dice, the sample space consists of all possible ordered pairs of outcomes from the two dice. Each outcome is represented as $(d_1, d_2)$, where $d_1$ is the result on the first die and $d_2$ is the result on the second die.

The sample space S has $6 \times 6 = 36$ possible outcomes:

$S = \{(1,1), (1,2), ..., (1,6),$

$\phantom{S = \{}(2,1), (2,2), ..., (2,6),$

$\phantom{S = \{}\dots$

$\phantom{S = \{}(6,1), (6,2), ..., (6,6)\}$

An elementary event is an event that contains only a single outcome from the sample space.

Let's examine the given options:

(A) Getting a sum of 12: The outcomes that sum to 12 is the set $\{(6, 6)\}$. This event contains only one outcome.

(B) Getting a sum of 5: The outcomes that sum to 5 are $\{(1, 4), (2, 3), (3, 2), (4, 1)\}$. This event contains 4 outcomes.

(C) Getting a specific pair like (3, 4): This refers to the event $\{(3, 4)\}$. This event contains only one outcome.

(D) Getting doubles: The outcomes that are doubles are $\{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\}$. This event contains 6 outcomes.

Both options (A) and (C) describe events that contain only one outcome from the sample space, making them elementary events according to the definition. However, option (C) directly names one of the distinct elements of the sample space, which is a specific ordered pair. Option (A) describes a property (sum=12) that happens to correspond to a single outcome. In the context of listing or identifying basic outcomes in the sample space, a specific pair like (3, 4) is a direct representation of an elementary event.

Therefore, "Getting a specific pair like (3, 4)" is a clear example of an elementary event.

The correct answer is (A) $\frac{4}{8}$.

Solution:

The wheel has 8 equally likely sectors numbered from 1 to 8.

The total number of possible outcomes when spinning the wheel is the number of sectors, which is 8.

Total number of outcomes = 8.

We are interested in the event of getting an odd number.

The odd numbers between 1 and 8 are 1, 3, 5, and 7.

The favorable outcomes are $\{1, 3, 5, 7\}$.

The number of favorable outcomes is 4.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{getting an odd number}) = \frac{\text{Number of odd numbers}}{\text{Total number of sectors}}$

$P(\text{getting an odd number}) = \frac{4}{8}$

This fraction can be simplified to $\frac{1}{2}$ or 0.5. Option (A) provides the probability in the unsimplified form $\frac{4}{8}$.

The correct answer is (B) $\frac{3}{8}$.

Solution:

The spinning wheel has 8 equally likely sectors numbered 1 to 8.

Total number of possible outcomes = 8.

We are interested in the event of getting a number greater than 5.

The numbers greater than 5 in the set {1, 2, 3, 4, 5, 6, 7, 8} are 6, 7, and 8.

The favorable outcomes are $\{6, 7, 8\}$.

The number of favorable outcomes is 3.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{getting a number > 5}) = \frac{\text{Number of numbers > 5}}{\text{Total number of sectors}}$

$P(\text{getting a number > 5}) = \frac{3}{8}$

This matches option (B).

The correct answer is (A) 0.

Explanation:

When two standard dice are thrown, the total number of possible outcomes is $6 \times 6 = 36$. Each outcome is a pair $(d_1, d_2)$ where $d_1$ is the result on the first die and $d_2$ is the result on the second die, and $d_1, d_2 \in \{1, 2, 3, 4, 5, 6\}$.

The possible sums of the numbers on the two faces range from the minimum sum ($1+1=2$) to the maximum sum ($6+6=12$).

We are interested in the event that the sum of the numbers is 1.

Let E be the event that the sum of the numbers is 1. We need to find pairs $(d_1, d_2)$ such that $d_1 + d_2 = 1$.

Since the minimum value on each die is 1, the smallest possible sum is $1 + 1 = 2$. It is impossible to get a sum of 1 when rolling two standard dice.

Therefore, the number of favorable outcomes for event E is 0.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{sum is 1}) = \frac{0}{36}$

$P(\text{sum is 1}) = 0$

This is an impossible event, and its probability is 0.

The correct answer is (B) $\frac{2}{52}$.

Explanation:

A standard deck of playing cards contains 52 cards.

Total number of possible outcomes = 52.

We are interested in the event of drawing a red King.

The Kings in a deck are King of Spades, King of Hearts, King of Diamonds, and King of Clubs.

The red suits are Hearts and Diamonds.

The red Kings are the King of Hearts and the King of Diamonds.

Number of favorable outcomes (getting a red King) = 2.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{getting a red King}) = \frac{\text{Number of red Kings}}{\text{Total number of cards}}$

$P(\text{getting a red King}) = \frac{2}{52}$

This probability can be simplified to $\frac{\cancel{2}^{1}}{\cancel{52}_{26}} = \frac{1}{26}$. However, option (B) is provided in the unsimplified form $\frac{2}{52}$.

The correct answer is (C) $1 - P(E)$.

Explanation:

The event "not E" is the complement of the event E. It represents all outcomes in the sample space that are not in event E.

For any event E, its complement is denoted by $E'$ or $\bar{E}$.

The sum of the probability of an event and the probability of its complement is always 1.

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

In this question, "not E" represents the complementary event, so we can write:

$P(E) + P(\text{not E}) = 1$

To find $P(\text{not E})$, we subtract $P(E)$ from both sides of the equation:

$P(\text{not E}) = 1 - P(E)$

This relationship is a fundamental rule in probability theory.

The correct answer is (D) Theoretical probability is based on observed frequencies.

Explanation:

Let's analyze each statement:

(A) The sum of probabilities of all elementary events is 1. This statement is true. Elementary events are disjoint and their union forms the sample space. The sum of probabilities of all possible disjoint outcomes must equal the probability of the entire sample space, which is 1.

(B) The probability of an event is always between 0 and 1 (inclusive). This statement is true. The range of probability for any event E is $0 \leq P(E) \leq 1$.

(C) An event with probability 0 is called an impossible event. This statement is true. An impossible event has no chance of occurring, and its probability is 0.

(D) Theoretical probability is based on observed frequencies. This statement is false. Theoretical probability (also called classical probability) is calculated based on the number of favorable outcomes and the total number of equally likely outcomes, without conducting an experiment. Probability based on observed frequencies from experiments is called experimental probability or empirical probability.

Therefore, the false statement is (D).

The correct answer is (B) 10.

Solution:

Let the number of red pens be $n_R = 5$.

Let the number of blue pens be $n_B$. We need to find $n_B$.

The total number of pens in the box is $N = n_R + n_B = 5 + n_B$.

The probability of drawing a red pen is:

$P(\text{Red}) = \frac{\text{Number of red pens}}{\text{Total number of pens}} = \frac{5}{5 + n_B}$

The probability of drawing a blue pen is:

$P(\text{Blue}) = \frac{\text{Number of blue pens}}{\text{Total number of pens}} = \frac{n_B}{5 + n_B}$

According to the problem, the probability of drawing a blue pen is double that of a red pen.

$P(\text{Blue}) = 2 \times P(\text{Red})$

Substitute the expressions for the probabilities:

$\frac{n_B}{5 + n_B} = 2 \times \frac{5}{5 + n_B}$

$\frac{n_B}{5 + n_B} = \frac{10}{5 + n_B}$

Since the denominators are the same and non-zero (as the total number of pens must be positive), we can equate the numerators:

$n_B = 10$

So, the number of blue pens is 10.

Verification:

If there are 10 blue pens, then the total number of pens is $5 + 10 = 15$.

$P(\text{Red}) = \frac{5}{15} = \frac{1}{3}$

$P(\text{Blue}) = \frac{10}{15} = \frac{2}{3}$

Is $P(\text{Blue}) = 2 \times P(\text{Red})$?

$\frac{2}{3} = 2 \times \frac{1}{3}$

$\frac{2}{3} = \frac{2}{3}$

The condition is satisfied.

The correct answer is (C) 4.

Explanation:

When a standard die is thrown, the possible outcomes are the numbers $\{1, 2, 3, 4, 5, 6\}$. The question asks to identify the composite number from the given options, which are a subset of these outcomes.

Let's define what a composite number is:

A composite number is a natural number greater than 1 that has at least one divisor other than 1 and itself.

Let's examine the given options:

• (A) 1: The number 1 is neither prime nor composite. It only has one divisor, which is 1.
• (B) 2: The number 2 is a prime number. Its only positive divisors are 1 and 2.
• (C) 4: The number 4 is a composite number. Its positive divisors are 1, 2, and 4. Since it has a divisor (2) other than 1 and 4, it is composite. Also, $4 > 1$.
• (D) 5: The number 5 is a prime number. Its only positive divisors are 1 and 5.

Among the given options, only 4 fits the definition of a composite number.

The correct answer is (D) $\frac{6}{36}$.

Solution:

When two standard dice are thrown simultaneously, the total number of possible outcomes is $6 \times 6 = 36$.

Total number of outcomes = 36.

We are interested in the event of getting a sum of at least 10. "At least 10" means the sum is 10 or 11 or 12.

Let's list the pairs of outcomes $(d_1, d_2)$ such that $d_1 + d_2 \geq 10$:

• Sum = 10: $\{(4, 6), (5, 5), (6, 4)\}$
• Sum = 11: $\{(5, 6), (6, 5)\}$
• Sum = 12: $\{(6, 6)\}$

The favorable outcomes for a sum of at least 10 are $\{(4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)\}$.

The number of favorable outcomes is 6.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{sum at least 10}) = \frac{\text{Number of outcomes with sum } \geq 10}{\text{Total number of outcomes}}$

$P(\text{sum at least 10}) = \frac{6}{36}$

This fraction can be simplified to $\frac{\cancel{6}^{1}}{\cancel{36}_{6}} = \frac{1}{6}$. However, option (D) is provided in the unsimplified form $\frac{6}{36}$.

The correct answer is (B) $\frac{2}{7}$.

Solution:

A leap year has 366 days.

We can express 366 days in terms of weeks and extra days.

$366 = 52 \times 7 + 2$

So, a leap year has 52 full weeks and 2 extra days.

Each of the 52 full weeks will contain exactly one Sunday. This accounts for 52 Sundays.

To have 53 Sundays in the leap year, one of the 2 extra days must be a Sunday.

The 2 extra days can form the following pairs:

• (Monday, Tuesday)
• (Tuesday, Wednesday)
• (Wednesday, Thursday)
• (Thursday, Friday)
• (Friday, Saturday)
• (Saturday, Sunday)
• (Sunday, Monday)

There are 7 possible outcomes for the pair of extra days, and these outcomes are equally likely.

Total number of possible outcomes = 7.

The event of having 53 Sundays occurs if one of the 2 extra days is a Sunday. This happens in the following pairs:

• (Saturday, Sunday)
• (Sunday, Monday)

The number of favorable outcomes (where one of the extra days is a Sunday) is 2.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{53 Sundays in a leap year}) = \frac{\text{Number of pairs with a Sunday}}{\text{Total number of pairs}}$

$P(\text{53 Sundays in a leap year}) = \frac{2}{7}$

The correct answer is (C) 144.

Solution:

The total number of pens in the mix is the sum of the faulty pens and the good pens.

Number of faulty pens = 12.

Number of good pens = 132.

Total number of pens = Number of faulty pens + Number of good pens

Total number of pens = $12 + 132$

$\begin{array}{cc} & 1 & 2 \\ + & 1 & 3 & 2 \\ \hline & 1 & 4 & 4 \\ \hline \end{array}$

Total number of pens = 144.

The correct answer is (B) $\frac{132}{144}$.

Solution:

From the case study in Question 39, we know the following:

Number of faulty pens = 12.

Number of good pens = 132.

Total number of pens = Number of faulty pens + Number of good pens

Total number of pens = $12 + 132 = 144$.

The total number of possible outcomes when drawing one pen at random is the total number of pens, which is 144.

Total number of outcomes = 144.

We are interested in the event of drawing a good pen.

The number of favorable outcomes for this event is the number of good pens, which is 132.

Number of favorable outcomes = 132.

The probability of an event E is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{drawing a good pen}) = \frac{\text{Number of good pens}}{\text{Total number of pens}}$

$P(\text{drawing a good pen}) = \frac{132}{144}$

This probability can be simplified by dividing the numerator and denominator by their greatest common divisor, which is 12:

$\frac{\cancel{132}^{11}}{\cancel{144}_{12}} = \frac{11}{12}$

However, option (B) is provided in the unsimplified form $\frac{132}{144}$.

Given:

Two functions $f(x)$ and $g(x)$ are given by:

$f(x) = x^2 - 2x + 1$

$g(x) = x - 1$

To Find:

1. The limit of the ratio $\frac{f(x)}{g(x)}$ as $x$ approaches 1.

2. The roots of the equation $f(x) = 0$.

3. Prime factorization of 120 (for demonstration).

4. LCM of 8, 12, and 18 (for demonstration).

5. Table of values for $f(x)$ (for demonstration).

6. Demonstrate a numbered inequality.

7. Demonstrate use of $\text{cosec}$, $\log$, and inverse trigonometric functions.

8. Demonstrate combined comment and equation number format.

Solution:

Part 1: Limit Evaluation

We need to evaluate $\lim\limits_{x \to 1} \frac{f(x)}{g(x)}$.

Substitute the expressions for $f(x)$ and $g(x)$:

$\lim\limits_{x \to 1} \frac{x^2 - 2x + 1}{x - 1}$

Notice that the numerator $x^2 - 2x + 1$ is a perfect square. It can be factored as $(x-1)^2$. We can also use the underbrace notation to show this structure:

$f(x) = x^2 - 2x + 1 = \underbrace{(x-1)^2}_{\text{Perfect Square}}$

So, the limit becomes:

$\lim\limits_{x \to 1} \frac{(x - 1)^2}{x - 1}$

For $x \neq 1$, we can cancel out the $(x-1)$ term from the numerator and the denominator.

$\lim\limits_{x \to 1} \frac{\cancel{(x - 1)}^2}{\cancel{x - 1}} = \lim\limits_{x \to 1} (x - 1)$

Now, we can substitute $x=1$ into the simplified expression:

$\lim\limits_{x \to 1} (x - 1) = 1 - 1 = 0$

So, the value of the limit is 0.

Part 2: Finding Roots of $f(x) = 0$

We need to find the values of $x$ for which $f(x) = 0$.

Set the function $f(x)$ equal to zero:

$x^2 - 2x + 1 = 0$

As we noted before, this is a perfect square trinomial:

$(x - 1)^2 = 0$

Taking the square root of both sides gives:

Solving for $x$:

Since the root $x=1$ appears twice (from $(x-1)^2 = 0$), it is a repeated root.

Part 3: Prime Factorization Demonstration

Let's find the prime factorization of 120.

So, the prime factorization of 120 is $2 \times 2 \times 2 \times 3 \times 5 = 2^3 \times 3 \times 5$.

Part 4: LCM Demonstration

Let's find the LCM of 8, 12, and 18.

The LCM is $2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2 = 8 \times 9 = 72$.

Part 5: Table Demonstration

Let's create a table for $f(x) = x^2 - 2x + 1$ for a few values of $x$. Note that $f(x) = (x-1)^2$, so the values are easy to calculate.

Part 6: Numbered Inequality Demonstration

Consider the inequality $2x + 3 > 7$.

The solution to the inequality is $x > 2$.

Part 7: $\text{cosec}$, $\log$, Inverse Trig Demonstration

We can use trigonometric functions like $\text{cosec}(\theta)$.

Example using log: $\log_{10}(100) = 2$ because $10^2 = 100$.

Example using an inverse trigonometric function: $\sin^{-1}(1) = \frac{\pi}{2}$ because $\sin(\frac{\pi}{2}) = 1$. Note the use of the inverse form $\sin^{-1}$ as required.

Part 8: Combined Comment and Number Demonstration

Let's assume we derived a length $L$.

This concludes the demonstration of the required formats within the answer block.

Given:

Two coins are tossed simultaneously.

To Find:

(i) The probability of getting exactly one head.

(ii) The probability of getting at most one tail.

Solution:

When two coins are tossed simultaneously, the possible outcomes are:

Head on the first coin and Head on the second coin (HH)

Head on the first coin and Tail on the second coin (HT)

Tail on the first coin and Head on the second coin (TH)

Tail on the first coin and Tail on the second coin (TT)

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

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

The total number of possible outcomes is the size of the sample space, $|S|$.

$|S| = 4$

Part (i): Probability of getting exactly one head

Let $E_1$ be the event of getting exactly one head.

The outcomes where there is exactly one head are HT and TH.

$E_1 = \{HT, TH\}$

The number of outcomes in event $E_1$ is $|E_1| = 2$.

The probability of event $E_1$ is given by the formula:

$P(E_1) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{|E_1|}{|S|}$

Thus, the probability of getting exactly one head is $\frac{1}{2}$.

Part (ii): Probability of getting at most one tail

"At most one tail" means the number of tails is less than or equal to one. This includes outcomes with 0 tails or 1 tail.

Let $E_2$ be the event of getting at most one tail.

Outcomes with 0 tails: HH

Outcomes with 1 tail: HT, TH

So, the outcomes in event $E_2$ are HH, HT, and TH.

$E_2 = \{HH, HT, TH\}$

The number of outcomes in event $E_2$ is $|E_2| = 3$.

The probability of event $E_2$ is:

$P(E_2) = \frac{|E_2|}{|S|}$

Thus, the probability of getting at most one tail is $\frac{3}{4}$.

Given:

A die is thrown once.

To Find:

(i) The probability of getting a prime number.

(ii) The probability of getting a number lying between 2 and 6.

Solution:

When a standard six-sided die is thrown once, the possible outcomes are the numbers 1, 2, 3, 4, 5, and 6.

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

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

The total number of possible outcomes is the size of the sample space, $|S|$.

$|S| = 6$

Part (i): Probability of getting a prime number

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

From the sample space $\{1, 2, 3, 4, 5, 6\}$, the prime numbers are 2, 3, and 5.

Let $E_1$ be the event of getting a prime number.

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

The number of outcomes in event $E_1$ is $|E_1| = 3$.

The probability of event $E_1$ is given by the formula:

$P(E_1) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{|E_1|}{|S|}$

Simplify the fraction:

Thus, the probability of getting a prime number is $\frac{1}{2}$.

Part (ii): Probability of getting a number lying between 2 and 6

We are looking for numbers in the sample space that are strictly greater than 2 and strictly less than 6.

From the sample space $\{1, 2, 3, 4, 5, 6\}$, the numbers lying between 2 and 6 are 3, 4, and 5.

Let $E_2$ be the event of getting a number lying between 2 and 6.

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

The number of outcomes in event $E_2$ is $|E_2| = 3$.

The probability of event $E_2$ is:

$P(E_2) = \frac{|E_2|}{|S|}$

Simplify the fraction:

Thus, the probability of getting a number lying between 2 and 6 is $\frac{1}{2}$.

Total number of cards in a well-shuffled deck = 52.

(i) Probability of getting a black face card:

Face cards are Jack, Queen, and King. There are 4 suits (Hearts, Diamonds, Clubs, Spades).

Black suits are Clubs and Spades.

Number of face cards in Clubs = 3 (Jack, Queen, King)

Number of face cards in Spades = 3 (Jack, Queen, King)

Total number of black face cards = Number of black face cards in Clubs + Number of black face cards in Spades

Total number of black face cards = $3 + 3 = 6$.

Number of favorable outcomes = 6.

Probability of getting a black face card = $\frac{\text{Number of black face cards}}{\text{Total number of cards}}$

Probability of getting a black face card = $\frac{6}{52}$

Simplifying the fraction, we get:

Probability of getting a black face card = $\frac{\cancel{6}^{3}}{\cancel{52}_{26}} = \frac{3}{26}$.

(ii) Probability of getting an ace of heart:

In a standard deck of 52 cards, there is only one Ace of Hearts.

Number of favorable outcomes = 1.

Probability of getting an ace of heart = $\frac{\text{Number of ace of heart}}{\text{Total number of cards}}$

Probability of getting an ace of heart = $\frac{1}{52}$.

Total number of $50 \text{ paise}$ coins = 100

Total number of $\textsf{₹} 1$ coins = 50

Total number of $\textsf{₹} 2$ coins = 20

Total number of $\textsf{₹} 5$ coins = 10

Total number of coins in the piggy bank = $100 + 50 + 20 + 10 = 180$.

(i) Probability that the coin will be a $50 \text{ paise}$ coin:

Number of favorable outcomes (getting a $50 \text{ paise}$ coin) = 100.

Total number of possible outcomes (total coins) = 180.

Probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Probability of getting a $50 \text{ paise}$ coin = $\frac{100}{180}$

Simplifying the fraction:

$\frac{\cancel{100}^{10}}{\cancel{180}_{18}} = \frac{\cancel{10}^{5}}{\cancel{18}_{9}} = \frac{5}{9}$.

So, the probability of getting a $50 \text{ paise}$ coin is $\frac{5}{9}$.

(ii) Probability that the coin will not be a $\textsf{₹} 5$ coin:

Number of $\textsf{₹} 5$ coins = 10.

Number of coins that are not $\textsf{₹} 5$ coins = Total coins - Number of $\textsf{₹} 5$ coins

Number of coins that are not $\textsf{₹} 5$ coins = $180 - 10 = 170$.

Number of favorable outcomes (not getting a $\textsf{₹} 5$ coin) = 170.

Total number of possible outcomes (total coins) = 180.

Probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Probability of not getting a $\textsf{₹} 5$ coin = $\frac{170}{180}$

Simplifying the fraction:

$\frac{\cancel{170}^{17}}{\cancel{180}_{18}} = \frac{17}{18}$.

So, the probability of not getting a $\textsf{₹} 5$ coin is $\frac{17}{18}$.

Total number of discs in the box = 90.

The discs are numbered from 1 to 90.

Total number of possible outcomes = 90.

(i) Probability of getting a two-digit number:

The numbers on the discs range from 1 to 90.

Single-digit numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9.

Number of single-digit numbers = 9.

Two-digit numbers are from 10 to 90.

Number of two-digit numbers = Total numbers - Number of single-digit numbers

Number of two-digit numbers = $90 - 9 = 81$.

Number of favorable outcomes (getting a two-digit number) = 81.

Probability of getting a two-digit number = $\frac{\text{Number of two-digit numbers}}{\text{Total number of discs}}$

Probability of getting a two-digit number = $\frac{81}{90}$.

Simplifying the fraction:

$\frac{\cancel{81}^{9}}{\cancel{90}_{10}} = \frac{9}{10}$.

So, the probability of getting a two-digit number is $\frac{9}{10}$.

(ii) Probability of getting a perfect square number:

The perfect square numbers from 1 to 90 are:

$1^2 = 1$

$2^2 = 4$

$3^2 = 9$

$4^2 = 16$

$5^2 = 25$

$6^2 = 36$

$7^2 = 49$

$8^2 = 64$

$9^2 = 81$

The next perfect square is $10^2 = 100$, which is greater than 90.

The perfect squares between 1 and 90 are 1, 4, 9, 16, 25, 36, 49, 64, 81.

Number of perfect square numbers = 9.

Number of favorable outcomes (getting a perfect square number) = 9.

Probability of getting a perfect square number = $\frac{\text{Number of perfect square numbers}}{\text{Total number of discs}}$

Probability of getting a perfect square number = $\frac{9}{90}$.

Simplifying the fraction:

$\frac{\cancel{9}^{1}}{\cancel{90}_{10}} = \frac{1}{10}$.

So, the probability of getting a perfect square number is $\frac{1}{10}$.

Given:

Total number of ball pens = 144.

Number of defective pens = 20.

Number of good pens = Total pens - Number of defective pens

Number of good pens = $144 - 20 = 124$.

To Find:

(i) The probability that Nuri will buy the pen.

(ii) The probability that Nuri will not buy the pen.

Solution:

Total number of pens from which one is drawn = 144.

Total number of possible outcomes = 144.

(i) Probability that Nuri will buy it:

Nuri will buy a pen if it is good.

Number of good pens = 124.

Number of favorable outcomes (getting a good pen) = 124.

Probability (Nuri will buy) = Probability (getting a good pen) = $\frac{\text{Number of good pens}}{\text{Total number of pens}}$

Probability (Nuri will buy) = $\frac{124}{144}$.

Simplifying the fraction:

$\frac{\cancel{124}^{31}}{\cancel{144}_{36}} = \frac{31}{36}$.

So, the probability that Nuri will buy the pen is $\frac{31}{36}$.

(ii) Probability that Nuri will not buy it:

Nuri will not buy a pen if it is defective.

Number of defective pens = 20.

Number of favorable outcomes (getting a defective pen) = 20.

Probability (Nuri will not buy) = Probability (getting a defective pen) = $\frac{\text{Number of defective pens}}{\text{Total number of pens}}$

Probability (Nuri will not buy) = $\frac{20}{144}$.

Simplifying the fraction:

$\frac{\cancel{20}^{5}}{\cancel{144}_{36}} = \frac{5}{36}$.

So, the probability that Nuri will not buy the pen is $\frac{5}{36}$.

Alternate Solution for (ii):

The event that Nuri will not buy the pen is the complement of the event that Nuri will buy the pen.

Probability (Nuri will not buy) = $1$ - Probability (Nuri will buy)

Probability (Nuri will not buy) = $1 - \frac{31}{36}$

Probability (Nuri will not buy) = $\frac{36}{36} - \frac{31}{36} = \frac{36-31}{36} = \frac{5}{36}$.

This confirms the previous result.

Total number of cards in a well-shuffled deck = 52.

Total number of possible outcomes = 52.

(i) Probability of getting a card of spades or an ace:

Let A be the event of getting a card of spades.

Number of spades = 13.

$P(A) = \frac{13}{52}$.

Let B be the event of getting an ace.

Number of aces = 4.

$P(B) = \frac{4}{52}$.

The event A $\cap$ B is getting a card that is both a spade and an ace, which is the Ace of Spades.

Number of cards that are both spades and aces = 1.

$P(A \cap B) = \frac{1}{52}$.

The probability of getting a card of spades or an ace is $P(A \cup B)$.

Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:

$P(A \cup B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52}$

$P(A \cup B) = \frac{13 + 4 - 1}{52} = \frac{16}{52}$.

Simplifying the fraction:

$P(\text{spade or ace}) = \frac{\cancel{16}^{4}}{\cancel{52}_{13}} = \frac{4}{13}$.

Alternatively, by counting favorable outcomes:

Number of spades = 13.

Number of aces that are not spades (Ace of Hearts, Ace of Diamonds, Ace of Clubs) = 3.

Total number of favorable outcomes = Number of spades + Number of aces (not spades) = $13 + 3 = 16$.

Probability of getting a card of spades or an ace = $\frac{16}{52} = \frac{4}{13}$.

(ii) Probability of getting a red king:

Red suits are Hearts and Diamonds.

There is one King in the Hearts suit and one King in the Diamonds suit.

Number of red kings = 2.

Number of favorable outcomes = 2.

Probability of getting a red king = $\frac{\text{Number of red kings}}{\text{Total number of cards}}$

Probability of getting a red king = $\frac{2}{52}$.

Simplifying the fraction:

$\frac{\cancel{2}^{1}}{\cancel{52}_{26}} = \frac{1}{26}$.

When three coins are tossed simultaneously, the possible outcomes are:

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

The sample space S is \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}.

Total number of possible outcomes = $|S| = 8$.

(i) Probability of getting at least two heads:

Let E be the event of getting at least two heads.

At least two heads means getting exactly two heads or exactly three heads.

The outcomes with at least two heads are: HHH, HHT, HTH, THH.

Number of favorable outcomes for event E = 4.

Probability of event E = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{at least two heads}) = \frac{4}{8}$

Simplifying the fraction:

$P(\text{at least two heads}) = \frac{\cancel{4}^{1}}{\cancel{8}_{2}} = \frac{1}{2}$.

(ii) Probability of getting no tails:

Let F be the event of getting no tails.

No tails means getting all heads.

The outcome with no tails is: HHH.

Number of favorable outcomes for event F = 1.

Probability of event F = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

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

When a die is thrown twice, the total number of possible outcomes is $6 \times 6 = 36$.

The sample space consists of all ordered pairs $(a, b)$, where $a$ is the result of the first throw and $b$ is the result of the second throw, and $a, b \in \{1, 2, 3, 4, 5, 6\}$.

Total number of possible outcomes = 36.

(i) Probability that 5 will not come up either time:

Let A be the event that 5 does not come up on the first throw, and 5 does not come up on the second throw.

For the first throw, there are 5 possible outcomes (1, 2, 3, 4, 6) where 5 does not come up.

For the second throw, there are 5 possible outcomes (1, 2, 3, 4, 6) where 5 does not come up.

The number of outcomes where 5 does not come up either time is the product of the possibilities for each throw.

Number of favorable outcomes for event A = $5 \times 5 = 25$.

The favorable outcomes are pairs $(a, b)$ where $a \neq 5$ and $b \neq 5$.

Probability (5 will not come up either time) = $\frac{\text{Number of outcomes with no 5}}{\text{Total number of outcomes}}$

Probability (5 will not come up either time) = $\frac{25}{36}$.

(ii) Probability that 5 will come up at least once:

Let B be the event that 5 will come up at least once.

This means 5 comes up on the first throw, or on the second throw, or on both throws.

The outcomes where 5 comes up at least once are:

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

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

(Note that (5, 5) is included only once).

Number of favorable outcomes for event B = $6 + 6 - 1 = 11$. (Outcomes with 5 on first throw + Outcomes with 5 on second throw - Outcome with 5 on both throws to avoid double counting).

Alternatively, the outcomes are: (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5), (5, 5).

Total number of favorable outcomes = 11.

Probability (5 will come up at least once) = $\frac{\text{Number of outcomes with at least one 5}}{\text{Total number of outcomes}}$

Probability (5 will come up at least once) = $\frac{11}{36}$.

Alternate Solution for (ii):

The event "5 will come up at least once" is the complement of the event "5 will not come up either time".

Let B be the event that 5 will come up at least once.

Let A be the event that 5 will not come up either time.

$P(B) = 1 - P(A)$

From part (i), $P(A) = \frac{25}{36}$.

$P(B) = 1 - \frac{25}{36}$

$P(B) = \frac{36}{36} - \frac{25}{36} = \frac{36 - 25}{36} = \frac{11}{36}$.

Given:

Number of red balls in the bag = 3.

Number of black balls in the bag = 5.

Total number of balls in the bag = Number of red balls + Number of black balls = $3 + 5 = 8$.

To Find:

(i) The probability that the ball drawn (first draw) is red.

(ii) The probability that the second ball is black, given the first ball drawn was red and not replaced.

Solution:

Total number of balls in the bag = 8.

When one ball is drawn at random from the bag, the total number of possible outcomes is 8.

(i) Probability that the ball drawn is red:

The number of favorable outcomes (getting a red ball) is the number of red balls in the bag, which is 3.

Probability (drawing a red ball) = $\frac{\text{Number of red balls}}{\text{Total number of balls}}$

Probability (drawing a red ball) = $\frac{3}{8}$.

(ii) Probability that the second ball is black, given the first red ball was not replaced:

After drawing one red ball and not replacing it in the bag, the composition of the bag changes.

Number of red balls remaining in the bag = $3 - 1 = 2$.

Number of black balls remaining in the bag = 5 (as a red ball was drawn).

Total number of balls remaining in the bag = $8 - 1 = 7$.

Now, a second ball is drawn from these 7 balls.

The number of favorable outcomes for the second draw (getting a black ball) is the number of black balls remaining, which is 5.

The total number of possible outcomes for the second draw is the total number of balls remaining, which is 7.

Probability (drawing a black ball in the second draw | first was red and not replaced) = $\frac{\text{Number of black balls remaining}}{\text{Total number of balls remaining}}$

Probability (second ball is black) = $\frac{5}{7}$.

Given:

Number of defective pens = 12.

Number of good pens = 132.

To Find:

The probability that the pen taken out is a good one.

Solution:

Total number of pens in the lot = Number of defective pens + Number of good pens

Total number of pens = $12 + 132 = 144$.

When one pen is taken out at random from the lot, the total number of possible outcomes is the total number of pens, which is 144.

The number of favorable outcomes (getting a good pen) is the number of good pens, which is 132.

Probability of getting a good pen = $\frac{\text{Number of good pens}}{\text{Total number of pens}}$

Probability (Good pen) = $\frac{132}{144}$.

Now, we simplify the fraction $\frac{132}{144}$. Both numbers are divisible by 12.

$\frac{\cancel{132}^{11}}{\cancel{144}_{12}} = \frac{11}{12}$.

Thus, the probability that the pen taken out is a good one is $\frac{11}{12}$.

Given:

Number of red marbles in the box = 5.

Number of white marbles in the box = 8.

Number of green marbles in the box = 4.

To Find:

The probability that the marble taken out will be (i) white (ii) not red.

Solution:

Total number of marbles in the box = Number of red marbles + Number of white marbles + Number of green marbles

Total number of marbles = $5 + 8 + 4 = 17$.

When one marble is taken out at random from the box, the total number of possible outcomes is the total number of marbles, which is 17.

(i) Probability that the marble taken out will be white:

The number of favorable outcomes (getting a white marble) is the number of white marbles in the box, which is 8.

Probability (drawing a white marble) = $\frac{\text{Number of white marbles}}{\text{Total number of marbles}}$

Probability (white) = $\frac{8}{17}$.

Since 8 and 17 have no common factors other than 1, the fraction is already in its simplest form.

So, the probability that the marble taken out will be white is $\frac{8}{17}$.

(ii) Probability that the marble taken out will be not red:

A marble that is not red must be either white or green.

Number of non-red marbles = Number of white marbles + Number of green marbles

Number of non-red marbles = $8 + 4 = 12$.

The number of favorable outcomes (getting a non-red marble) is 12.

Probability (drawing a not red marble) = $\frac{\text{Number of non-red marbles}}{\text{Total number of marbles}}$

Probability (not red) = $\frac{12}{17}$.

Since 12 and 17 have no common factors other than 1, the fraction is already in its simplest form.

So, the probability that the marble taken out will be not red is $\frac{12}{17}$.

Alternate Solution for (ii):

The event "not red" is the complement of the event "red".

First, calculate the probability of drawing a red marble:

Number of red marbles = 5.

Total number of marbles = 17.

Probability (drawing a red marble) = $\frac{\text{Number of red marbles}}{\text{Total number of marbles}} = \frac{5}{17}$.

Probability (not red) = $1$ - Probability (red)

Probability (not red) = $1 - \frac{5}{17}$

Probability (not red) = $\frac{17}{17} - \frac{5}{17} = \frac{17 - 5}{17} = \frac{12}{17}$.

This confirms the previous result.

Given:

A spinner with numbers 1, 2, 3, 4, 5, 6, 7, 8.

The outcomes are equally likely.

To Find:

(i) The probability that it will point at 8.

(ii) The probability that it will point at an odd number.

Solution:

The numbers on the spinner are 1, 2, 3, 4, 5, 6, 7, 8.

Total number of possible outcomes = 8.

(i) Probability that it will point at 8:

Let E be the event that the arrow points at 8.

There is only one outcome where the arrow points at 8.

Number of favorable outcomes for event E = 1.

Probability of event E = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{pointing at 8}) = \frac{1}{8}$.

(ii) Probability that it will point at an odd number:

Let F be the event that the arrow points at an odd number.

The odd numbers among 1, 2, 3, 4, 5, 6, 7, 8 are 1, 3, 5, 7.

Number of favorable outcomes for event F (getting an odd number) = 4.

Probability of event F = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{pointing at an odd number}) = \frac{4}{8}$.

Simplifying the fraction:

$P(\text{pointing at an odd number}) = \frac{\cancel{4}^{1}}{\cancel{8}_{2}} = \frac{1}{2}$.

Given:

A die is thrown once.

To Find:

The probability of getting (i) a number greater than 4 (ii) a number less than or equal to 4.

Solution:

When a die is thrown once, the possible outcomes are the numbers on its faces: 1, 2, 3, 4, 5, 6.

Total number of possible outcomes = 6.

(i) Probability of getting a number greater than 4:

Let A be the event of getting a number greater than 4.

The numbers greater than 4 on a die are 5 and 6.

Number of favorable outcomes for event A = 2.

Probability of event A = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{number} > 4) = \frac{2}{6}$

Simplifying the fraction:

$P(\text{number} > 4) = \frac{\cancel{2}^{1}}{\cancel{6}_{3}} = \frac{1}{3}$.

So, the probability of getting a number greater than 4 is $\frac{1}{3}$.

(ii) Probability of getting a number less than or equal to 4:

Let B be the event of getting a number less than or equal to 4.

The numbers less than or equal to 4 on a die are 1, 2, 3, 4.

Number of favorable outcomes for event B = 4.

Probability of event B = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{number} \leq 4) = \frac{4}{6}$

Simplifying the fraction:

$P(\text{number} \leq 4) = \frac{\cancel{4}^{2}}{\cancel{6}_{3}} = \frac{2}{3}$.

So, the probability of getting a number less than or equal to 4 is $\frac{2}{3}$.

Alternate Solution for (ii):

The event "getting a number less than or equal to 4" is the complement of the event "getting a number greater than 4".

Probability (number $\leq$ 4) = $1$ - Probability (number > 4)

From part (i), Probability (number > 4) = $\frac{1}{3}$.

Probability (number $\leq$ 4) = $1 - \frac{1}{3}$

Probability (number $\leq$ 4) = $\frac{3}{3} - \frac{1}{3} = \frac{3-1}{3} = \frac{2}{3}$.

This confirms the result obtained earlier.

Given:

A well-shuffled pack of 52 playing cards.

To Find:

The probability that the card drawn is (i) a queen (ii) a card of diamond.

Solution:

Total number of cards in a well-shuffled pack = 52.

Total number of possible outcomes = 52.

(i) Probability of getting a queen:

In a standard pack of 52 playing cards, there are 4 queens (one for each suit: Hearts, Diamonds, Clubs, Spades).

Number of favorable outcomes (getting a queen) = 4.

Probability of getting a queen = $\frac{\text{Number of queens}}{\text{Total number of cards}}$

Probability (queen) = $\frac{4}{52}$.

Simplifying the fraction:

Probability (queen) = $\frac{\cancel{4}^{1}}{\cancel{52}_{13}} = \frac{1}{13}$.

So, the probability of getting a queen is $\frac{1}{13}$.

(ii) Probability of getting a card of diamond:

In a standard pack of 52 playing cards, there are 13 cards of each suit.

The number of cards of the diamond suit is 13.

Number of favorable outcomes (getting a diamond) = 13.

Probability of getting a card of diamond = $\frac{\text{Number of diamonds}}{\text{Total number of cards}}$

Probability (diamond) = $\frac{13}{52}$.

Simplifying the fraction:

Probability (diamond) = $\frac{\cancel{13}^{1}}{\cancel{52}_{4}} = \frac{1}{4}$.

So, the probability of getting a card of diamond is $\frac{1}{4}$.

Given:

Total number of bulbs in the lot = 20.

Number of defective bulbs = 4.

Number of good bulbs = Total bulbs - Number of defective bulbs = $20 - 4 = 16$.

To Find:

(i) The probability that the first bulb drawn is defective.

(ii) The probability that the second bulb drawn is not defective, given the first bulb was defective and not replaced.

Solution:

When one bulb is drawn at random from the initial lot, the total number of possible outcomes is 20.

(i) Probability that the first bulb is defective:

Let A be the event that the first bulb drawn is defective.

Number of favorable outcomes for event A (getting a defective bulb) is the number of defective bulbs, which is 4.

Probability of event A = $\frac{\text{Number of defective bulbs}}{\text{Total number of bulbs}}$

$P(\text{1st bulb is defective}) = \frac{4}{20}$.

Simplifying the fraction:

$P(\text{1st bulb is defective}) = \frac{\cancel{4}^{1}}{\cancel{20}_{5}} = \frac{1}{5}$.

So, the probability that the first bulb drawn is defective is $\frac{1}{5}$.

(ii) Probability that the second bulb is not defective (given the first was defective and not replaced):

Given that the first bulb drawn was defective and was not replaced, the composition of the remaining lot changes.

Total number of bulbs remaining = $20 - 1 = 19$.

Number of defective bulbs remaining = $4 - 1 = 3$.

Number of good bulbs remaining = 16 (since the removed bulb was defective).

Now, a second bulb is drawn at random from these 19 bulbs.

Let B be the event that the second bulb drawn is not defective.

The number of favorable outcomes for event B (getting a not defective bulb) is the number of good bulbs remaining, which is 16.

The total number of possible outcomes for the second draw is the total number of bulbs remaining, which is 19.

Probability of event B = $\frac{\text{Number of good bulbs remaining}}{\text{Total number of bulbs remaining}}$

$P(\text{2nd bulb is not defective | 1st was defective}) = \frac{16}{19}$.

Since 16 and 19 have no common factors other than 1, the fraction is already in its simplest form.

So, the probability that the second bulb is not defective, given the first was defective and not replaced, is $\frac{16}{19}$.

Given:

Two dice are thrown simultaneously.

To Find:

The probability that the sum of the numbers appearing on the two dice is (i) 8 (ii) less than 5.

Solution:

When two dice are thrown simultaneously, the total number of possible outcomes is the product of the number of outcomes for each die. Each die has 6 faces (1, 2, 3, 4, 5, 6).

Total number of possible outcomes = $6 \times 6 = 36$.

The sample space S consists of 36 ordered pairs $(a, b)$, where $a$ is the result of the first die and $b$ is the result of the second die, with $a, b \in \{1, 2, 3, 4, 5, 6\}$.

(i) Probability that the sum of the numbers is 8:

Let E be the event that the sum of the numbers appearing on the two dice is 8.

The outcomes where the sum is 8 are:

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

Number of favorable outcomes for event E = 5.

Probability of event E = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{sum is 8}) = \frac{5}{36}$.

Since 5 and 36 have no common factors other than 1, the fraction is in its simplest form.

So, the probability that the sum of the numbers is 8 is $\frac{5}{36}$.

(ii) Probability that the sum of the numbers is less than 5:

Let F be the event that the sum of the numbers appearing on the two dice is less than 5.

A sum less than 5 means the sum can be 2, 3, or 4.

Outcomes where the sum is 2: (1, 1).

Outcomes where the sum is 3: (1, 2), (2, 1).

Outcomes where the sum is 4: (1, 3), (2, 2), (3, 1).

Number of outcomes with sum 2 = 1.

Number of outcomes with sum 3 = 2.

Number of outcomes with sum 4 = 3.

Total number of favorable outcomes for event F = (Outcomes with sum 2) + (Outcomes with sum 3) + (Outcomes with sum 4)

Total number of favorable outcomes = $1 + 2 + 3 = 6$.

Probability of event F = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{sum is less than 5}) = \frac{6}{36}$.

Simplifying the fraction:

$P(\text{sum is less than 5}) = \frac{\cancel{6}^{1}}{\cancel{36}_{6}} = \frac{1}{6}$.

So, the probability that the sum of the numbers is less than 5 is $\frac{1}{6}$.

Given:

A well-shuffled pack of 52 playing cards.

To Find:

The probability that the card drawn is (i) a red card (ii) a face card.

Solution:

Total number of cards in a well-shuffled pack = 52.

Total number of possible outcomes = 52.

(i) Probability of getting a red card:

In a standard pack of 52 playing cards, there are two red suits: Hearts and Diamonds.

Each red suit contains 13 cards.

Total number of red cards = Number of Hearts + Number of Diamonds = $13 + 13 = 26$.

Number of favorable outcomes (getting a red card) = 26.

Probability of getting a red card = $\frac{\text{Number of red cards}}{\text{Total number of cards}}$

Probability (red card) = $\frac{26}{52}$.

Simplifying the fraction:

Probability (red card) = $\frac{\cancel{26}^{1}}{\cancel{52}_{2}} = \frac{1}{2}$.

So, the probability of getting a red card is $\frac{1}{2}$.

(ii) Probability of getting a face card:

In a standard pack of 52 playing cards, the face cards are King, Queen, and Jack.

There are 4 suits, and each suit has 3 face cards (King, Queen, Jack).

Total number of face cards = Number of face cards per suit $\times$ Number of suits = $3 \times 4 = 12$.

Number of favorable outcomes (getting a face card) = 12.

Probability of getting a face card = $\frac{\text{Number of face cards}}{\text{Total number of cards}}$

Probability (face card) = $\frac{12}{52}$.

Simplifying the fraction:

Probability (face card) = $\frac{\cancel{12}^{3}}{\cancel{52}_{13}} = \frac{3}{13}$.

So, the probability of getting a face card is $\frac{3}{13}$.

Given:

A bag contains tickets numbered from 11 to 30.

To Find:

The probability that the ticket taken out is a multiple of 5.

Solution:

The numbers on the tickets are 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30.

To find the total number of tickets, we can count them or use the formula: Last Number - First Number + 1.

Total number of tickets = $30 - 11 + 1 = 20$.

Total number of possible outcomes = 20.

We need to find the numbers in this range that are multiples of 5.

A multiple of 5 is a number that can be divided by 5 with no remainder.

The multiples of 5 starting from 11 are 15, 20, 25, and 30.

The multiples of 5 within the range 11 to 30 are: 15, 20, 25, 30.

Number of favorable outcomes (getting a multiple of 5) = 4.

The probability of an event is given by the formula:

Probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Probability (multiple of 5) = $\frac{4}{20}$.

Simplifying the fraction:

Probability (multiple of 5) = $\frac{\cancel{4}^{1}}{\cancel{20}_{5}} = \frac{1}{5}$.

So, the probability that the ticket taken out is a multiple of 5 is $\frac{1}{5}$.

Given:

The word is 'ASSASSINATION'.

To Find:

The probability that a randomly chosen letter from the word is a vowel.

Solution:

First, list all the letters in the word 'ASSASSINATION'.

A, S, S, A, S, S, I, N, A, T, I, O, N

Total number of letters in the word = 13.

Total number of possible outcomes when choosing one letter = 13.

Next, identify the vowels in the word. The vowels are A, E, I, O, U.

From the letters of the word, the vowels are: A, A, I, A, I, O.

Count the number of vowels in the word:

• Number of A's = 3
• Number of E's = 0
• Number of I's = 2
• Number of O's = 1
• Number of U's = 0

Total number of vowels = $3 + 0 + 2 + 1 + 0 = 6$.

Number of favorable outcomes (getting a vowel) = 6.

The probability of choosing a vowel is given by:

Probability = $\frac{\text{Number of vowels}}{\text{Total number of letters}}$

Probability (vowel) = $\frac{6}{13}$.

The fraction $\frac{6}{13}$ cannot be simplified further.

So, the probability that the letter chosen is a vowel is $\frac{6}{13}$.

Given:

Total number of marbles in the jar = 24.

The marbles are either green or blue.

Probability of drawing a green marble = $P(\text{Green}) = \frac{2}{3}$.

To Find:

The number of blue marbles in the jar.

Solution:

Let the number of green marbles be $n_g$.

Let the number of blue marbles be $n_b$.

The total number of marbles is $n_g + n_b = 24$.

The probability of drawing a green marble is given by the formula:

$P(\text{Green}) = \frac{\text{Number of green marbles}}{\text{Total number of marbles}}$

To find the number of green marbles ($n_g$), we can rearrange the equation:

So, there are 16 green marbles in the jar.

We know that the total number of marbles is 24 and they are either green or blue.

Total marbles = Number of green marbles + Number of blue marbles

Substitute the value of $n_g$ we found:

Now, solve for $n_b$:

Therefore, there are 8 blue marbles in the jar.

Verification:

Number of green marbles = 16

Number of blue marbles = 8

Total marbles = $16 + 8 = 24$ (Matches the given total)

Probability of green = $\frac{\text{Number of green marbles}}{\text{Total marbles}} = \frac{16}{24} = \frac{\cancel{16}^{2}}{\cancel{24}_{3}} = \frac{2}{3}$ (Matches the given probability)

The calculation is correct.

Given:

Total number of balls in the box initially = 12.

Number of black balls initially = $x$.

Number of non-black balls initially = $12 - x$.

6 more black balls are added to the box.

The new probability of drawing a black ball is double the initial probability.

To Find:

The value of $x$.

Solution:

Initial Probability:

When one ball is drawn at random from the initial box, the total number of possible outcomes is 12.

The number of favorable outcomes (getting a black ball) is the initial number of black balls, which is $x$.

Initial probability of drawing a black ball, let's call it $P_1$:

Probability after adding 6 black balls:

After adding 6 more black balls:

New number of black balls = $x + 6$.

New total number of balls = Initial total balls + Added black balls = $12 + 6 = 18$.

When one ball is drawn at random from the new box, the total number of possible outcomes is 18.

The number of favorable outcomes (getting a black ball) is the new number of black balls, which is $x + 6$.

New probability of drawing a black ball, let's call it $P_2$:

Using the given relationship:

We are given that the new probability ($P_2$) is double the initial probability ($P_1$).

Substitute the expressions for $P_1$ and $P_2$ from equations (i) and (ii) into this relationship:

Simplify the right side of the equation:

Further simplification of the fraction on the right side:

Now, solve for $x$ by cross-multiplying or multiplying both sides by the LCM of 18 and 6, which is 18.

Multiplying both sides by 18:

Rearrange the equation to isolate $x$ terms:

Divide by 2:

The value of $x$ must be a non-negative integer, and less than or equal to the initial total number of balls (12), which $x=3$ satisfies.

Therefore, the value of $x$ is 3.

Given:

Number of red balls in the bag = 5.

Number of blue balls in the bag = $y$ (Let $y$ be the number of blue balls).

The probability of drawing a blue ball is double that of drawing a red ball.

To Find:

The number of blue balls in the bag ($y$).

Solution:

Let the number of blue balls be $y$.

Total number of balls in the bag = Number of red balls + Number of blue balls

Total number of balls = $5 + y$.

When a ball is drawn at random from the bag, the total number of possible outcomes is $5+y$.

The probability of drawing a red ball is:

The probability of drawing a blue ball is:

We are given that the probability of drawing a blue ball is double that of a red ball.

Substitute the expressions for $P(\text{Blue})$ and $P(\text{Red})$ from equations (i) and (ii) into this relationship:

Since the denominators are the same and are non-zero (as $y$ must be non-negative), we can equate the numerators:

The number of blue balls must be a non-negative integer, which $y=10$ satisfies.

Therefore, the number of blue balls in the bag is 10.

Verification:

If the number of blue balls is 10, then the total number of balls is $5 + 10 = 15$.

Probability of drawing a red ball = $\frac{5}{15} = \frac{1}{3}$.

Probability of drawing a blue ball = $\frac{10}{15} = \frac{2}{3}$.

Is the probability of drawing a blue ball double that of a red ball?

$\frac{2}{3} = 2 \times \frac{1}{3}$

$\frac{2}{3} = \frac{2}{3}$

The relationship holds true. The calculation is correct.

Given:

A one-rupee coin is tossed 3 times.

Hanif wins if the outcome is three heads (HHH) or three tails (TTT).

Hanif loses otherwise.

To Calculate:

The probability that Hanif will lose the game.

Solution:

When a coin is tossed 3 times, the total possible outcomes are:

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

The sample space S is \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}.

Total number of possible outcomes = $|S| = 8$.

Hanif wins if the outcome is three heads (HHH) or three tails (TTT).

Let W be the event that Hanif wins.

The favorable outcomes for W are \{HHH, TTT\}.

Number of favorable outcomes for W = 2.

The probability that Hanif wins is $P(W) = \frac{\text{Number of favorable outcomes for W}}{\text{Total number of possible outcomes}}$

Simplifying the fraction:

Hanif loses if the outcome is not three heads and not three tails.

Let L be the event that Hanif loses.

The event L is the complement of the event W (Hanif wins).

The favorable outcomes for L are the outcomes in the sample space that are not in W:

\{HHT, HTH, THH, HTT, THT, TTH\}.

Number of favorable outcomes for L = 6.

The probability that Hanif loses is $P(L) = \frac{\text{Number of favorable outcomes for L}}{\text{Total number of possible outcomes}}$

Simplifying the fraction:

Thus, the probability that Hanif will lose the game is $\frac{3}{4}$.

Alternate Solution using Complement Rule:

The event that Hanif loses is the complement of the event that Hanif wins.

$P(\text{Loses}) = 1 - P(\text{Wins})$

From equation (i), we found $P(\text{Wins}) = \frac{1}{4}$.

Both methods yield the same result.

Given:

A well-shuffled deck of 52 playing cards.

To Find:

The probability that the card drawn is (i) a face card or a red card (ii) a black card which is not a face card.

Solution:

Total number of cards in a well-shuffled deck = 52.

Total number of possible outcomes = 52.

(i) Probability that the card drawn is a face card or a red card:

Let A be the event that the card drawn is a face card.

There are 12 face cards in a deck (3 in each of the 4 suits).

Number of face cards = 12.

$P(A) = \frac{\text{Number of face cards}}{\text{Total number of cards}} = \frac{12}{52}$.

Let B be the event that the card drawn is a red card.

There are 2 red suits (Hearts and Diamonds), each with 13 cards.

Number of red cards = $13 + 13 = 26$.

$P(B) = \frac{\text{Number of red cards}}{\text{Total number of cards}} = \frac{26}{52}$.

The event "A and B" (A $\cap$ B) is getting a card that is both a face card and a red card (Red Face Card).

The red face cards are the King, Queen, and Jack of Hearts and the King, Queen, and Jack of Diamonds.

Number of red face cards = $3 + 3 = 6$.

$P(A \cap B) = \frac{\text{Number of red face cards}}{\text{Total number of cards}} = \frac{6}{52}$.

The probability of getting a face card or a red card is $P(A \cup B)$.

Using the formula for the probability of the union of two events:

Simplifying the fraction:

So, the probability that the card drawn is a face card or a red card is $\frac{8}{13}$.

(ii) Probability that the card drawn is a black card which is not a face card:

We need to find the number of black cards that are not face cards.

Total number of black cards = 26 (13 Clubs + 13 Spades).

Number of black face cards = 6 (King, Queen, Jack of Clubs and King, Queen, Jack of Spades).

Number of black cards which are not face cards = Total black cards - Number of black face cards

Number of black cards which are not face cards = $26 - 6 = 20$.

These 20 cards are the Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10 of Clubs and the Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10 of Spades.

Number of favorable outcomes = 20.

Total number of possible outcomes = 52.

Probability (black card which is not a face card) = $\frac{\text{Number of black cards which are not face cards}}{\text{Total number of cards}}$

Simplifying the fraction:

So, the probability that the card drawn is a black card which is not a face card is $\frac{5}{13}$.

Given:

Two dice are thrown simultaneously.

To Find:

The probability that the sum of the two numbers appearing on the top of the dice is (i) 9 (ii) 13 (iii) less than or equal to 12.

Solution:

When two dice are thrown simultaneously, the total number of possible outcomes is the product of the number of outcomes for each die. Each die has 6 faces (1, 2, 3, 4, 5, 6).

Total number of possible outcomes = $6 \times 6 = 36$.

The sample space S consists of 36 ordered pairs $(a, b)$, where $a$ is the result of the first die and $b$ is the result of the second die, with $a, b \in \{1, 2, 3, 4, 5, 6\}$.

(i) Probability that the sum of the numbers is 9:

Let E be the event that the sum of the numbers appearing on the two dice is 9.

The favorable outcomes are the pairs $(a, b)$ such that $a+b=9$. These are: (3, 6), (4, 5), (5, 4), (6, 3).

Number of favorable outcomes for event E = 4.

Probability of event E = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{sum is 9}) = \frac{4}{36}$.

Simplifying the fraction:

$P(\text{sum is 9}) = \frac{\cancel{4}^{1}}{\cancel{36}_{9}} = \frac{1}{9}$.

So, the probability that the sum of the numbers is 9 is $\frac{1}{9}$.

(ii) Probability that the sum of the numbers is 13:

Let F be the event that the sum of the numbers appearing on the two dice is 13.

The maximum possible sum when throwing two standard dice is $6 + 6 = 12$.

A sum of 13 is not possible with two standard dice.

Number of favorable outcomes for event F = 0.

Probability of event F = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{sum is 13}) = \frac{0}{36} = 0$.

So, the probability that the sum of the numbers is 13 is 0.

(iii) Probability that the sum of the numbers is less than or equal to 12:

Let G be the event that the sum of the numbers appearing on the two dice is less than or equal to 12.

The maximum possible sum when throwing two dice is 12.

All possible outcomes in the sample space have a sum that is less than or equal to 12. (The sums range from $1+1=2$ to $6+6=12$).

The favorable outcomes are all the outcomes in the sample space.

Number of favorable outcomes for event G = 36.

Probability of event G = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{sum} \leq 12) = \frac{36}{36} = 1$.

So, the probability that the sum of the numbers is less than or equal to 12 is 1.

Given:

A box contains 90 discs numbered from 1 to 90.

To Find:

The probability that a randomly drawn disc bears (i) a perfect square number (ii) a number divisible by 5 (iii) a two-digit number divisible by both 2 and 5.

Solution:

Total number of discs in the box = 90.

The numbers on the discs are $\{1, 2, 3, \dots, 90\}$.

Total number of possible outcomes when drawing one disc = 90.

(i) Probability of getting a perfect square number:

A perfect square number is an integer that is the square of an integer.

The perfect square numbers between 1 and 90 are:

$1^2 = 1$

$2^2 = 4$

$3^2 = 9$

$4^2 = 16$

$5^2 = 25$

$6^2 = 36$

$7^2 = 49$

$8^2 = 64$

$9^2 = 81$

The next perfect square is $10^2 = 100$, which is greater than 90.

The perfect square numbers on the discs are $\{1, 4, 9, 16, 25, 36, 49, 64, 81\}$.

Number of favorable outcomes (getting a perfect square) = 9.

Probability of getting a perfect square number = $\frac{\text{Number of perfect square numbers}}{\text{Total number of discs}}$

Probability (perfect square) = $\frac{9}{90}$.

Simplifying the fraction:

Probability (perfect square) = $\frac{\cancel{9}^{1}}{\cancel{90}_{10}} = \frac{1}{10}$.

So, the probability of getting a perfect square number is $\frac{1}{10}$.

(ii) Probability of getting a number divisible by 5:

A number is divisible by 5 if its unit digit is 0 or 5.

The numbers divisible by 5 in the range 1 to 90 are:

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90.

To count these numbers, we can use the formula: $(\text{Last multiple} - \text{First multiple}) / \text{Divisor} + 1$

Number of multiples of 5 = $(90 - 5) / 5 + 1 = 85 / 5 + 1 = 17 + 1 = 18$.

Number of favorable outcomes (getting a number divisible by 5) = 18.

Probability of getting a number divisible by 5 = $\frac{\text{Number of numbers divisible by 5}}{\text{Total number of discs}}$

Probability (divisible by 5) = $\frac{18}{90}$.

Simplifying the fraction:

Probability (divisible by 5) = $\frac{\cancel{18}^{1}}{\cancel{90}_{5}} = \frac{1}{5}$.

So, the probability of getting a number divisible by 5 is $\frac{1}{5}$.

(iii) Probability of getting a two-digit number divisible by both 2 and 5:

A number divisible by both 2 and 5 is divisible by their least common multiple, which is LCM(2, 5) = 10.

We need to find the two-digit numbers between 1 and 90 that are divisible by 10.

The two-digit numbers start from 10.

The numbers divisible by 10 in the range 1 to 90 are:

10, 20, 30, 40, 50, 60, 70, 80, 90.

All these numbers are two-digit numbers and are within the range 1 to 90.

Number of such numbers = 9.

Number of favorable outcomes (getting a two-digit number divisible by both 2 and 5) = 9.

Probability of getting a two-digit number divisible by both 2 and 5 = $\frac{\text{Number of two-digit numbers divisible by 10}}{\text{Total number of discs}}$

Probability (two-digit and divisible by 10) = $\frac{9}{90}$.

Simplifying the fraction:

Probability (two-digit and divisible by 10) = $\frac{\cancel{9}^{1}}{\cancel{90}_{10}} = \frac{1}{10}$.

So, the probability of getting a two-digit number divisible by both 2 and 5 is $\frac{1}{10}$.

Given:

The cards are the ten, jack, queen, king, and ace of diamonds.

These 5 cards are well-shuffled.

To Find:

(i) The probability that the first card picked is the queen.

(ii) If the queen is drawn and put aside, the probability that the second card picked is (a) an ace (b) a queen.

Solution:

The initial set of cards is \{Ten of Diamonds, Jack of Diamonds, Queen of Diamonds, King of Diamonds, Ace of Diamonds\}.

Total number of cards initially = 5.

Total number of possible outcomes for the first draw = 5.

(i) Probability that the card is the queen:

Let A be the event that the first card picked is the queen.

There is only one Queen of Diamonds in the initial set of 5 cards.

Number of favorable outcomes for event A = 1.

Probability of event A = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{first card is queen}) = \frac{1}{5}$.

So, the probability that the card is the queen is $\frac{1}{5}$.

(ii) If the queen is drawn and put aside:

Assume that the first card drawn was indeed the Queen of Diamonds, and it is not replaced in the set.

The remaining cards in the set are \{Ten of Diamonds, Jack of Diamonds, King of Diamonds, Ace of Diamonds\}.

Number of cards remaining in the set = $5 - 1 = 4$.

Total number of possible outcomes for the second draw = 4.

(ii) (a) Probability that the second card picked up is an ace:

Let B be the event that the second card picked is an ace, given the first was the queen and not replaced.

In the remaining set of 4 cards, there is one Ace of Diamonds.

Number of favorable outcomes for event B = 1.

Probability of event B = $\frac{\text{Number of aces remaining}}{\text{Total number of cards remaining}}$

$P(\text{second card is ace}) = \frac{1}{4}$.

So, the probability that the second card picked up is an ace is $\frac{1}{4}$.

(ii) (b) Probability that the second card picked up is a queen:

Let C be the event that the second card picked is a queen, given the first was the queen and not replaced.

Since the Queen of Diamonds was drawn first and put aside, there are no queens left in the remaining set of 4 cards.

Number of favorable outcomes for event C = 0.

Probability of event C = $\frac{\text{Number of queens remaining}}{\text{Total number of cards remaining}}$

$P(\text{second card is queen}) = \frac{0}{4} = 0$.

So, the probability that the second card picked up is a queen is 0.

Given:

A bag contains 15 tickets numbered from 1 to 15.

To Find:

The probability that the ticket has (i) an even number (ii) a number divisible by 3 (iii) a number which is a prime number (iv) a number divisible by 3 and 5.

Solution:

The numbers on the tickets are $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}$.

Total number of tickets = 15.

Total number of possible outcomes when drawing one ticket = 15.

(i) Probability that the ticket has an even number:

Let E be the event that the ticket has an even number.

The even numbers among 1 to 15 are: 2, 4, 6, 8, 10, 12, 14.

Number of favorable outcomes for event E = 7.

Probability of event E = $\frac{\text{Number of even numbers}}{\text{Total number of tickets}}$

$P(\text{even number}) = \frac{7}{15}$.

The fraction $\frac{7}{15}$ cannot be simplified further.

(ii) Probability that the ticket has a number divisible by 3:

Let F be the event that the ticket has a number divisible by 3.

The numbers divisible by 3 among 1 to 15 are: 3, 6, 9, 12, 15.

Number of favorable outcomes for event F = 5.

Probability of event F = $\frac{\text{Number of numbers divisible by 3}}{\text{Total number of tickets}}$

$P(\text{divisible by 3}) = \frac{5}{15}$.

Simplifying the fraction:

$P(\text{divisible by 3}) = \frac{\cancel{5}^{1}}{\cancel{15}_{3}} = \frac{1}{3}$.

(iii) Probability that the ticket has a number which is a prime number:

Let G be the event that the ticket has a prime number.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

The prime numbers among 1 to 15 are: 2, 3, 5, 7, 11, 13.

Number of favorable outcomes for event G = 6.

Probability of event G = $\frac{\text{Number of prime numbers}}{\text{Total number of tickets}}$

$P(\text{prime number}) = \frac{6}{15}$.

Simplifying the fraction:

$P(\text{prime number}) = \frac{\cancel{6}^{2}}{\cancel{15}_{5}} = \frac{2}{5}$.

(iv) Probability that the ticket has a number divisible by 3 and 5:

Let H be the event that the ticket has a number divisible by both 3 and 5.

A number divisible by both 3 and 5 is divisible by their least common multiple, which is LCM(3, 5) = 15.

We need to find the numbers among 1 to 15 that are divisible by 15.

The only number divisible by 15 in the range 1 to 15 is 15 itself.

Number of favorable outcomes for event H = 1.

Probability of event H = $\frac{\text{Number of numbers divisible by 15}}{\text{Total number of tickets}}$

$P(\text{divisible by 3 and 5}) = \frac{1}{15}$.

The fraction $\frac{1}{15}$ cannot be simplified further.

Given:

The jar contains red, white, and orange marbles.

Probability of selecting a red marble, $P(\text{Red}) = \frac{1}{4}$.

Probability of selecting a white marble, $P(\text{White}) = \frac{1}{3}$.

Number of orange marbles = 10.

To Find:

The total number of marbles in the jar.

Solution:

Let $P(\text{Orange})$ be the probability of selecting an orange marble.

Since the marbles are only red, white, or orange, these events are mutually exclusive and exhaustive.

The sum of the probabilities of these events must be equal to 1.

$P(\text{Red}) + P(\text{White}) + P(\text{Orange}) = 1$

Substitute the given probabilities:

$\frac{1}{4} + \frac{1}{3} + P(\text{Orange}) = 1$

To find $P(\text{Orange})$, subtract the sum of $P(\text{Red})$ and $P(\text{White})$ from 1.

$P(\text{Orange}) = 1 - \left( \frac{1}{4} + \frac{1}{3} \right)$

Find a common denominator for $\frac{1}{4}$ and $\frac{1}{3}$, which is 12.

$\frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12}$

$\frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12}$

So, $\frac{1}{4} + \frac{1}{3} = \frac{3}{12} + \frac{4}{12} = \frac{3+4}{12} = \frac{7}{12}$.

Now, calculate $P(\text{Orange})$:

$P(\text{Orange}) = 1 - \frac{7}{12}$

$P(\text{Orange}) = \frac{12}{12} - \frac{7}{12} = \frac{12-7}{12} = \frac{5}{12}$.

Let $N$ be the total number of marbles in the jar.

The probability of selecting an orange marble is also given by the ratio of the number of orange marbles to the total number of marbles:

$P(\text{Orange}) = \frac{\text{Number of orange marbles}}{\text{Total number of marbles}}$

To find $N$, we can cross-multiply:

Divide both sides by 5:

The total number of marbles in the jar is 24.

Verification:

Total marbles = 24.

$P(\text{Red}) = \frac{1}{4}$. Number of red marbles = $24 \times \frac{1}{4} = 6$. (Not given directly, but implies consistency)

$P(\text{White}) = \frac{1}{3}$. Number of white marbles = $24 \times \frac{1}{3} = 8$. (Not given directly, but implies consistency)

Number of orange marbles = 10 (Given).

Total marbles = Red + White + Orange = $6 + 8 + 10 = 24$. (Matches the calculated total)

$P(\text{Orange}) = \frac{10}{24} = \frac{\cancel{10}^{5}}{\cancel{24}_{12}} = \frac{5}{12}$. (Matches the calculated probability of orange)

Given:

A number is selected at random from the numbers 1 to 30.

The set of possible numbers is $\{1, 2, 3, \dots, 30\}$.

To Find:

The probability that the selected number is (i) a prime number (ii) a multiple of 4 (iii) a multiple of 3 or 5 (iv) a factor of 24.

Solution:

Total number of possible outcomes = 30.

When a number is selected at random from the numbers 1 to 30, the total number of possible outcomes is 30.

(i) Probability that the selected number is a prime number:

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

The prime numbers among 1 to 30 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

Number of favorable outcomes (getting a prime number) = 10.

Probability of getting a prime number = $\frac{\text{Number of prime numbers}}{\text{Total number of numbers}}$

$P(\text{prime number}) = \frac{10}{30}$.

Simplifying the fraction:

$P(\text{prime number}) = \frac{\cancel{10}^{1}}{\cancel{30}_{3}} = \frac{1}{3}$.

So, the probability of getting a prime number is $\frac{1}{3}$.

(ii) Probability that the selected number is a multiple of 4:

A multiple of 4 is a number that can be obtained by multiplying 4 by an integer.

The multiples of 4 among 1 to 30 are: 4, 8, 12, 16, 20, 24, 28.

Number of favorable outcomes (getting a multiple of 4) = 7.

Probability of getting a multiple of 4 = $\frac{\text{Number of multiples of 4}}{\text{Total number of numbers}}$

$P(\text{multiple of 4}) = \frac{7}{30}$.

The fraction $\frac{7}{30}$ cannot be simplified further.

So, the probability of getting a multiple of 4 is $\frac{7}{30}$.

(iii) Probability that the selected number is a multiple of 3 or 5:

Let A be the event that the number is a multiple of 3.

Multiples of 3 among 1 to 30: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30. Number of multiples of 3 = 10.

$P(A) = \frac{10}{30}$.

Let B be the event that the number is a multiple of 5.

Multiples of 5 among 1 to 30: 5, 10, 15, 20, 25, 30. Number of multiples of 5 = 6.

$P(B) = \frac{6}{30}$.

The event "A and B" (A $\cap$ B) is that the number is a multiple of both 3 and 5, which means it is a multiple of 15 (LCM of 3 and 5).

Multiples of 15 among 1 to 30: 15, 30. Number of multiples of 15 = 2.

$P(A \cap B) = \frac{2}{30}$.

The probability of getting a multiple of 3 or 5 is $P(A \cup B)$.

Using the principle of inclusion-exclusion: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

$P(\text{multiple of 3 or 5}) = \frac{10}{30} + \frac{6}{30} - \frac{2}{30}$.

$P(\text{multiple of 3 or 5}) = \frac{10 + 6 - 2}{30} = \frac{14}{30}$.

Simplifying the fraction:

$P(\text{multiple of 3 or 5}) = \frac{\cancel{14}^{7}}{\cancel{30}_{15}} = \frac{7}{15}$.

So, the probability of getting a multiple of 3 or 5 is $\frac{7}{15}$.

(iv) Probability that the selected number is a factor of 24:

The factors of 24 are the numbers that divide 24 exactly.

The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24.

All these factors are within the range 1 to 30.

Number of favorable outcomes (getting a factor of 24) = 8.

Probability of getting a factor of 24 = $\frac{\text{Number of factors of 24}}{\text{Total number of numbers}}$

$P(\text{factor of 24}) = \frac{8}{30}$.

Simplifying the fraction:

$P(\text{factor of 24}) = \frac{\cancel{8}^{4}}{\cancel{30}_{15}} = \frac{4}{15}$.

So, the probability that the selected number is a factor of 24 is $\frac{4}{15}$.

Given:

A die is thrown once.

To Find:

The probability of getting (i) a number greater than 3 (ii) a number less than or equal to 1 (iii) a multiple of 3 (iv) an even prime number.

Solution:

When a standard die is thrown once, the possible outcomes are the integers from 1 to 6.

The sample space is $\{1, 2, 3, 4, 5, 6\}$.

Total number of possible outcomes = 6.

(i) Probability of getting a number greater than 3:

Let A be the event of getting a number greater than 3.

The numbers greater than 3 in the sample space are \{4, 5, 6\}.

Number of favorable outcomes for event A = 3.

Probability of event A = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{number} > 3) = \frac{3}{6}$.

Simplifying the fraction:

$P(\text{number} > 3) = \frac{\cancel{3}^{1}}{\cancel{6}_{2}} = \frac{1}{2}$.

So, the probability of getting a number greater than 3 is $\frac{1}{2}$.

(ii) Probability of getting a number less than or equal to 1:

Let B be the event of getting a number less than or equal to 1.

The numbers less than or equal to 1 in the sample space are \{1\}.

Number of favorable outcomes for event B = 1.

Probability of event B = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

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

The fraction $\frac{1}{6}$ cannot be simplified further.

So, the probability of getting a number less than or equal to 1 is $\frac{1}{6}$.

(iii) Probability of getting a multiple of 3:

Let C be the event of getting a multiple of 3.

The multiples of 3 in the sample space are \{3, 6\}.

Number of favorable outcomes for event C = 2.

Probability of event C = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{multiple of 3}) = \frac{2}{6}$.

Simplifying the fraction:

$P(\text{multiple of 3}) = \frac{\cancel{2}^{1}}{\cancel{6}_{3}} = \frac{1}{3}$.

So, the probability of getting a multiple of 3 is $\frac{1}{3}$.

(iv) Probability of getting an even prime number:

Let D be the event of getting an even prime number.

The prime numbers in the sample space are \{2, 3, 5\}.

The even numbers in the sample space are \{2, 4, 6\}.

The number that is both even and prime is 2.

The favorable outcomes for event D is \{2\}.

Number of favorable outcomes for event D = 1.

Probability of event D = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$P(\text{even prime number}) = \frac{1}{6}$.

The fraction $\frac{1}{6}$ cannot be simplified further.

So, the probability of getting an even prime number is $\frac{1}{6}$.

Given:

A well-shuffled deck of 52 playing cards.

To Find:

The probability that the card drawn is (i) a card of heart (ii) a queen (iii) the ace of spades (iv) the king of red colour.

Solution:

Total number of cards in a well-shuffled deck = 52.

Total number of possible outcomes = 52.

(i) Probability of getting a card of heart:

Let H be the event of drawing a card of heart.

In a standard deck of 52 cards, there are 13 cards of the heart suit.

Number of favorable outcomes for event H = 13.

Probability of event H = $\frac{\text{Number of heart cards}}{\text{Total number of cards}}$

$P(\text{heart}) = \frac{13}{52}$.

Simplifying the fraction:

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

So, the probability of getting a card of heart is $\frac{1}{4}$.

(ii) Probability of getting a queen:

Let Q be the event of drawing a queen.

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

Number of favorable outcomes for event Q = 4.

Probability of event Q = $\frac{\text{Number of queens}}{\text{Total number of cards}}$

$P(\text{queen}) = \frac{4}{52}$.

Simplifying the fraction:

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

So, the probability of getting a queen is $\frac{1}{13}$.

(iii) Probability of getting the ace of spades:

Let A be the event of drawing the ace of spades.

In a standard deck of 52 cards, there is only one Ace of Spades.

Number of favorable outcomes for event A = 1.

Probability of event A = $\frac{\text{Number of ace of spades}}{\text{Total number of cards}}$

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

The fraction $\frac{1}{52}$ cannot be simplified further.

So, the probability of getting the ace of spades is $\frac{1}{52}$.

(iv) Probability of getting the king of red colour:

Let K be the event of drawing a king of red colour.

The red suits are Hearts and Diamonds.

There is one King in the Hearts suit and one King in the Diamonds suit.

Number of kings of red colour = 2.

Number of favorable outcomes for event K = 2.

Probability of event K = $\frac{\text{Number of red kings}}{\text{Total number of cards}}$

$P(\text{red king}) = \frac{2}{52}$.

Simplifying the fraction:

$P(\text{red king}) = \frac{\cancel{2}^{1}}{\cancel{52}_{26}} = \frac{1}{26}$.

So, the probability of getting the king of red colour is $\frac{1}{26}$.