Chapter 14 Probability

Given:

Experiment of rolling a die.

Event A: 'getting a prime number'

Event B: 'getting an odd number'

To Find:

The sets representing the events A or B, A and B, A but not B, and not A.

Solution:

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

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

Event A is the event of 'getting a prime number'. The prime numbers in the sample space S are 2, 3, and 5.

So, the set representing event A is:

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

Event B is the event of 'getting an odd number'. The odd numbers in the sample space S are 1, 3, and 5.

So, the set representing event B is:

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

(i) A or B

The event 'A or B' corresponds to the union of the sets A and B. It includes all outcomes that are in A, or in B, or in both.

This is denoted by $A \cup B$.

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

To find the union, we combine all unique elements from both sets:

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

(ii) A and B

The event 'A and B' corresponds to the intersection of the sets A and B. It includes only the outcomes that are common to both A and B.

This is denoted by $A \cap B$.

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

The common elements are 3 and 5.

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

(iii) A but not B

The event 'A but not B' corresponds to the set difference $A - B$. It includes all outcomes that are in A but are not in B.

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

We remove the elements of B from A. The element 2 is in A but not in B. Elements 3 and 5 are in both.

$A - B = \{2\}$

Alternate approach: This event can also be represented as the intersection of A and the complement of B, $A \cap B'$. First, find the complement of B, $B'$. The complement of B contains all outcomes in the sample space S that are not in B.

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

$B' = \{2, 4, 6\}$

Now find the intersection of A and $B'$.

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

The common element is 2.

$A \cap B' = \{2\}$

(iv) ‘not A’

The event 'not A' corresponds to the complement of set A. It includes all outcomes in the sample space S that are not in A.

This is denoted by $A'$.

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

The elements in S that are not in A are 1, 4, and 6.

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

Given:

Experiment: Throwing two dice.

Events A, B, C, D are defined based on the sum of the numbers on the dice.

To Find:

Pairs of mutually exclusive events among A, B, C, and D.

Solution:

When two dice are thrown, the possible outcomes are ordered pairs $(i, j)$ where $i$ is the result of the first die and $j$ is the result of the second die, with $i, j \in \{1, 2, 3, 4, 5, 6\}$. The sum of the numbers ranges from $1+1=2$ to $6+6=12$. The possible sums are $\{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$.

Let's list the sums corresponding to each event:

Event A: 'the sum is even'

$A = \{2, 4, 6, 8, 10, 12\}$

Event B: 'the sum is a multiple of 3'

$B = \{3, 6, 9, 12\}$

Event C: 'the sum is less than 4' (i.e., sum is 2 or 3)

$C = \{2, 3\}$

Event D: 'the sum is greater than 11' (i.e., sum is 12)

$D = \{12\}$

Two events are mutually exclusive if they cannot occur simultaneously. In terms of sets, this means their intersection is the empty set ($\emptyset$). We need to find pairs of sets whose intersection is empty.

Intersection of A and B:

$A \cap B = \{2, 4, 6, 8, 10, 12\} \cap \{3, 6, 9, 12\}$

$A \cap B = \{6, 12\}$

Since $A \cap B \neq \emptyset$, events A and B are not mutually exclusive.

Intersection of A and C:

$A \cap C = \{2, 4, 6, 8, 10, 12\} \cap \{2, 3\}$

$A \cap C = \{2\}$

Since $A \cap C \neq \emptyset$, events A and C are not mutually exclusive.

Intersection of A and D:

$A \cap D = \{2, 4, 6, 8, 10, 12\} \cap \{12\}$

$A \cap D = \{12\}$

Since $A \cap D \neq \emptyset$, events A and D are not mutually exclusive.

Intersection of B and C:

$B \cap C = \{3, 6, 9, 12\} \cap \{2, 3\}$

$B \cap C = \{3\}$

Since $B \cap C \neq \emptyset$, events B and C are not mutually exclusive.

Intersection of B and D:

$B \cap D = \{3, 6, 9, 12\} \cap \{12\}$

$B \cap D = \{12\}$

Since $B \cap D \neq \emptyset$, events B and D are not mutually exclusive.

Intersection of C and D:

$C \cap D = \{2, 3\} \cap \{12\}$

$C \cap D = \emptyset$

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

The only pair of events with an empty intersection is C and D.

Therefore, the pair of events that is mutually exclusive is C and D.

Given:

Experiment: A coin is tossed three times.

Event A: ‘No head appears’

Event B: ‘Exactly one head appears’

Event C: ‘Atleast two heads appear’

To Check:

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

Solution:

The sample space S for the experiment of tossing a coin three times is the set of all possible outcomes:

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

Now, let's write the sets corresponding to the given events:

Event A: ‘No head appears’

This means all three tosses are tails.

$A = \{TTT\}$

Event B: ‘Exactly one head appears’

This means one toss is head and the other two are tails.

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

Event C: ‘Atleast two heads appear’

This means either two heads and one tail, or three heads.

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

Check for Mutually Exclusive Events:

Events are mutually exclusive if the occurrence of one event excludes the occurrence of any other event. This means the intersection of any two events is the empty set.

Consider the intersection of A and B:

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

Consider the intersection of A and C:

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

Consider the intersection of B and C:

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

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

Check for Exhaustive Events:

Events are exhaustive if at least one of the events must occur whenever the experiment is performed. This means the union of all events is equal to the sample space.

Consider the union of A, B, and C:

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

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

Comparing the union with the sample space S:

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

We observe that $A \cup B \cup C = S$.

Since the union of events A, B, and C covers the entire sample space S, the events A, B, and C are exhaustive.

Conclusion:

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

Given:

Experiment: A die is rolled.

Event E: “die shows 4”

Event F: “die shows even number”

To Check:

If events E and F are mutually exclusive.

Solution:

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

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

Event E: “die shows 4”

The set representing event E is:

$E = \{4\}$

Event F: “die shows even number”

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

The set representing event F is:

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

Two events are mutually exclusive if they cannot occur at the same time, meaning their intersection is the empty set ($\emptyset$).

Consider the intersection of events E and F:

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

The common element in both sets is 4.

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

Since the intersection $E \cap F$ is not empty ($E \cap F = \{4\}$), the events E and F can occur at the same time (specifically, when the die shows 4).

Therefore, events E and F are not mutually exclusive.

Given:

Experiment: A die is thrown.

Events A, B, C, D, E, F are defined based on the outcome.

To Describe and Find:

Describe the sets for events A, B, C, D, E, F.

Find $A \cup B, A \cap B, B \cup C, E \cap F, D \cap E, A – C, D – E, E \cap F′, F′$.

Solution:

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

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

Let's describe each event as a set:

(i) A: a number less than 7

All outcomes in the sample space are less than 7.

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

(ii) B: a number greater than 7

There are no outcomes in the sample space that are greater than 7.

$B = \emptyset$ (the empty set)

(iii) C: a multiple of 3

The multiples of 3 in the sample space are 3 and 6.

$C = \{3, 6\}$

(iv) D: a number less than 4

The numbers less than 4 in the sample space are 1, 2, and 3.

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

(v) E: an even number greater than 4

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

The numbers greater than 4 are 5, 6.

The numbers that are both even and greater than 4 is 6.

$E = \{6\}$

(vi) F: a number not less than 3

This means a number greater than or equal to 3.

The numbers $\geq 3$ in the sample space are 3, 4, 5, and 6.

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

Now, let's find the requested set operations:

$A \cup B$ (A union B)

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

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

$A \cap B$ (A intersection B)

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

$A \cap B = \emptyset$

$B \cup C$ (B union C)

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

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

$E \cap F$ (E intersection F)

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

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

$D \cap E$ (D intersection E)

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

$D \cap E = \emptyset$

$A - C$ (A minus C)

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

$A - C = \{1, 2, 4, 5\}$

$D - E$ (D minus E)

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

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

$F′$ (Complement of F)

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

$F′ = \{1, 2\}$

$E \cap F′$ (E intersection with Complement of F)

$E \cap F′ = \{6\} \cap \{1, 2\}$

$E \cap F′ = \emptyset$

Given:

Experiment: Rolling a pair of dice and recording the numbers.

Events A, B, C are defined.

To Describe and Find:

Describe the sets for events A, B, C.

Identify which pairs of these events are mutually exclusive.

Solution:

The sample space S for rolling a pair of dice consists of all possible ordered pairs $(i, j)$ where $i, j \in \{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is $6 \times 6 = 36$.

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

Let's describe each event as a set of outcomes:

Event A: the sum is greater than 8

Possible sums greater than 8 are 9, 10, 11, 12.

Sum = 9: $(3,6), (4,5), (5,4), (6,3)$

Sum = 10: $(4,6), (5,5), (6,4)$

Sum = 11: $(5,6), (6,5)$

Sum = 12: $(6,6)$

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

Event B: 2 occurs on either die

This means the first die shows 2 or the second die shows 2 (or both).

Outcomes with 2 on the first die: $(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)$

Outcomes with 2 on the second die: $(1,2), (3,2), (4,2), (5,2), (6,2)$ (excluding (2,2) already listed)

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

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

Sums that are multiples of 3 are 3, 6, 9, 12.

Sums that are at least 7 are 7, 8, 9, 10, 11, 12.

We need the sum to be $\geq 7$ AND a multiple of 3. The possible sums are 9 and 12.

Sum = 9: $(3,6), (4,5), (5,4), (6,3)$

Sum = 12: $(6,6)$

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

Now, let's check for mutually exclusive pairs. Two events are mutually exclusive if their intersection is the empty set.

Intersection of A and B ($A \cap B$):

We look for outcomes that are in both A and B. Outcomes in A have a sum greater than 8. Outcomes in B have a 2 on at least one die.

Outcomes in B and their sums:

$(2,1)$ sum = 3

$(2,2)$ sum = 4

$(2,3)$ sum = 5

$(2,4)$ sum = 6

$(2,5)$ sum = 7

$(2,6)$ sum = 8

$(1,2)$ sum = 3

$(3,2)$ sum = 5

$(4,2)$ sum = 6

$(5,2)$ sum = 7

$(6,2)$ sum = 8

None of these sums (3, 4, 5, 6, 7, 8) are greater than 8. So, there are no outcomes common to A and B.

$A \cap B = \emptyset$

Therefore, events A and B are mutually exclusive.

Intersection of A and C ($A \cap C$):

We look for outcomes that are in both A and C. Outcomes in A have a sum greater than 8. Outcomes in C have a sum of 9 or 12.

If an outcome is in C, its sum is 9 or 12. Both 9 and 12 are greater than 8. So, every outcome in C is also in A.

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

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

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

Since $A \cap C \neq \emptyset$, events A and C are not mutually exclusive.

Intersection of B and C ($B \cap C$):

We look for outcomes that are in both B and C. Outcomes in B have a 2 on at least one die. Outcomes in C have a sum of 9 or 12.

Let's check the outcomes in B to see if any have a sum of 9 or 12:

$(2,1)$ sum = 3

$(2,2)$ sum = 4

$(2,3)$ sum = 5

$(2,4)$ sum = 6

$(2,5)$ sum = 7

$(2,6)$ sum = 8

$(1,2)$ sum = 3

$(3,2)$ sum = 5

$(4,2)$ sum = 6

$(5,2)$ sum = 7

$(6,2)$ sum = 8

None of the outcomes in B have a sum of 9 or 12. So, there are no outcomes common to B and C.

$B \cap C = \emptyset$

Therefore, events B and C are mutually exclusive.

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

Given:

Experiment: Three coins are tossed once.

Event A: ‘three heads show’

Event B: ‘two heads and one tail show’

Event C: ‘three tails show’

Event D: ‘a head shows on the first coin’

To Find:

(i) Mutually exclusive pairs of events.

(ii) Simple events.

(iii) Compound events.

Solution:

The sample space S for the experiment of tossing three coins once is the set of all possible outcomes:

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

Let's describe each event as a set of outcomes:

Event A: ‘three heads show’

$A = \{HHH\}$

Event B: ‘two heads and one tail show’

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

Event C: ‘three tails show’

$C = \{TTT\}$

Event D: ‘a head shows on the first coin’

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

(i) Mutually exclusive events:

Two events are mutually exclusive if their intersection is the empty set ($\emptyset$).

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

Events A and B are mutually exclusive.

$A \cap C = \{HHH\} \cap \{TTT\} = \emptyset$

Events A and C are mutually exclusive.

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

Since the intersection is not empty ($\{HHH\}$), events A and D are not mutually exclusive.

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

Events B and C are mutually exclusive.

$B \cap D = \{HHT, HTH, THH\} \cap \{HHH, HHT, HTH, HTT\} = \{HHT, HTH\}$

Since the intersection is not empty ($\{HHT, HTH\}$), events B and D are not mutually exclusive.

$C \cap D = \{TTT\} \cap \{HHH, HHT, HTH, HTT\} = \emptyset$

Events C and D are mutually exclusive.

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

(ii) Simple events:

A simple event is an event that contains only one outcome from the sample space.

Event A has one outcome: $\{HHH\}$. So, A is a simple event.

Event B has three outcomes: $\{HHT, HTH, THH\}$. So, B is not a simple event.

Event C has one outcome: $\{TTT\}$. So, C is a simple event.

Event D has four outcomes: $\{HHH, HHT, HTH, HTT\}$. So, D is not a simple event.

The simple events are A and C.

(iii) Compound events:

A compound event is an event that contains more than one outcome from the sample space.

Event A has one outcome. So, A is not a compound event.

Event B has three outcomes. So, B is a compound event.

Event C has one outcome. So, C is not a compound event.

Event D has four outcomes. So, D is a compound event.

The compound events are B and D.

Given:

Experiment: Three coins are tossed once.

To Describe:

Events based on the given criteria.

Solution:

The sample space S for the experiment of tossing three coins once is the set of all possible outcomes:

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

(i) Two events which are mutually exclusive:

Two events are mutually exclusive if they cannot occur at the same time, meaning their intersection is the empty set ($\emptyset$).

Consider the events:

Event 1: “Getting at least two heads”

This event includes outcomes with 2 or 3 heads: $\{HHT, HTH, THH, HHH\}$.

Event 2: “Getting at least two tails”

This event includes outcomes with 2 or 3 tails: $\{HTT, THT, TTH, TTT\}$.

Let's find the intersection of these two events:

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

Since the intersection is the empty set, these two events are mutually exclusive.

So, two mutually exclusive events are “Getting at least two heads” and “Getting at least two tails”.

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

Three events are mutually exclusive if the intersection of any pair is empty. They are exhaustive if their union equals the sample space S.

Consider the events:

Event 1: “Getting no heads”

This event is $\{TTT\}$.

Event 2: “Getting exactly one head”

This event is $\{HTT, THT, TTH\}$.

Event 3: “Getting at least two heads”

This event includes outcomes with 2 or 3 heads: $\{HHT, HTH, THH, HHH\}$.

Check for mutual exclusivity (pairwise):

Event 1 $\cap$ Event 2 = $\{TTT\} \cap \{HTT, THT, TTH\} = \emptyset$

Event 1 $\cap$ Event 3 = $\{TTT\} \cap \{HHT, HTH, THH, HHH\} = \emptyset$

Event 2 $\cap$ Event 3 = $\{HTT, THT, TTH\} \cap \{HHT, HTH, THH, HHH\} = \emptyset$

The pairwise intersections are empty, so the events are mutually exclusive.

Check for exhaustiveness:

Event 1 $\cup$ Event 2 $\cup$ Event 3 = $\{TTT\} \cup \{HTT, THT, TTH\} \cup \{HHT, HTH, THH, HHH\}$

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

= $S$ (The sample space)

Their union is the sample space S, so they are exhaustive.

So, three mutually exclusive and exhaustive events are “Getting no heads”, “Getting exactly one head”, and “Getting at least two heads”.

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

Two events are not mutually exclusive if their intersection is not the empty set.

Consider the events:

Event 1: “Getting at most two tails”

This event includes outcomes with 0, 1, or 2 tails: $\{HHH, HHT, HTH, THH, HTT, THT, TTH\}$.

Event 2: “Getting exactly two tails”

This event is $\{HTT, THT, TTH\}$.

Let's find the intersection of these two events:

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

Since the intersection is $\{HTT, THT, TTH\}$, which is not the empty set, these two events are not mutually exclusive.

So, two events which are not mutually exclusive are “Getting at most two tails” and “Getting exactly two tails”.

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

Mutually exclusive: Intersection is empty.

Not exhaustive: Union does not equal the sample space S.

Consider the events:

Event 1: “Getting exactly one head”

This event is $\{HTT, THT, TTH\}$.

Event 2: “Getting exactly two heads”

This event is $\{HHT, HTH, THH\}$.

Check for mutual exclusivity:

Event 1 $\cap$ Event 2 = $\{HTT, THT, TTH\} \cap \{HHT, HTH, THH\} = \emptyset$

The intersection is empty, so they are mutually exclusive.

Check for exhaustiveness:

Event 1 $\cup$ Event 2 = $\{HTT, THT, TTH\} \cup \{HHT, HTH, THH\}$

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

This union is not equal to the sample space S (missing outcomes like HHH and TTT), so they are not exhaustive.

So, two events which are mutually exclusive but not exhaustive are “Getting exactly one head” and “Getting exactly two heads”.

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

Mutually exclusive: Pairwise intersections are empty.

Not exhaustive: Union does not equal the sample space S.

Consider the events:

Event 1: “Getting exactly one tail”

This event is $\{HHT, HTH, THH\}$.

Event 2: “Getting exactly two tails”

This event is $\{HTT, THT, TTH\}$.

Event 3: “Getting exactly three tails”

This event is $\{TTT\}$.

Check for mutual exclusivity (pairwise):

Event 1 $\cap$ Event 2 = $\{HHT, HTH, THH\} \cap \{HTT, THT, TTH\} = \emptyset$

Event 1 $\cap$ Event 3 = $\{HHT, HTH, THH\} \cap \{TTT\} = \emptyset$

Event 2 $\cap$ Event 3 = $\{HTT, THT, TTH\} \cap \{TTT\} = \emptyset$

The pairwise intersections are empty, so the events are mutually exclusive.

Check for exhaustiveness:

Event 1 $\cup$ Event 2 $\cup$ Event 3 = $\{HHT, HTH, THH\} \cup \{HTT, THT, TTH\} \cup \{TTT\}$

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

This union is not equal to the sample space S (missing the outcome HHH), so they are not exhaustive.

So, three events which are mutually exclusive but not exhaustive are “Getting exactly one tail”, “Getting exactly two tails”, and “Getting exactly three tails”.

Given:

Experiment: Throwing two dice.

Event A: 'getting an even number on the first die'

Event B: 'getting an odd number on the first die'

Event C: 'getting the sum of the numbers on the dice $\leq 5$'

To Describe:

The sets for the events $A′, not B, A \cup B, A \cap B, A – C, B \cup C, B \cap C, A \cap B′ \cap C′$.

Solution:

The sample space S for the experiment of rolling two dice consists of all 36 possible ordered pairs $(i, j)$, where $i$ is the result on the first die and $j$ is the result on the second die ($i, j \in \{1, 2, 3, 4, 5, 6\}$).

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

Let's write the sets corresponding to the given events A, B, and C:

Event A: 'getting an even number on the first die'

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

Event B: 'getting an odd number on the first die'

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

Event C: 'getting the sum of the numbers on the dice $\leq 5$'

Possible sums are 2, 3, 4, or 5.

Sum 2: (1,1)

Sum 3: (1,2), (2,1)

Sum 4: (1,3), (2,2), (3,1)

Sum 5: (1,4), (2,3), (3,2), (4,1)

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

Now, let's describe the required events:

(i) A′ (Complement of A)

A′ is the event 'not getting an even number on the first die', which means 'getting an odd number on the first die'. This is exactly event B.

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

(ii) not B (Complement of B, $B′$)

not B is the event 'not getting an odd number on the first die', which means 'getting an even number on the first die'. This is exactly event A.

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

(iii) A or B ($A \cup B$)

A or B is the event that the first die is even or the first die is odd. Since the first die must be either even or odd, this event covers all possible outcomes in the sample space.

$A \cup B = A \cup B = S = \{(1,1), (1,2), ..., (6,6)\}$

(iv) A and B ($A \cap B$)

A and B is the event that the first die is both even and odd. This is impossible.

$A \cap B = A \cap B = \emptyset$

(v) A but not C ($A - C$)

A but not C is the event that the first die is even and the sum of the numbers on the dice is not $\leq 5$ (i.e., the sum is $> 5$).

We take the outcomes in A and remove those that are also in C.

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

(vi) B or C ($B \cup C$)

B or C is the event that the first die is odd or the sum of the numbers on the dice is $\leq 5$. This is the union of the sets B and C.

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

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

(vii) B and C ($B \cap C$)

B and C is the event that the first die is odd and the sum of the numbers on the dice is $\leq 5$. This is the intersection of the sets B and C.

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

The outcomes common to both sets are those in C where the first die is odd.

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

(viii) A ∩ B′ ∩ C′

We know that $B′$ is the event 'not B', which is the same as event A.

So, $A \cap B′ \cap C′ = A \cap A \cap C′ = A \cap C′$.

$C′$ is the complement of C, meaning the sum of the numbers is not $\leq 5$, so the sum is $> 5$.

The event $A \cap C′$ is the event where the first die is even (event A) AND the sum of the numbers on the dice is greater than 5 (event C').

Outcomes in A where the first die is 2 and the sum is $> 5$: (2,4), (2,5), (2,6)

Outcomes in A where the first die is 4 and the sum is $> 5$: (4,2), (4,3), (4,4), (4,5), (4,6)

Outcomes in A where the first die is 6 and the sum is $> 5$: (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

Combining these, the set of outcomes for $A \cap B′ \cap C′$ is:

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

This is the same set of outcomes as $A \setminus C$, which we found in part (v).

Given:

Experiment: Throwing two dice.

Sample Space $S = \{(i, j) | i, j \in \{1, 2, 3, 4, 5, 6\}\}$

Event A: 'getting an even number on the first die'.

$A = \{(2,j), (4,j), (6,j) \text{ for } j=1..6\}$

Event B: 'getting an odd number on the first die'.

$B = \{(1,j), (3,j), (5,j) \text{ for } j=1..6\}$

Event C: 'getting the sum of the numbers on the dice $\leq 5$'.

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

(i) A and B are mutually exclusive

Two events are mutually exclusive if their intersection is empty.

$A \cap B$ contains outcomes where the first die is both even and odd, which is impossible.

$A \cap B = \emptyset$

Reason: Events A and B have no outcomes in common as the first die cannot be both even and odd simultaneously.

Statement is True.

(ii) A and B are mutually exclusive and exhaustive

From (i), A and B are mutually exclusive.

Events are exhaustive if their union is the sample space.

$A \cup B$ contains outcomes where the first die is even or the first die is odd. This covers all possible outcomes in the sample space S.

$A \cup B = S$

Reason: A and B cover all possible outcomes based on the first die's result (even or odd), and they have no outcomes in common.

Statement is True.

(iii) A = B'

B' is the complement of B, which means 'not getting an odd number on the first die'. This is equivalent to 'getting an even number on the first die', which is the definition of event A.

$B' = S - B = A$

Reason: The complement of getting an odd number on the first die is getting an even number on the first die, which is event A.

Statement is True.

(iv) A and C are mutually exclusive

Two events are mutually exclusive if their intersection is empty.

$A \cap C$ contains outcomes that are in both A (first die is even) and C (sum $\leq 5$).

Outcomes in C where the first die is even are: $(2,1), (2,2), (2,3), (4,1)$.

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

Reason: The intersection of A and C is not empty. For example, the outcome (2,1) is in both A (first die is 2, which is even) and C (sum is $2+1=3 \leq 5$).

Statement is False.

(v) A and B′ are mutually exclusive.

From (iii), we know that $B′ = A$. The statement is equivalent to asking if A and A are mutually exclusive.

$A \cap B′ = A \cap A = A$

Since event A is not empty (it contains 18 outcomes), $A \cap B′ \neq \emptyset$.

Reason: Since $B′=A$, the statement is asking if A is mutually exclusive with itself. This is only true if A is the empty set, which it is not here.

Statement is False.

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

We know that $A′ = B$ and $B′ = A$. The set of events is $\{B, A, C\}$.

For events to be mutually exclusive, the pairwise intersection of all events must be empty.

$A \cap C = \{(2,1), (2,2), (2,3), (4,1)\} \neq \emptyset$ (as shown in iv)

$B \cap C = \{(1,1), (1,2), (1,3), (1,4), (3,1), (3,2)\} \neq \emptyset$ (as shown in Q6 vii)

Reason: The set of events $\{A′, B′, C\}$ which is $\{B, A, C\}$ is not mutually exclusive because the intersection of A and C is not empty, and the intersection of B and C is not empty.

Statement is False.

Given:

Sample space $S = \{\omega_1, \omega_2, \omega_3, \omega_4, \omega_5, \omega_6\}$.

Several assignments of probabilities $P(\omega_i)$ to each outcome $\omega_i$.

To Check:

Which of the given assignments of probabilities are valid.

Solution:

For an assignment of probabilities to the outcomes of a sample space to be valid, two conditions must be satisfied:

1. The probability of each outcome must be a non-negative real number between 0 and 1, inclusive. That is, $0 \leq P(\omega_i) \leq 1$ for all $\omega_i \in S$.

2. The sum of the probabilities of all outcomes in the sample space must be equal to 1. That is, $\sum_{i=1}^{6} P(\omega_i) = P(\omega_1) + P(\omega_2) + P(\omega_3) + P(\omega_4) + P(\omega_5) + P(\omega_6) = 1$.

Let's examine each assignment:

(a) $P(\omega_1) = \frac{1}{6}, P(\omega_2) = \frac{1}{6}, P(\omega_3) = \frac{1}{6}, P(\omega_4) = \frac{1}{6}, P(\omega_5) = \frac{1}{6}, P(\omega_6) = \frac{1}{6}$

Check 1: All $P(\omega_i) = \frac{1}{6}$. Since $0 \leq \frac{1}{6} \leq 1$, this condition is satisfied for all outcomes.

Check 2: Sum of probabilities = $P(\omega_1) + P(\omega_2) + P(\omega_3) + P(\omega_4) + P(\omega_5) + P(\omega_6) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = 6 \times \frac{1}{6} = 1$.

Since both conditions are satisfied, this assignment is valid.

(b) $P(\omega_1) = 1, P(\omega_2) = 0, P(\omega_3) = 0, P(\omega_4) = 0, P(\omega_5) = 0, P(\omega_6) = 0$

Check 1: All $P(\omega_i)$ are either 0 or 1. Since $0 \leq P(\omega_i) \leq 1$ for all outcomes, this condition is satisfied.

Check 2: Sum of probabilities = $P(\omega_1) + P(\omega_2) + P(\omega_3) + P(\omega_4) + P(\omega_5) + P(\omega_6) = 1 + 0 + 0 + 0 + 0 + 0 = 1$.

Since both conditions are satisfied, this assignment is valid.

(c) $P(\omega_1) = \frac{1}{8}, P(\omega_2) = \frac{2}{3}, P(\omega_3) = \frac{1}{3}, P(\omega_4) = \frac{1}{3}, P(\omega_5) = -\frac{1}{4}, P(\omega_6) = -\frac{1}{3}$

Check 1: We have $P(\omega_5) = -\frac{1}{4}$ and $P(\omega_6) = -\frac{1}{3}$. These probabilities are negative.

The condition $0 \leq P(\omega_i) \leq 1$ is not satisfied because probabilities must be non-negative.

Since the first condition is not satisfied, this assignment is not valid.

(d) $P(\omega_1) = \frac{1}{12}, P(\omega_2) = \frac{1}{12}, P(\omega_3) = \frac{1}{6}, P(\omega_4) = \frac{1}{6}, P(\omega_5) = \frac{1}{6}, P(\omega_6) = \frac{3}{2}$

Check 1: We have $P(\omega_6) = \frac{3}{2} = 1.5$. This probability is greater than 1.

The condition $0 \leq P(\omega_i) \leq 1$ is not satisfied.

Since the first condition is not satisfied, this assignment is not valid.

(e) $P(\omega_1) = 0.1, P(\omega_2) = 0.2, P(\omega_3) = 0.3, P(\omega_4) = 0.4, P(\omega_5) = 0.5, P(\omega_6) = 0.6$

Check 1: All $P(\omega_i)$ are between 0 and 1. Since $0 \leq P(\omega_i) \leq 1$ for all outcomes, this condition is satisfied.

Check 2: Sum of probabilities = $P(\omega_1) + P(\omega_2) + P(\omega_3) + P(\omega_4) + P(\omega_5) + P(\omega_6) = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 = 2.1$.

The sum of probabilities is 2.1, which is not equal to 1.

Since the second condition is not satisfied, this assignment is not valid.

Based on the checks, the valid assignments of probabilities are (a) and (b).

Given:

Experiment: Drawing one card from a well-shuffled deck of 52 cards.

Each outcome is equally likely.

To Calculate:

The probabilities of the events: (i) a diamond, (ii) not an ace, (iii) a black card, (iv) not a diamond, (v) not a black card.

Solution:

The total number of possible outcomes in the sample space S is the total number of cards in the deck, which is 52.

Since each outcome is equally likely, the probability of any single outcome is $\frac{1}{52}$.

The probability of an event E is given by:

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

$P(E) = \frac{n(E)}{n(S)}$

Here, $n(S) = 52$.

(i) Probability of drawing a diamond

Let A be the event that the card is a diamond.

There are 13 diamond cards in a deck (Ace, 2, 3, ..., 10, Jack, Queen, King of Diamonds).

$n(A) = 13$

The probability of event A is:

$P(A) = \frac{n(A)}{n(S)} = \frac{13}{52} = \frac{1}{4}$

(ii) Probability of not drawing an ace

Let B be the event that the card is an ace.

There are 4 ace cards in a deck (Ace of Hearts, Diamonds, Clubs, Spades).

$n(B) = 4$

The probability of drawing an ace is $P(B) = \frac{n(B)}{n(S)} = \frac{4}{52} = \frac{1}{13}$.

The event 'not an ace' is the complement of event B, denoted by $B′$.

The probability of the complement of an event is given by $P(B′) = 1 - P(B)$.

$P(\text{not an ace}) = P(B′) = 1 - \frac{1}{13} = \frac{13 - 1}{13} = \frac{12}{13}$

Alternate approach: The number of cards that are not aces is the total number of cards minus the number of aces: $52 - 4 = 48$.

Let $B′$ be the event 'not an ace'. $n(B′) = 48$.

$P(B′) = \frac{n(B′)}{n(S)} = \frac{48}{52} = \frac{\cancel{48}^{12}}{\cancel{52}_{13}} = \frac{12}{13}$.

(iii) Probability of drawing a black card (a club or, a spade)

Let C be the event that the card is a black card.

There are 13 club cards and 13 spade cards. Clubs and spades are black.

Number of black cards = Number of clubs + Number of spades = $13 + 13 = 26$.

$n(C) = 26$

The probability of event C is:

$P(C) = \frac{n(C)}{n(S)} = \frac{26}{52} = \frac{1}{2}$

(iv) Probability of not drawing a diamond

Let A be the event that the card is a diamond (from part i). The event 'not a diamond' is the complement of event A, denoted by $A′$.

$P(\text{not a diamond}) = P(A′) = 1 - P(A) = 1 - \frac{1}{4} = \frac{4 - 1}{4} = \frac{3}{4}$

Alternate approach: The number of cards that are not diamonds is the total number of cards minus the number of diamonds: $52 - 13 = 39$.

Let $A′$ be the event 'not a diamond'. $n(A′) = 39$.

$P(A′) = \frac{n(A′)}{n(S)} = \frac{39}{52} = \frac{\cancel{39}^{3}}{\cancel{52}_{4}} = \frac{3}{4}$.

(v) Probability of not drawing a black card

Let C be the event that the card is a black card (from part iii). The event 'not a black card' is the complement of event C, denoted by $C′$.

$P(\text{not a black card}) = P(C′) = 1 - P(C) = 1 - \frac{1}{2} = \frac{1}{2}$

Alternate approach: The cards that are not black are the red cards (hearts and diamonds).

Number of red cards = Number of hearts + Number of diamonds = $13 + 13 = 26$.

Let $C′$ be the event 'not a black card'. $n(C′) = 26$.

$P(C′) = \frac{n(C′)}{n(S)} = \frac{26}{52} = \frac{1}{2}$.

Given:

A bag contains 9 discs.

Number of red discs = 4

Number of blue discs = 3

Number of yellow discs = 2

Total number of discs = $4 + 3 + 2 = 9$.

A disc is drawn at random.

To Calculate:

The probabilities of drawing a red, yellow, blue, not blue, either red or blue disc.

Solution:

The total number of possible outcomes in the sample space S is the total number of discs in the bag, which is 9.

$n(S) = 9$

Since the discs are similar in shape and size and a disc is drawn at random, each outcome (drawing a specific disc) is equally likely.

The probability of an event E is given by:

$P(E) = \frac{\text{Number of outcomes favourable to E}}{\text{Total number of possible outcomes}} = \frac{n(E)}{n(S)}$

(i) Probability of drawing a red disc

Let R be the event that the disc drawn is red.

The number of red discs is 4.

$n(R) = 4$

The probability of event R is:

$P(R) = \frac{n(R)}{n(S)} = \frac{4}{9}$

(ii) Probability of drawing a yellow disc

Let Y be the event that the disc drawn is yellow.

The number of yellow discs is 2.

$n(Y) = 2$

The probability of event Y is:

$P(Y) = \frac{n(Y)}{n(S)} = \frac{2}{9}$

(iii) Probability of drawing a blue disc

Let B be the event that the disc drawn is blue.

The number of blue discs is 3.

$n(B) = 3$

The probability of event B is:

$P(B) = \frac{n(B)}{n(S)} = \frac{3}{9} = \frac{1}{3}$

(iv) Probability of not drawing a blue disc

Let B be the event that the disc is blue (from part iii). The event 'not blue' is the complement of event B, denoted by $B′$.

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

Alternate approach: The number of discs that are not blue is the total number of discs minus the number of blue discs: $9 - 3 = 6$. These are the red and yellow discs.

Number of not blue discs = Number of red discs + Number of yellow discs = $4 + 2 = 6$.

Let $B′$ be the event 'not blue'. $n(B′) = 6$.

$P(B′) = \frac{n(B′)}{n(S)} = \frac{6}{9} = \frac{\cancel{6}^{2}}{\cancel{9}_{3}} = \frac{2}{3}$.

(v) Probability of drawing either red or blue disc

Let R be the event 'red' and B be the event 'blue'. The event 'either red or blue' is the union of events R and B, $R \cup B$.

Since a single disc is drawn, it cannot be both red and blue at the same time. Thus, events R and B are mutually exclusive ($R \cap B = \emptyset$).

For mutually exclusive events, the probability of their union is the sum of their individual probabilities:

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

From parts (i) and (iii): $P(R) = \frac{4}{9}$ and $P(B) = \frac{3}{9}$.

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

Alternate approach: The number of discs that are either red or blue is the sum of the number of red discs and the number of blue discs.

Number of red or blue discs = Number of red discs + Number of blue discs = $4 + 3 = 7$.

Let $E$ be the event 'either red or blue'. $n(E) = 7$.

$P(E) = \frac{n(E)}{n(S)} = \frac{7}{9}$.

Given:

Let A be the event that Anil qualifies the examination.

Let B be the event that Ashima qualifies the examination.

Probability that Anil qualifies, $P(A) = 0.05$

Probability that Ashima qualifies, $P(B) = 0.10$

Probability that both qualify, $P(A \cap B) = 0.02$

To Find:

(a) Probability that both Anil and Ashima will not qualify the examination, $P(A' \cap B')$

(b) Probability that at least one of them will not qualify the examination, $P(A' \cup B')$

(c) Probability that only one of them will qualify the examination, $P((A \cap B') \cup (A' \cap B))$

Solution:

First, let's find the probability that at least one of them qualifies the examination using the formula for the union of two events:

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

Substitute the given values:

$P(A \cup B) = 0.05 + 0.10 - 0.02$

$P(A \cup B) = 0.15 - 0.02$

$P(A \cup B) = 0.13$

(a) Probability that both Anil and Ashima will not qualify the examination.

This is represented by the event $A' \cap B'$.

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

The probability of the complement of an event is 1 minus the probability of the event.

Therefore, $P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$

Substitute the calculated value of $P(A \cup B)$:

$P(A' \cap B') = 1 - 0.13$

$P(A' \cap B') = 0.87$

(b) Probability that at least one of them will not qualify the examination.

This is represented by the event $A' \cup B'$.

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

Therefore, $P(A' \cup B') = P((A \cap B)') = 1 - P(A \cap B)$

Substitute the given value of $P(A \cap B)$:

$P(A' \cup B') = 1 - 0.02$

$P(A' \cup B') = 0.98$

(c) Probability that only one of them will qualify the examination.

This means either Anil qualifies and Ashima does not (event $A \cap B'$) OR Anil does not qualify and Ashima does (event $A' \cap B$).

These two events ($A \cap B'$ and $A' \cap B$) are mutually exclusive because they cannot happen at the same time.

Therefore, the probability that only one qualifies is the sum of their individual probabilities:

$P(\text{Only one qualifies}) = P(A \cap B') + P(A' \cap B)$

We know that the probability of an event can be expressed as the sum of the probabilities of its intersections with another event and its complement:

$P(A) = P(A \cap B) + P(A \cap B')$

Rearranging to find $P(A \cap B')$:

$P(A \cap B') = P(A) - P(A \cap B)$

$P(A \cap B') = 0.05 - 0.02 = 0.03$

Similarly, for event B:

$P(B) = P(B \cap A) + P(B \cap A')$ which is the same as $P(B) = P(A \cap B) + P(A' \cap B)$

Rearranging to find $P(A' \cap B)$:

$P(A' \cap B) = P(B) - P(A \cap B)$

$P(A' \cap B) = 0.10 - 0.02 = 0.08$

Now, sum these probabilities to find the probability that only one qualifies:

$P(\text{Only one qualifies}) = P(A \cap B') + P(A' \cap B) = 0.03 + 0.08$

$P(\text{Only one qualifies}) = 0.11$

Given:

Number of men = 2

Number of women = 2

Total number of people = $2 + 2 = 4$

Committee size = 2 persons

To Find:

(a) Probability that the committee has no man.

(b) Probability that the committee has one man.

(c) Probability that the committee has two men.

Solution:

The total number of ways to select a committee of 2 persons from 4 people is given by the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$, where $n$ is the total number of items and $r$ is the number of items to choose.

Total number of possible committees = $\binom{4}{2}$

$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{24}{4} = 6$

So, the total number of outcomes is 6.

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

This means the committee consists of 0 men and 2 women.

Number of ways to select 0 men from 2 = $\binom{2}{0} = 1$

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

Number of ways to select a committee with no man = $\binom{2}{0} \times \binom{2}{2} = 1 \times 1 = 1$

Probability (no man) = $\frac{\text{Number of committees with no man}}{\text{Total number of committees}} = \frac{1}{6}$

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

This means the committee consists of 1 man and 1 woman.

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

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

Number of ways to select a committee with one man = $\binom{2}{1} \times \binom{2}{1} = 2 \times 2 = 4$

Probability (one man) = $\frac{\text{Number of committees with one man}}{\text{Total number of committees}} = \frac{4}{6} = \frac{2}{3}$

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

This means the committee consists of 2 men and 0 women.

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

Number of ways to select 0 women from 2 = $\binom{2}{0} = 1$

Number of ways to select a committee with two men = $\binom{2}{2} \times \binom{2}{0} = 1 \times 1 = 1$

Probability (two men) = $\frac{\text{Number of committees with two men}}{\text{Total number of committees}} = \frac{1}{6}$

Given:

Sample Space S = $\{\omega_1, \omega_2, \omega_3, \omega_4, \omega_5, \omega_6, \omega_7\}$

Conditions for a Valid Probability Assignment:

For any probability assignment $P(\omega_i)$ for each outcome $\omega_i$ in the sample space S, the following conditions must be met:

1. The probability of each outcome must be non-negative and less than or equal to 1: $0 \leq P(\omega_i) \leq 1$ for all $\omega_i \in S$.

2. The sum of the probabilities of all outcomes in the sample space must be exactly 1: $\sum_{i=1}^{7} P(\omega_i) = 1$.

Check Assignments:

We will check each assignment against the two conditions.

(a) Assignment:

Probabilities are: $P(\omega_1) = 0.1, P(\omega_2) = 0.01, P(\omega_3) = 0.05, P(\omega_4) = 0.03, P(\omega_5) = 0.01, P(\omega_6) = 0.2, P(\omega_7) = 0.6$

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

Condition 2: Sum of probabilities = $0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1.0$

The sum is 1. This condition is satisfied.

Conclusion: Assignment (a) is valid.

(b) Assignment:

Probabilities are: $P(\omega_1) = \frac{1}{7}, P(\omega_2) = \frac{1}{7}, P(\omega_3) = \frac{1}{7}, P(\omega_4) = \frac{1}{7}, P(\omega_5) = \frac{1}{7}, P(\omega_6) = \frac{1}{7}, P(\omega_7) = \frac{1}{7}$

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

Condition 2: Sum of probabilities = $\sum_{i=1}^{7} \frac{1}{7} = 7 \times \frac{1}{7} = 1$

The sum is 1. This condition is satisfied.

Conclusion: Assignment (b) is valid.

(c) Assignment:

Probabilities are: $P(\omega_1) = 0.1, P(\omega_2) = 0.2, P(\omega_3) = 0.3, P(\omega_4) = 0.4, P(\omega_5) = 0.5, P(\omega_6) = 0.6, P(\omega_7) = 0.7$

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

Condition 2: Sum of probabilities = $0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8$

The sum is 2.8, which is not equal to 1. This condition is not satisfied.

Conclusion: Assignment (c) is not valid.

(d) Assignment:

Probabilities are: $P(\omega_1) = -0.1, P(\omega_2) = 0.2, P(\omega_3) = 0.3, P(\omega_4) = 0.4, P(\omega_5) = -0.2, P(\omega_6) = 0.1, P(\omega_7) = 0.3$

Condition 1: Probabilities must be non-negative. The probabilities $P(\omega_1) = -0.1$ and $P(\omega_5) = -0.2$ are negative. This condition is not satisfied.

Conclusion: Assignment (d) is not valid.

(e) Assignment:

Probabilities are: $P(\omega_1) = \frac{1}{14}, P(\omega_2) = \frac{2}{14}, P(\omega_3) = \frac{3}{14}, P(\omega_4) = \frac{4}{14}, P(\omega_5) = \frac{5}{14}, P(\omega_6) = \frac{6}{14}, P(\omega_7) = \frac{15}{14}$

Condition 1: Probabilities must be less than or equal to 1. The probability $P(\omega_7) = \frac{15}{14} = 1.07...$ which is greater than 1. This condition is not satisfied.

Conclusion: Assignment (e) is not valid.

Based on the checks, the assignments that cannot be valid assignments of probabilities for the outcomes of sample space S are those that fail either Condition 1 or Condition 2.

The assignments that are not valid are:

(c), (d), and (e)

Given:

A coin is tossed twice.

To Find:

The probability that at least one tail occurs.

Solution:

When a coin is tossed twice, the sample space S, which is the set of all possible outcomes, is:

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

The total number of possible outcomes is $|S| = 4$.

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

The outcomes included in the event E are those with one tail or two tails:

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

The number of favorable outcomes is $|E| = 3$.

Assuming the coin is fair and the tosses are independent, each outcome in the sample space has a probability of $\frac{1}{4}$.

The probability of the event E is given by:

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

$P(\text{at least one tail}) = \frac{3}{4}$

Alternate Solution using Complement:

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

The complement event E' is that no tail occurs. This means both tosses are heads (HH).

The outcome for the event E' is: $\{HH\}$

The number of outcomes for E' is $|E'| = 1$.

The probability of E' is:

$P(E') = \frac{\text{Number of outcomes with no tail}}{\text{Total number of outcomes}} = \frac{1}{4}$

The probability of the event E (at least one tail) is $P(E) = 1 - P(E')$.

$P(\text{at least one tail}) = 1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4}$

Given:

A fair die is thrown once.

The sample space S for this experiment is the set of possible outcomes when a die is thrown: $S = \{1, 2, 3, 4, 5, 6\}$.

The total number of possible outcomes is $|S| = 6$.

To Find:

The probability of the following events:

(i) A prime number will appear.

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

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

(iv) A number more than 6 will appear.

(v) A number less than 6 will appear.

Solution:

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

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

(i) A prime number will appear.

Let E1 be the event that a prime number appears.

The prime numbers in the sample space S are $\{2, 3, 5\}$.

The favorable outcomes for E1 are $\{2, 3, 5\}$.

The number of favorable outcomes is $|E1| = 3$.

The probability of E1 is:

$P(E1) = \frac{|E1|}{|S|} = \frac{3}{6} = \frac{1}{2}$

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

Let E2 be the event that a number greater than or equal to 3 appears.

The numbers in S that are greater than or equal to 3 are $\{3, 4, 5, 6\}$.

The favorable outcomes for E2 are $\{3, 4, 5, 6\}$.

The number of favorable outcomes is $|E2| = 4$.

The probability of E2 is:

$P(E2) = \frac{|E2|}{|S|} = \frac{4}{6} = \frac{2}{3}$

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

Let E3 be the event that a number less than or equal to one appears.

The numbers in S that are less than or equal to 1 are $\{1\}$.

The favorable outcomes for E3 are $\{1\}$.

The number of favorable outcomes is $|E3| = 1$.

The probability of E3 is:

$P(E3) = \frac{|E3|}{|S|} = \frac{1}{6}$

(iv) A number more than 6 will appear.

Let E4 be the event that a number more than 6 appears.

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

The favorable outcomes for E4 are $\emptyset$ (the empty set).

The number of favorable outcomes is $|E4| = 0$.

The probability of E4 is:

$P(E4) = \frac{|E4|}{|S|} = \frac{0}{6} = 0$

(v) A number less than 6 will appear.

Let E5 be the event that a number less than 6 appears.

The numbers in S that are less than 6 are $\{1, 2, 3, 4, 5\}$.

The favorable outcomes for E5 are $\{1, 2, 3, 4, 5\}$.

The number of favorable outcomes is $|E5| = 5$.

The probability of E5 is:

$P(E5) = \frac{|E5|}{|S|} = \frac{5}{6}$

Given:

A standard deck of 52 playing cards.

One card is selected from the deck.

To Find:

(a) Number of points in the sample space.

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

(c) Probability that the card is (i) an ace (ii) black card.

Solution:

The sample space S is the set of all possible outcomes when selecting one card from a standard deck of 52 cards.

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

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

Total number of cards in a standard deck = 52.

So, the number of points in the sample space is 52.

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

Let E1 be the event that the selected card is an ace of spades.

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

Number of favorable outcomes for E1 = 1.

Total number of outcomes = 52.

The probability of E1 is:

$P(E1) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{52}$

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

(i) Probability that the card is an ace.

Let E2 be the event that the selected card is an ace.

There are four aces in a standard deck: Ace of Spades, Ace of Hearts, Ace of Diamonds, and Ace of Clubs.

Number of favorable outcomes for E2 = 4.

Total number of outcomes = 52.

The probability of E2 is:

$P(E2) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{52} = \frac{1}{13}$

(ii) Probability that the card is a black card.

Let E3 be the event that the selected card is a black card.

There are two black suits in a standard deck: Spades and Clubs.

Each suit has 13 cards.

Number of black cards = Number of cards in Spades + Number of cards in Clubs = $13 + 13 = 26$.

Number of favorable outcomes for E3 = 26.

Total number of outcomes = 52.

The probability of E3 is:

$P(E3) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{26}{52} = \frac{1}{2}$

Given:

A fair coin with faces marked 1 and 6.

A fair standard die with faces marked 1, 2, 3, 4, 5, 6.

Both the coin and the die are tossed simultaneously.

To Find:

The probability that the sum of the numbers that turn up is:

(i) 3

(ii) 12

Solution:

Let the outcome of the coin toss be C and the outcome of the die toss be D.

The possible outcomes for the coin are $\{1, 6\}$.

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

The sample space S for the combined experiment consists of pairs (C, D), where C is the result of the coin and D is the result of the die.

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

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

The total number of possible outcomes is $|S| = 2 \times 6 = 12$.

Since both the coin and the die are fair, each outcome in the sample space is equally likely, with a probability of $\frac{1}{12}$.

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

Let E1 be the event that the sum of the numbers is 3.

We need to find the pairs (C, D) from the sample space such that C + D = 3.

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

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

So, the only favorable outcome for E1 is (1, 2).

The number of favorable outcomes is $|E1| = 1$.

The probability of E1 is:

$P(E1) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{|E1|}{|S|} = \frac{1}{12}$

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

Let E2 be the event that the sum of the numbers is 12.

We need to find the pairs (C, D) from the sample space such that C + D = 12.

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

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

So, the only favorable outcome for E2 is (6, 6).

The number of favorable outcomes is $|E2| = 1$.

The probability of E2 is:

$P(E2) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{|E2|}{|S|} = \frac{1}{12}$

Given:

Number of men on the city council = 4

Number of women on the city council = 6

To Find:

The probability that a randomly selected council member is a woman.

Solution:

The total number of people on the city council is the sum of the number of men and the number of women.

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

When one council member is selected at random, the total number of possible outcomes is the total number of council members, which is 10.

The event we are interested in is selecting a woman.

The number of favorable outcomes for this event is the number of women on the council, which is 6.

The probability of selecting a woman is given by:

Probability (Woman) = $\frac{\text{Number of women}}{\text{Total number of council members}}$

Probability (Woman) = $\frac{6}{10}$

Simplifying the fraction:

Probability (Woman) = $\frac{3}{5}$

The probability that the selected council member is a woman is $\frac{3}{5}$, or 0.6.

Given:

A fair coin is tossed four times.

Win $\textsf{₹}$ 1 for each head.

Lose $\textsf{₹}$ 1.50 for each tail.

To Find:

The number of different amounts of money possible after four tosses and the probability of having each of these amounts.

Solution:

When a fair coin is tossed four times, the total number of possible outcomes is $2^4 = 16$. The sample space S consists of all possible sequences of 4 heads (H) and tails (T).

The amount of money obtained depends on the number of heads and tails in the four tosses.

Let $n_H$ be the number of heads and $n_T$ be the number of tails in 4 tosses. Then $n_H + n_T = 4$.

The amount of money earned is $n_H \times \textsf{₹} 1$ and the amount lost is $n_T \times \textsf{₹} 1.50$.

The net amount of money M is given by:

$M = n_H \times 1 + n_T \times (-1.50)$

Since $n_T = 4 - n_H$, we can write M in terms of $n_H$:

$M = n_H - 1.5(4 - n_H)$

$M = n_H - 6 + 1.5 n_H$

$M = 2.5 n_H - 6$

The possible values for the number of heads ($n_H$) in 4 tosses are 0, 1, 2, 3, or 4.

Let's calculate the amount of money M for each possible number of heads:

If $n_H = 4$, then $n_T = 0$. $M = 2.5(4) - 6 = 10 - 6 = \textsf{₹} 4$. (Outcome: HHHH)

If $n_H = 3$, then $n_T = 1$. $M = 2.5(3) - 6 = 7.5 - 6 = \textsf{₹} 1.50$. (e.g., HHHT)

If $n_H = 2$, then $n_T = 2$. $M = 2.5(2) - 6 = 5 - 6 = \textsf{₹} -1$. (e.g., HHTT)

If $n_H = 1$, then $n_T = 3$. $M = 2.5(1) - 6 = 2.5 - 6 = \textsf{₹} -3.50$. (e.g., HTTT)

If $n_H = 0$, then $n_T = 4$. $M = 2.5(0) - 6 = 0 - 6 = \textsf{₹} -6$. (Outcome: TTTT)

The different amounts of money possible are $\textsf{₹} 4$, $\textsf{₹} 1.50$, $\textsf{₹} -1$, $\textsf{₹} -3.50$, and $\textsf{₹} -6$. There are 5 different possible amounts.

Now, let's calculate the probability of obtaining each of these amounts.

The number of ways to get $n_H$ heads in 4 tosses is given by the binomial coefficient $\binom{4}{n_H}$.

Total number of outcomes = 16.

Amount: $\textsf{₹}$ 4

This occurs when $n_H = 4$.

Number of outcomes with 4 heads = $\binom{4}{4} = 1$. (HHHH)

Probability($\textsf{₹}$ 4) = $\frac{\text{Number of outcomes with 4 heads}}{\text{Total number of outcomes}} = \frac{1}{16}$

Amount: $\textsf{₹}$ 1.50

This occurs when $n_H = 3$.

Number of outcomes with 3 heads = $\binom{4}{3} = 4$. (HHHT, HHTH, HTHH, THHH)

Probability($\textsf{₹}$ 1.50) = $\frac{\text{Number of outcomes with 3 heads}}{\text{Total number of outcomes}} = \frac{4}{16} = \frac{1}{4}$

Amount: $\textsf{₹}$ -1

This occurs when $n_H = 2$.

Number of outcomes with 2 heads = $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$. (HHTT, HTHT, HTTH, THHT, THTH, TTHH)

Probability($\textsf{₹}$ -1) = $\frac{\text{Number of outcomes with 2 heads}}{\text{Total number of outcomes}} = \frac{6}{16} = \frac{3}{8}$

Amount: $\textsf{₹}$ -3.50

This occurs when $n_H = 1$.

Number of outcomes with 1 head = $\binom{4}{1} = 4$. (HTTT, THTT, TTHT, TTTH)

Probability($\textsf{₹}$ -3.50) = $\frac{\text{Number of outcomes with 1 head}}{\text{Total number of outcomes}} = \frac{4}{16} = \frac{1}{4}$

Amount: $\textsf{₹}$ -6

This occurs when $n_H = 0$.

Number of outcomes with 0 heads = $\binom{4}{0} = 1$. (TTTT)

Probability($\textsf{₹}$ -6) = $\frac{\text{Number of outcomes with 0 heads}}{\text{Total number of outcomes}} = \frac{1}{16}$

Summary of possible amounts and their probabilities:

Given:

Three fair coins are tossed once.

To Find:

The probability of the following events:

(i) 3 heads

(ii) 2 heads

(iii) at least 2 heads

(iv) at most 2 heads

(v) no head

(vi) 3 tails

(vii) exactly two tails

(viii) no tail

(ix) at most two tails

Solution:

When three fair coins are tossed, the sample space S, which is the set of all possible outcomes, is:

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

The total number of possible outcomes is $|S| = 8$.

Since the coins are fair, each outcome in the sample space is equally likely, with a probability of $\frac{1}{8}$.

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

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

(i) Probability of getting 3 heads.

Let E1 be the event of getting 3 heads.

The favorable outcome is $\{HHH\}$.

Number of favorable outcomes is $|E1| = 1$.

$P(\text{3 heads}) = \frac{|E1|}{|S|} = \frac{1}{8}$

(ii) Probability of getting 2 heads.

Let E2 be the event of getting 2 heads.

The favorable outcomes are $\{HHT, HTH, THH\}$.

Number of favorable outcomes is $|E2| = 3$.

$P(\text{2 heads}) = \frac{|E2|}{|S|} = \frac{3}{8}$

(iii) Probability of getting at least 2 heads.

Let E3 be the event of getting at least 2 heads. This means getting 2 heads or 3 heads.

The favorable outcomes are $\{HHT, HTH, THH\}$ (2 heads) and $\{HHH\}$ (3 heads).

So, the favorable outcomes for E3 are $\{HHT, HTH, THH, HHH\}$.

Number of favorable outcomes is $|E3| = 4$.

$P(\text{at least 2 heads}) = \frac{|E3|}{|S|} = \frac{4}{8} = \frac{1}{2}$

(iv) Probability of getting at most 2 heads.

Let E4 be the event of getting at most 2 heads. This means getting 0 heads, 1 head, or 2 heads.

This is the complement of the event of getting 3 heads.

The favorable outcomes are $\{HHT, HTH, THH, HTT, THT, TTH, TTT\}$.

Number of favorable outcomes is $|E4| = 7$.

$P(\text{at most 2 heads}) = \frac{|E4|}{|S|} = \frac{7}{8}$

Alternatively, using the complement rule:

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

(v) Probability of getting no head.

Let E5 be the event of getting no head. This means getting 3 tails.

The favorable outcome is $\{TTT\}$.

Number of favorable outcomes is $|E5| = 1$.

$P(\text{no head}) = \frac{|E5|}{|S|} = \frac{1}{8}$

(vi) Probability of getting 3 tails.

Let E6 be the event of getting 3 tails.

The favorable outcome is $\{TTT\}$.

Number of favorable outcomes is $|E6| = 1$.

$P(\text{3 tails}) = \frac{|E6|}{|S|} = \frac{1}{8}$

Note that this is the same as the probability of getting no head, which makes sense.

(vii) Probability of getting exactly two tails.

Let E7 be the event of getting exactly two tails. This means getting 1 head and 2 tails.

The favorable outcomes are $\{HTT, THT, TTH\}$.

Number of favorable outcomes is $|E7| = 3$.

$P(\text{exactly two tails}) = \frac{|E7|}{|S|} = \frac{3}{8}$

Note that this is the same as the probability of getting exactly two heads, which makes sense due to the symmetry of heads and tails.

(viii) Probability of getting no tail.

Let E8 be the event of getting no tail. This means getting 3 heads.

The favorable outcome is $\{HHH\}$.

Number of favorable outcomes is $|E8| = 1$.

$P(\text{no tail}) = \frac{|E8|}{|S|} = \frac{1}{8}$

Note that this is the same as the probability of getting 3 heads.

(ix) Probability of getting at most two tails.

Let E9 be the event of getting at most two tails. This means getting 0 tails, 1 tail, or 2 tails.

This is the complement of the event of getting 3 tails.

The favorable outcomes are $\{HHH, HHT, HTH, THH, HTT, THT, TTH\}$.

Number of favorable outcomes is $|E9| = 7$.

$P(\text{at most two tails}) = \frac{|E9|}{|S|} = \frac{7}{8}$

Alternatively, using the complement rule:

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

Note that this is the same as the probability of getting at most two heads.

Given:

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

To Find:

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

Solution:

The probability of the complement of an event A is given by the formula:

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

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

$P(A') = 1 - \frac{2}{11}$

To subtract the fraction from 1, we can write 1 as a fraction with the same denominator:

$1 = \frac{11}{11}$

So, the equation becomes:

$P(A') = \frac{11}{11} - \frac{2}{11}$

$P(A') = \frac{11 - 2}{11}$

$P(A') = \frac{9}{11}$

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

Given:

The word is 'ASSASSINATION'.

The letters in the word are A, S, S, A, S, S, I, N, A, T, I, O, N.

To Find:

The probability that the randomly chosen letter is:

(i) a vowel

(ii) a consonant

Solution:

First, let's determine the total number of letters in the word 'ASSASSINATION'.

Counting the letters: A (3 times), S (4 times), I (2 times), N (2 times), T (1 time), O (1 time).

Total number of letters = $3 + 4 + 2 + 2 + 1 + 1 = 13$.

The total number of outcomes in the sample space when a letter is chosen at random is 13.

(i) Probability that the letter is a vowel.

The vowels in the English alphabet are A, E, I, O, U.

From the word 'ASSASSINATION', the vowels are A, A, A, I, I, O.

The number of vowels in the word is 6.

Let E1 be the event that the chosen letter is a vowel.

Number of favorable outcomes for E1 = 6.

Total number of outcomes = 13.

The probability of event E1 is:

$P(\text{a vowel}) = \frac{\text{Number of vowels}}{\text{Total number of letters}}$

$P(\text{a vowel}) = \frac{6}{13}$

(ii) Probability that the letter is a consonant.

The consonants in the word 'ASSASSINATION' are S, S, S, S, N, N, T.

The number of consonants in the word is 7.

Let E2 be the event that the chosen letter is a consonant.

Number of favorable outcomes for E2 = 7.

Total number of outcomes = 13.

The probability of event E2 is:

$P(\text{a consonant}) = \frac{\text{Number of consonants}}{\text{Total number of letters}}$

$P(\text{a consonant}) = \frac{7}{13}$

Alternatively, since a letter is either a vowel or a consonant (and these are mutually exclusive and exhaustive events within the context of this word's letters), the probability of choosing a consonant is also $1 - P(\text{a vowel})$.

$P(\text{a consonant}) = 1 - P(\text{a vowel}) = 1 - \frac{6}{13} = \frac{13}{13} - \frac{6}{13} = \frac{13 - 6}{13} = \frac{7}{13}$

Given:

Total natural numbers available = 20 (from 1 to 20).

Number of distinct natural numbers chosen by a person = 6.

Number of distinct natural numbers fixed by the lottery committee = 6.

The order of the numbers chosen does not matter for winning.

To Find:

The probability of winning the prize.

Solution:

Since the order of the numbers does not matter, the total number of possible ways to choose 6 distinct numbers from 20 is given by the combination formula $\binom{n}{r}$.

Here, $n = 20$ (total numbers available) and $r = 6$ (numbers to be chosen).

Total number of outcomes in the sample space = $\binom{20}{6}$

We calculate the value of $\binom{20}{6}$:

$\binom{20}{6} = \frac{20!}{6!(20-6)!} = \frac{20!}{6!14!}$

$\binom{20}{6} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$

$\binom{20}{6} = 38760$

So, the total number of possible combinations of 6 numbers a person can choose is 38760.

For a person to win the prize, their chosen set of 6 numbers must exactly match the specific set of 6 numbers fixed by the lottery committee. There is only one such matching set.

Number of favorable outcomes = 1.

The probability of winning the prize is the ratio of the number of favorable outcomes to the total number of outcomes.

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

$P(\text{Winning}) = \frac{1}{\binom{20}{6}}$

$P(\text{Winning}) = \frac{1}{38760}$

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

Given:

Probabilities for events A and B and their intersection/union are provided in two cases.

To Check:

Whether the given probabilities P(A) and P(B) are consistently defined for each case.

Conditions for Consistent Probabilities:

For probabilities $P(A)$, $P(B)$, $P(A \cap B)$, and $P(A \cup B)$ to be consistently defined, they must satisfy the following conditions:

1. $0 \leq P(A) \leq 1$, $0 \leq P(B) \leq 1$, $0 \leq P(A \cap B) \leq 1$, $0 \leq P(A \cup B) \leq 1$.

2. $P(A \cap B) \leq P(A)$ and $P(A \cap B) \leq P(B)$ (since $A \cap B$ is a subset of both A and B).

3. $P(A \cup B) \geq P(A)$ and $P(A \cup B) \geq P(B)$ (since A and B are subsets of $A \cup B$).

4. The Addition Rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Checking Case (i):

Given: $P(A) = 0.5$, $P(B) = 0.7$, $P(A \cap B) = 0.6$.

Check Condition 1:

$0 \leq 0.5 \leq 1$ (True)

$0 \leq 0.7 \leq 1$ (True)

$0 \leq 0.6 \leq 1$ (True)

All given probabilities are within the valid range [0, 1].

Check Condition 2:

$P(A \cap B) \leq P(A)$: $0.6 \leq 0.5$ (False)

$P(A \cap B) \leq P(B)$: $0.6 \leq 0.7$ (True)

Since $P(A \cap B) > P(A)$, this violates a fundamental property of probability (the probability of a subset cannot be greater than the probability of the set). Therefore, the probabilities in case (i) are not consistently defined.

We can also check Condition 4 by calculating $P(A \cup B)$: $P(A \cup B) = 0.5 + 0.7 - 0.6 = 0.6$. This calculated $P(A \cup B)$ is valid ($0 \leq 0.6 \leq 1$). However, the violation of Condition 2 is sufficient to deem the definition inconsistent.

Checking Case (ii):

Given: $P(A) = 0.5$, $P(B) = 0.4$, $P(A \cup B) = 0.8$.

Check Condition 1:

$0 \leq 0.5 \leq 1$ (True)

$0 \leq 0.4 \leq 1$ (True)

$0 \leq 0.8 \leq 1$ (True)

All given probabilities are within the valid range [0, 1].

Check Condition 4 (Addition Rule) to find the implied $P(A \cap B)$:

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

$0.8 = 0.5 + 0.4 - P(A \cap B)$

$0.8 = 0.9 - P(A \cap B)$

$P(A \cap B) = 0.9 - 0.8 = 0.1$

Now, check if this implied $P(A \cap B) = 0.1$ is consistent with the other probabilities using Conditions 1, 2, and 3.

Check Condition 1 for $P(A \cap B)$:

$0 \leq 0.1 \leq 1$ (True)

Check Condition 2:

$P(A \cap B) \leq P(A)$: $0.1 \leq 0.5$ (True)

$P(A \cap B) \leq P(B)$: $0.1 \leq 0.4$ (True)

Check Condition 3:

$P(A \cup B) \geq P(A)$: $0.8 \geq 0.5$ (True)

$P(A \cup B) \geq P(B)$: $0.8 \geq 0.4$ (True)

All conditions are satisfied with the given probabilities and the implied value of $P(A \cap B)$. Therefore, the probabilities in case (ii) are consistently defined.

To fill in the blanks in the table, we use the Addition Rule for probabilities:

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

(i) Finding $P(A \cup B)$

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

Using the Addition Rule:

$P(A \cup B) = \frac{1}{3} + \frac{1}{5} - \frac{1}{15}$

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

$P(A \cup B) = \frac{1 \times 5}{3 \times 5} + \frac{1 \times 3}{5 \times 3} - \frac{1}{15}$

$P(A \cup B) = \frac{5}{15} + \frac{3}{15} - \frac{1}{15}$

$P(A \cup B) = \frac{5 + 3 - 1}{15} = \frac{7}{15}$

(ii) Finding $P(B)$

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

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

Rearrange the formula to solve for $P(B)$:

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

Substitute the given values:

$P(B) = 0.6 - 0.35 + 0.25$

$P(B) = 0.25 + 0.25$

$P(B) = 0.50$

(iii) Finding $P(A \cap B)$

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

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

Rearrange the formula to solve for $P(A \cap B)$:

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

Substitute the given values:

$P(A \cap B) = 0.5 + 0.35 - 0.7$

$P(A \cap B) = 0.85 - 0.7$

$P(A \cap B) = 0.15$

The completed table is:

Given:

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

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

Events A and B are mutually exclusive.

To Find:

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

Solution:

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

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

Since events A and B are mutually exclusive, their intersection is the empty set ($\emptyset$), meaning they cannot occur simultaneously.

Therefore, the Addition Rule for mutually exclusive events simplifies to:

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

Substitute the given probabilities into this simplified formula:

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

Add the fractions:

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

$P(A \cup B) = \frac{4}{5}$

Given:

Probability of event E, $P(E) = \frac{1}{4}$

Probability of event F, $P(F) = \frac{1}{2}$

Probability of event E and F (intersection), $P(E \cap F) = \frac{1}{8}$

To Find:

(i) Probability of event E or F (union), $P(E \cup F)$

(ii) Probability of event not E and not F, $P(E' \cap F')$

Solution:

(i) Find P(E or F) = P(E $\cup$ F)

We use the Addition Rule for probabilities:

$P(E \cup F) = P(E) + P(F) - P(E \cap F)$

Substitute the given values into the formula:

$P(E \cup F) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8}$

To add and subtract the fractions, find a common denominator, which is 8.

$P(E \cup F) = \frac{1 \times 2}{4 \times 2} + \frac{1 \times 4}{2 \times 4} - \frac{1}{8}$

$P(E \cup F) = \frac{2}{8} + \frac{4}{8} - \frac{1}{8}$

$P(E \cup F) = \frac{2 + 4 - 1}{8}$

$P(E \cup F) = \frac{6 - 1}{8}$

$P(E \cup F) = \frac{5}{8}$

(ii) Find P(not E and not F) = P(E' $\cap$ F')

We use De Morgan's Law, which states that the complement of the union of two events is the intersection of their complements:

$E' \cap F' = (E \cup F)'$

The probability of the complement of an event is 1 minus the probability of the event:

$P((E \cup F)') = 1 - P(E \cup F)$

From part (i), we found that $P(E \cup F) = \frac{5}{8}$.

Substitute this value into the formula:

$P(E' \cap F') = 1 - \frac{5}{8}$

To subtract the fraction from 1, write 1 as a fraction with the same denominator:

$P(E' \cap F') = \frac{8}{8} - \frac{5}{8}$

$P(E' \cap F') = \frac{8 - 5}{8}$

$P(E' \cap F') = \frac{3}{8}$

Given:

Events E and F are given.

The probability of 'not E or not F' is $P(\text{not E or not F}) = 0.25$. This can be written as $P(E' \cup F') = 0.25$.

To State:

Whether E and F are mutually exclusive.

Solution:

Events E and F are said to be mutually exclusive if their intersection is an empty set, meaning they cannot occur simultaneously. In terms of probability, this means $P(E \cap F) = 0$.

We are given $P(E' \cup F') = 0.25$.

Using De Morgan's Law, we know that $(E \cap F)' = E' \cup F'$.

This means that the event 'not E or not F' is the complement of the event 'E and F'.

Therefore, $P(E' \cup F') = P((E \cap F)')$.

The probability of the complement of an event is 1 minus the probability of the event.

$P((E \cap F)') = 1 - P(E \cap F)$

Substitute the given probability:

$0.25 = 1 - P(E \cap F)$

Now, solve for $P(E \cap F)$:

$P(E \cap F) = 1 - 0.25$

$P(E \cap F) = 0.75$

For events E and F to be mutually exclusive, $P(E \cap F)$ must be equal to 0.

In this case, we found $P(E \cap F) = 0.75$.

Since $0.75 \neq 0$, the events E and F are not mutually exclusive.

Conclusion:

Based on the given probability, $P(E \cap F) = 0.75$. Since this is not equal to 0, events E and F are not mutually exclusive.

Given:

Probability of event A, $P(A) = 0.42$

Probability of event B, $P(B) = 0.48$

Probability of event A and B (intersection), $P(A \cap B) = 0.16$

To Determine:

(i) $P(\text{not A})$

(ii) $P(\text{not B})$

(iii) $P(A \text{ or } B)$

Solution:

(i) Determine P(not A)

The probability of the complement of an event A (not A, denoted as $A'$) is given by the formula:

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

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

$P(A') = 1 - 0.42$

$P(A') = 0.58$

(ii) Determine P(not B)

Similarly, the probability of the complement of an event B (not B, denoted as $B'$) is:

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

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

$P(B') = 1 - 0.48$

$P(B') = 0.52$

(iii) Determine P(A or B)

The probability of the union of two events A and B (A or B, denoted as $A \cup B$) is given by the Addition Rule:

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

Substitute the given values of $P(A)$, $P(B)$, and $P(A \cap B)$:

$P(A \cup B) = 0.42 + 0.48 - 0.16$

$P(A \cup B) = 0.90 - 0.16$

$P(A \cup B) = 0.74$

Given:

Percentage of students studying Mathematics = 40%

Percentage of students studying Biology = 30%

Percentage of students studying both Mathematics and Biology = 10%

Let M be the event that a randomly selected student studies Mathematics.

Let B be the event that a randomly selected student studies Biology.

From the given percentages, the probabilities are:

$P(M) = 40\% = 0.40$

$P(B) = 30\% = 0.30$

$P(M \text{ and } B) = P(M \cap B) = 10\% = 0.10$

To Find:

The probability that a randomly selected student studies Mathematics or Biology, i.e., $P(M \text{ or } B) = P(M \cup B)$.

Solution:

We use the Addition Rule for probabilities for any two events M and B:

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

Substitute the given probabilities into the formula:

$P(M \cup B) = 0.40 + 0.30 - 0.10$

$P(M \cup B) = 0.70 - 0.10$

$P(M \cup B) = 0.60$

The probability that a randomly selected student studies Mathematics or Biology is 0.60 or 60%.

Given:

Let A be the event that a randomly chosen student passes the first examination.

Let B be the event that a randomly chosen student passes the second examination.

Probability of passing the first examination, $P(A) = 0.8$

Probability of passing the second examination, $P(B) = 0.7$

Probability of passing at least one of them (A or B), $P(A \cup B) = 0.95$

To Find:

The probability of passing both examinations (A and B), $P(A \cap B)$.

Solution:

We use the Addition Rule for probabilities, which relates the probabilities of the union and intersection of two events:

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

We are given $P(A)$, $P(B)$, and $P(A \cup B)$, and we need to find $P(A \cap B)$.

Substitute the given values into the Addition Rule formula:

$0.95 = 0.8 + 0.7 - P(A \cap B)$

Combine the probabilities on the right side:

$0.95 = 1.5 - P(A \cap B)$

Now, rearrange the equation to solve for $P(A \cap B)$:

$P(A \cap B) = 1.5 - 0.95$

Perform the subtraction:

$P(A \cap B) = 0.55$

The probability of passing both examinations is 0.55.

Given:

Let E be the event that a student passes the English examination.

Let H be the event that a student passes the Hindi examination.

Probability of passing both English and Hindi, $P(E \cap H) = 0.5$

Probability of passing neither English nor Hindi, $P(\text{neither E nor H}) = 0.1$

This means the probability of the student not passing English AND not passing Hindi is 0.1.

$P(E' \cap H') = 0.1$

Probability of passing the English examination, $P(E) = 0.75$

To Find:

The probability of passing the Hindi examination, $P(H)$.

Solution:

We are given the probability of 'not E and not H', which is $P(E' \cap H')$.

Using De Morgan's Law, we know that the event $(E' \cap H')$ is the complement of the event $(E \cup H)$.

$E' \cap H' = (E \cup H)'$

So, $P(E' \cap H') = P((E \cup H)')$

The probability of the complement of an event is 1 minus the probability of the event.

$P((E \cup H)') = 1 - P(E \cup H)$

Substitute the given value of $P(E' \cap H')$:

$0.1 = 1 - P(E \cup H)$

Rearrange the equation to find $P(E \cup H)$:

$P(E \cup H) = 1 - 0.1$

$P(E \cup H) = 0.9$

Now, we use the Addition Rule for the union of two events:

$P(E \cup H) = P(E) + P(H) - P(E \cap H)$

We know $P(E \cup H) = 0.9$, $P(E) = 0.75$, and $P(E \cap H) = 0.5$. We want to find $P(H)$.

Substitute the known values into the formula:

$0.9 = 0.75 + P(H) - 0.5$

Combine the known values on the right side:

$0.9 = (0.75 - 0.5) + P(H)$

$0.9 = 0.25 + P(H)$

Subtract 0.25 from both sides to solve for $P(H)$:

$P(H) = 0.9 - 0.25$

$P(H) = 0.65$

The probability of passing the Hindi examination is 0.65.

Given:

Total number of students in the class = 60

Number of students who opted for NCC, $n(\text{NCC}) = 30$

Number of students who opted for NSS, $n(\text{NSS}) = 32$

Number of students who opted for both NCC and NSS, $n(\text{NCC} \cap \text{NSS}) = 24$

Let N be the event that a randomly selected student opted for NCC.

Let S be the event that a randomly selected student opted for NSS.

The total number of possible outcomes is the total number of students, which is 60.

The probabilities of the individual events and their intersection are:

$P(\text{N}) = \frac{n(\text{NCC})}{\text{Total students}} = \frac{30}{60} = \frac{1}{2}$

$P(\text{S}) = \frac{n(\text{NSS})}{\text{Total students}} = \frac{32}{60} = \frac{8}{15}$

$P(\text{N} \cap \text{S}) = \frac{n(\text{NCC} \cap \text{NSS})}{\text{Total students}} = \frac{24}{60} = \frac{2}{5}$

To Find:

(i) The probability that the student opted for NCC or NSS, $P(\text{N} \cup \text{S})$.

(ii) The probability that the student has opted neither NCC nor NSS, $P(\text{N'} \cap \text{S'})$.

(iii) The probability that the student has opted NSS but not NCC, $P(\text{S} \cap \text{N'})$.

Solution:

(i) Probability that the student opted for NCC or NSS.

This is the probability of the union of events N and S, $P(\text{N} \cup \text{S})$.

Using the Addition Rule for probabilities:

$P(\text{N} \cup \text{S}) = P(\text{N}) + P(\text{S}) - P(\text{N} \cap \text{S})$

Substitute the calculated probabilities:

$P(\text{N} \cup \text{S}) = \frac{1}{2} + \frac{8}{15} - \frac{2}{5}$

Find a common denominator (LCM of 2, 15, 5 is 30):

$P(\text{N} \cup \text{S}) = \frac{1 \times 15}{2 \times 15} + \frac{8 \times 2}{15 \times 2} - \frac{2 \times 6}{5 \times 6}$

$P(\text{N} \cup \text{S}) = \frac{15}{30} + \frac{16}{30} - \frac{12}{30}$

$P(\text{N} \cup \text{S}) = \frac{15 + 16 - 12}{30} = \frac{31 - 12}{30} = \frac{19}{30}$

(ii) Probability that the student has opted neither NCC nor NSS.

This event is the complement of the student opting for NCC or NSS, i.e., $(\text{N} \cup \text{S})'$.

Using the complement rule:

$P(\text{neither N nor S}) = P((\text{N} \cup \text{S})') = 1 - P(\text{N} \cup \text{S})$

Substitute the result from part (i):

$P(\text{neither N nor S}) = 1 - \frac{19}{30} = \frac{30}{30} - \frac{19}{30} = \frac{30 - 19}{30} = \frac{11}{30}$

Alternatively, using the number of students:

Number of students who opted for NCC or NSS = $n(\text{NCC} \cup \text{NSS}) = n(\text{NCC}) + n(\text{NSS}) - n(\text{NCC} \cap \text{NSS}) = 30 + 32 - 24 = 38$.

Number of students who opted neither = Total students - Number of students who opted for NCC or NSS = $60 - 38 = 22$.

$P(\text{neither N nor S}) = \frac{\text{Number of students who opted neither}}{\text{Total students}} = \frac{22}{60} = \frac{11}{30}$

(iii) Probability that the student has opted NSS but not NCC.

This event is the set of students who are in NSS but not in the intersection of NCC and NSS. It is represented as $\text{S} \cap \text{N}'$.

The probability of this event can be found by subtracting the probability of the intersection from the probability of NSS:

$P(\text{S} \cap \text{N}') = P(\text{S}) - P(\text{S} \cap \text{N})$

Since $\text{S} \cap \text{N}$ is the same as $\text{N} \cap \text{S}$, we have:

$P(\text{S} \cap \text{N}') = P(\text{S}) - P(\text{N} \cap \text{S})$

Substitute the probabilities $P(\text{S}) = \frac{8}{15}$ and $P(\text{N} \cap \text{S}) = \frac{2}{5}$:

$P(\text{S} \cap \text{N}') = \frac{8}{15} - \frac{2}{5}$

Find a common denominator (LCM of 15 and 5 is 15):

$P(\text{S} \cap \text{N}') = \frac{8}{15} - \frac{2 \times 3}{5 \times 3}$

$P(\text{S} \cap \text{N}') = \frac{8}{15} - \frac{6}{15} = \frac{8 - 6}{15} = \frac{2}{15}$

Alternatively, using the number of students:

Number of students who opted for NSS only = Number of students who opted for NSS - Number of students who opted for both = $32 - 24 = 8$.

$P(\text{S} \text{ but not } \text{N}) = \frac{\text{Number of students who opted for NSS only}}{\text{Total students}} = \frac{8}{60} = \frac{2}{15}$

Given:

Veena visits four distinct cities: A, B, C, and D.

She visits these cities in a random order.

To Find:

The probability of the specified events regarding the order of visiting the cities.

Solution:

The total number of possible orders in which Veena can visit the four distinct cities is the number of permutations of 4 items.

Total number of possible orders = $4! = 4 \times 3 \times 2 \times 1 = 24$.

Each of these 24 possible orders (permutations) is equally likely.

The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is $4!$.

$4! = 4 \times 3 \times 2 \times 1 = 24$.

Therefore, $n(S) = 24$, where S is the sample space.

Since the number of elements in the sample space of the experiment is 24, all of these outcomes are considered to be equally likely.

A sample space for the experiment is:

$S = \{ABCD, ABDC, ACBD, ACDB, ADBC, ADCB,$

$\phantom{= }BACD, BADC, BDAC, BDCA, BCAD, BCDA,$

$\phantom{= }CABD, CADB, CBDA, CBAD, CDAB, CDBA,$

$\phantom{= \{}DABC, DACB, DBCA, DBAC, DCAB, DCBA \}$

(i) Probability that she visits A before B?

Let the event ‘she visits A before B’ be denoted by E.

Therefore, $E = \{ABCD, CABD, DABC, ABDC, CADB, DACB,$

$\phantom{= \{ }ACBD, ACDB, ADBC, CDAB, DCAB, ADCB\}$

Thus, the number of favorable outcomes is $n(E) = 12$.

The probability of event E is:

$P(E) = \frac{n(E)}{n(S)}$

$P(E) = \frac{12}{24} = \frac{1}{2}$

(ii) Probability that she visits A before B and B before C?

Let the event ‘Veena visits A before B and B before C’ be denoted by F.

Here $F = \{ABCD, DABC, ADBC, CABD\}$

Therefore, the number of favorable outcomes is $n(F) = 4$.

The probability of event F is:

$P(F) = \frac{n(F)}{n(S)}$

$P(F) = \frac{4}{24} = \frac{1}{6}$

(iii) Probability that she visits A first and B last?

Let the event ‘Veena visits A first and B last’ be denoted by G.

Here $G = \{ACDB, ADCB\}$

Therefore, the number of favorable outcomes is $n(G) = 2$.

The probability of event G is:

$P(G) = \frac{n(G)}{n(S)}$

$P(G) = \frac{2}{24} = \frac{1}{12}$

(iv) Probability that she visits A either first or second?

Let the event ‘Veena visits A either first or second’ be denoted by H.

This event is the union of the event 'A is first' and the event 'A is second'.

The outcomes where A is first are $\{ABCD, ABDC, ACBD, ACDB, ADBC, ADCB\}$.

The outcomes where A is second are $\{BACD, BADC, CABD, CADB, DABC, DACB\}$.

Therefore, $H = \{ABCD, ABDC, ACBD, ACDB, ADBC, ADCB,$

$\phantom{= \{}BACD, BADC, CABD, CADB, DABC, DACB \}$

Thus, the number of favorable outcomes is $n(H) = 6 + 6 = 12$.

The probability of event H is:

$P(H) = \frac{n(H)}{n(S)}$

$P(H) = \frac{12}{24} = \frac{1}{2}$

(v) Probability that she visits A just before B?

Let the event ‘Veena visits A just before B’ be denoted by K.

This means the cities A and B appear consecutively as AB in the visit sequence.

Here $K = \{ABCD, ABDC, CABD, DABC, CDAB, DCAB\}$

Therefore, the number of favorable outcomes is $n(K) = 6$.

The probability of event K is:

$P(K) = \frac{n(K)}{n(S)}$

$P(K) = \frac{6}{24} = \frac{1}{4}$

Given:

Total number of cards in a deck = 52.

Number of cards drawn to form a hand = 7.

Number of Kings in the deck = 4.

Number of non-Kings in the deck = $52 - 4 = 48$.

To Find:

The probability that the hand of 7 cards contains:

(i) exactly 4 Kings

(ii) exactly 3 Kings

(iii) at least 3 Kings.

Solution:

The total number of possible hands of 7 cards that can be drawn from 52 cards is the number of combinations of 52 items taken 7 at a time.

Total number of possible hands = $n(S) = \binom{52}{7}$

$n(S) = \frac{52!}{7!45!} = 133784560$.

(i) Probability that the hand contains all Kings.

This event means the hand has exactly 4 Kings and $7 - 4 = 3$ non-Kings.

Number of ways to choose 4 Kings from 4 = $\binom{4}{4} = 1$.

Number of ways to choose 3 non-Kings from 48 = $\binom{48}{3} = 17296$.

Number of favorable outcomes (4 Kings and 3 non-Kings) = $\binom{4}{4} \times \binom{48}{3} = 1 \times 17296 = 17296$.

The probability of getting all Kings is:

$P(\text{4 Kings}) = \frac{17296}{133784560} = \frac{1}{7735}$.

(ii) Probability that the hand contains 3 Kings.

This event means the hand has exactly 3 Kings and $7 - 3 = 4$ non-Kings.

Number of ways to choose 3 Kings from 4 = $\binom{4}{3} = 4$.

Number of ways to choose 4 non-Kings from 48 = $\binom{48}{4} = 194580$.

Number of favorable outcomes (3 Kings and 4 non-Kings) = $\binom{4}{3} \times \binom{48}{4} = 4 \times 194580 = 778320$.

The probability of getting 3 Kings is:

$P(\text{3 Kings}) = \frac{778320}{133784560} = \frac{9}{1547}$.

(iii) Probability that the hand contains at least 3 Kings.

This event means the hand contains either exactly 3 Kings or exactly 4 Kings.

These two cases are mutually exclusive.

$P(\text{at least 3 Kings}) = P(\text{exactly 3 Kings}) + P(\text{exactly 4 Kings})$

$P(\text{at least 3 Kings}) = \frac{778320}{133784560} + \frac{17296}{133784560} = \frac{795616}{133784560}$.

Alternatively, using the simplified probabilities from (i) and (ii):

$P(\text{at least 3 Kings}) = \frac{9}{1547} + \frac{1}{7735}$.

Finding a common denominator (LCM of 1547 and 7735 is 7735):

$P(\text{at least 3 Kings}) = \frac{9 \times 5}{1547 \times 5} + \frac{1}{7735} = \frac{45}{7735} + \frac{1}{7735} = \frac{46}{7735}$.

Simplifying the fraction $\frac{46}{7735}$ further, we find the greatest common divisor (GCD) is 23.

$\frac{46 \div 23}{7735 \div 23} = \frac{2}{335}$.

$P(\text{at least 3 Kings}) = \frac{2}{335}$.

To Prove:

For three events A, B, and C associated with a random experiment,

$P(A \cup B \cup C) = P(A) + P(B) + P(C) − P(A \cap B) − P(A \cap C) – P ( B \cap C) + P ( A \cap B \cap C)$

Proof:

We know the Addition Rule for two events X and Y:

$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$

Consider the union of the three events $A \cup B \cup C$ as the union of two events, say $(A \cup B)$ and $C$.

Let $X = (A \cup B)$ and $Y = C$. Applying the Addition Rule for $X \cup Y$:

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

Now, we apply the Addition Rule to the term $P(A \cup B)$:

Next, we apply the Addition Rule to the term $P((A \cap C) \cup (B \cap C))$. Let $U = (A \cap C)$ and $V = (B \cap C)$.

$P(U \cup V) = P(U) + P(V) - P(U \cap V)$

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

The intersection of the two events $(A \cap C)$ and $(B \cap C)$ is the intersection of A, C, B, and C, which simplifies to $A \cap B \cap C$.

So, the equation becomes:

Now, substitute the expressions from equation (ii) and equation (iii) into equation (i):

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

Remove the brackets and distribute the negative sign:

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

Rearrange the terms to match the desired form:

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

This proves the Addition Rule for three events.

Given:

There are five teams: A, B, C, D, and E.

The teams finish a relay race in a random order, with all finishing orders being equally likely.

To Find:

(a) The probability that A, B, and C finish first, second, and third, respectively.

(b) The probability that A, B, and C are the first three to finish (in any order).

Solution:

The total number of possible finishing orders for the five teams is the number of permutations of 5 distinct items.

Total number of finishing orders = $5!$

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

This is the total number of outcomes in the sample space.

(a) Probability that A, B and C finish first, second and third, respectively.

Let E1 be the event that A finishes 1st, B finishes 2nd, and C finishes 3rd.

The order of the first three teams is fixed as A, B, C.

The first position must be A (1 choice).

The second position must be B (1 choice).

The third position must be C (1 choice).

The remaining two teams (D and E) can finish in the 4th and 5th positions in any order.

The number of ways to arrange the remaining 2 teams in the remaining 2 positions is $2! = 2 \times 1 = 2$.

The possible finishing orders are ABCDE and ABCED.

Number of favorable outcomes for E1 = $1 \times 1 \times 1 \times 2! = 2$.

The probability of E1 is:

$P(\text{A is 1st, B is 2nd, C is 3rd}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{120} = \frac{1}{60}$

(b) Probability that A, B and C are the first three to finish (in any order).

Let E2 be the event that A, B, and C are the first three to finish.

This means that the teams finishing in the first three positions are A, B, and C, but their order among themselves can be any permutation of A, B, and C.

The set of teams in the first three positions is {A, B, C}.

The number of ways to arrange the teams A, B, and C in the first three positions is $3! = 3 \times 2 \times 1 = 6$.

The remaining two teams (D and E) must finish in the 4th and 5th positions.

The number of ways to arrange the remaining 2 teams in the remaining 2 positions is $2! = 2 \times 1 = 2$.

Number of favorable outcomes for E2 = (Number of ways to arrange A, B, C in first 3 places) $\times$ (Number of ways to arrange D, E in last 2 places)

Number of favorable outcomes for E2 = $3! \times 2! = 6 \times 2 = 12$.

The probability of E2 is:

$P(\text{A, B, C are first three}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{12}{120} = \frac{1}{10}$

Given:

Number of red marbles = 10.

Number of blue marbles = 20.

Number of green marbles = 30.

Total number of marbles in the box = $10 + 20 + 30 = 60$.

Number of marbles drawn from the box = 5.

To Find:

The probability that the 5 marbles drawn are:

(i) all blue

(ii) at least one green.

Solution:

The total number of possible ways to draw 5 marbles from the 60 marbles in the box is given by the combination $\binom{60}{5}$, since the order of drawing does not matter.

Total number of outcomes = $n(S) = \binom{60}{5}$

$n(S) = \frac{60!}{5!(60-5)!} = \frac{60!}{5!55!} = \frac{60 \times 59 \times 58 \times 57 \times 56}{5 \times 4 \times 3 \times 2 \times 1}$

$n(S) = 5,461,512$.

(i) Probability that all 5 will be blue?

Let E be the event that all 5 marbles drawn are blue.

To get 5 blue marbles, we must choose 5 marbles from the 20 available blue marbles.

Number of favorable outcomes for E = $n(E) = \binom{20}{5}$

$n(E) = \frac{20!}{5!(20-5)!} = \frac{20!}{5!15!} = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1}$

$n(E) = 15,504$.

The probability of drawing 5 blue marbles is:

$P(\text{all blue}) = \frac{n(E)}{n(S)} = \frac{15504}{5461512}$

Simplifying the fraction:

$P(\text{all blue}) = \frac{15504 \div 456}{5461512 \div 456} = \frac{34}{11977}$.

(ii) Probability that at least one will be green?

Let A be the event that at least one marble drawn is green.

It is easier to calculate the probability of the complement event, A', which is that none of the marbles drawn is green.

If none of the marbles drawn is green, then all 5 marbles must be drawn from the non-green marbles (red or blue).

Total number of non-green marbles = Number of red marbles + Number of blue marbles = $10 + 20 = 30$.

Number of ways to choose 5 non-green marbles from 30 = $n(A') = \binom{30}{5}$.

$n(A') = \frac{30!}{5!(30-5)!} = \frac{30!}{5!25!} = \frac{30 \times 29 \times 28 \times 27 \times 26}{5 \times 4 \times 3 \times 2 \times 1}$

$n(A') = 142,506$.

The probability that none of the marbles are green is:

$P(A') = \frac{n(A')}{n(S)} = \frac{142506}{5461512}$.

Simplifying the fraction:

$P(A') = \frac{142506 \div 1218}{5461512 \div 1218} = \frac{117}{4484}$.

The probability of at least one green marble is $P(A) = 1 - P(A')$.

$P(\text{at least one green}) = 1 - \frac{117}{4484} = \frac{4484}{4484} - \frac{117}{4484} = \frac{4484 - 117}{4484} = \frac{4367}{4484}$.

Given:

Total number of cards in a deck = 52.

Number of cards drawn = 4.

Number of cards in each suit (Diamonds, Hearts, Spades, Clubs) = 13.

To Find:

The probability of obtaining exactly 3 diamonds and exactly 1 spade in the hand of 4 cards.

Solution:

The total number of possible ways to draw a hand of 4 cards from a deck of 52 is given by the number of combinations of 52 items taken 4 at a time, as the order of cards in a hand does not matter.

Total number of possible hands = $n(S) = \binom{52}{4}$

$n(S) = \frac{52!}{4!(52-4)!} = \frac{52!}{4!48!} = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1}$

$n(S) = 270725$.

Now, we need to find the number of ways to obtain exactly 3 diamonds and 1 spade.

The number of ways to choose 3 diamonds from the 13 available diamonds is $\binom{13}{3}$.

Number of ways to choose 3 diamonds = $\binom{13}{3} = \frac{13!}{3!(13-3)!} = \frac{13!}{3!10!} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1}$

$\binom{13}{3} = 286$.

The number of ways to choose 1 spade from the 13 available spades is $\binom{13}{1}$.

Number of ways to choose 1 spade = $\binom{13}{1} = \frac{13!}{1!(13-1)!} = \frac{13!}{1!12!} = 13$.

The number of favorable outcomes (getting exactly 3 diamonds AND exactly 1 spade) is the product of the number of ways to choose 3 diamonds and the number of ways to choose 1 spade.

Number of favorable outcomes = $\binom{13}{3} \times \binom{13}{1} = 286 \times 13 = 3718$.

The probability of obtaining 3 diamonds and one spade is the ratio of the number of favorable outcomes to the total number of possible outcomes.

Probability (3 diamonds and 1 spade) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3718}{270725}$.

Simplifying the fraction by dividing both numerator and denominator by their greatest common divisor, which is 13:

$\frac{3718 \div 13}{270725 \div 13} = \frac{286}{20825}$.

The simplified probability is $\frac{286}{20825}$.

Given:

A die has 6 faces with the following numbers:

Two faces with number '1'.

Three faces with number '2'.

One face with number '3'.

The die is rolled once.

To Determine:

(i) $P(2)$

(ii) $P(1 \text{ or } 3)$

(iii) $P(\text{not } 3)$

Solution:

The total number of possible outcomes when the die is rolled once is the total number of faces.

Total number of outcomes = Number of faces with 1 + Number of faces with 2 + Number of faces with 3 = $2 + 3 + 1 = 6$.

The probability of a specific outcome is the number of faces with that outcome divided by the total number of faces.

(i) Determine P(2)

The event is rolling a 2.

Number of favorable outcomes (faces with 2) = 3.

Total number of outcomes = 6.

$P(2) = \frac{\text{Number of faces with 2}}{\text{Total number of faces}} = \frac{3}{6} = \frac{1}{2}$

(ii) Determine P(1 or 3)

The event is rolling a 1 or a 3.

This event occurs if the face shows a 1 or if it shows a 3. These are mutually exclusive events.

Number of faces with 1 = 2.

Number of faces with 3 = 1.

Number of favorable outcomes (faces with 1 or 3) = Number of faces with 1 + Number of faces with 3 = $2 + 1 = 3$.

Total number of outcomes = 6.

$P(1 \text{ or } 3) = \frac{\text{Number of faces with 1 or 3}}{\text{Total number of faces}} = \frac{3}{6} = \frac{1}{2}$

Alternatively, using the Addition Rule for mutually exclusive events:

$P(1 \text{ or } 3) = P(1) + P(3)$

$P(1) = \frac{\text{Number of faces with 1}}{\text{Total number of faces}} = \frac{2}{6}$

$P(3) = \frac{\text{Number of faces with 3}}{\text{Total number of faces}} = \frac{1}{6}$

$P(1 \text{ or } 3) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$

(iii) Determine P(not 3)

The event is not rolling a 3.

This means rolling a 1 or a 2.

Number of faces with 1 = 2.

Number of faces with 2 = 3.

Number of favorable outcomes (faces with 1 or 2) = Number of faces with 1 + Number of faces with 2 = $2 + 3 = 5$.

Total number of outcomes = 6.

$P(\text{not } 3) = \frac{\text{Number of faces with 1 or 2}}{\text{Total number of faces}} = \frac{5}{6}$

Alternatively, using the complement rule:

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

$P(3) = \frac{\text{Number of faces with 3}}{\text{Total number of faces}} = \frac{1}{6}$.

$P(\text{not } 3) = 1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6}$

Given:

Total number of tickets sold in a lottery = 10,000.

Number of equal prizes awarded = 10.

Number of winning tickets = 10.

Number of losing tickets = Total tickets - Number of winning tickets = $10000 - 10 = 9990$.

We assume that the selection of tickets is random and the tickets bought are distinct.

To Find:

The probability of not getting a prize if you buy:

(a) one ticket

(b) two tickets

(c) 10 tickets.

Solution:

The event of not getting a prize when buying a certain number of tickets means that all the tickets bought are losing tickets.

The total number of ways to choose $k$ tickets from $N$ tickets is $\binom{N}{k}$.

The number of ways to choose $k$ losing tickets from the 9990 losing tickets is $\binom{9990}{k}$.

The probability of not getting a prize when buying $k$ tickets is $\frac{\text{Number of ways to choose } k \text{ losing tickets}}{\text{Total number of ways to choose } k \text{ tickets}} = \frac{\binom{9990}{k}}{\binom{10000}{k}}$.

(a) Probability of not getting a prize if you buy one ticket.

Here, $k = 1$.

Total number of ways to choose 1 ticket from 10000 = $\binom{10000}{1} = 10000$.

Number of ways to choose 1 losing ticket from 9990 = $\binom{9990}{1} = 9990$.

$P(\text{not getting prize with 1 ticket}) = \frac{\binom{9990}{1}}{\binom{10000}{1}} = \frac{9990}{10000}$

$P(\text{not getting prize with 1 ticket}) = \frac{999}{1000} = 0.999$

(b) Probability of not getting a prize if you buy two tickets.

Here, $k = 2$.

Total number of ways to choose 2 tickets from 10000 = $\binom{10000}{2}$.

$\binom{10000}{2} = \frac{10000 \times 9999}{2 \times 1} = 5000 \times 9999 = 49995000$

Number of ways to choose 2 losing tickets from 9990 = $\binom{9990}{2}$.

$\binom{9990}{2} = \frac{9990 \times 9989}{2 \times 1} = 4995 \times 9989 = 49895055$

$P(\text{not getting prize with 2 tickets}) = \frac{\binom{9990}{2}}{\binom{10000}{2}} = \frac{49895055}{49995000}$

Alternatively, we can use the sequential probability argument:

$P(\text{not getting prize with 2 tickets}) = P(\text{1st ticket is losing}) \times P(\text{2nd ticket is losing | 1st was losing})$

$P(\text{not getting prize with 2 tickets}) = \frac{9990}{10000} \times \frac{9989}{9999}$

$P(\text{not getting prize with 2 tickets}) = \frac{99790110}{99990000} = \frac{9979011}{9999000}$

The fraction from the combinations is $\frac{49895055}{49995000}$. Let's check if they are the same by dividing the second fraction by 5: $\frac{49895055 \div 5}{49995000 \div 5} = \frac{9979011}{9999000}$. Yes, they are the same.

Using the product form is simpler:

$P(\text{not getting prize with 2 tickets}) = \frac{9990}{10000} \times \frac{9989}{9999}$

(c) Probability of not getting a prize if you buy 10 tickets.

Here, $k = 10$.

Total number of ways to choose 10 tickets from 10000 = $\binom{10000}{10}$.

Number of ways to choose 10 losing tickets from 9990 = $\binom{9990}{10}$.

$P(\text{not getting prize with 10 tickets}) = \frac{\binom{9990}{10}}{\binom{10000}{10}}$

Expanding the combinations:

$P(\text{not getting prize with 10 tickets}) = \frac{\frac{9990!}{10!9980!}}{\frac{10000!}{10!9990!}} = \frac{9990!}{10!9980!} \times \frac{10!9990!}{10000!} = \frac{9990! \times 9990!}{9980! \times 10000!}$

This simplifies to a product of fractions:

$P(\text{not getting prize with 10 tickets}) = \frac{9990}{10000} \times \frac{9989}{9999} \times \frac{9988}{9998} \times \dots \times \frac{9981}{9991}$

This product has 10 terms.

$P(\text{not getting prize with 10 tickets}) = \prod\limits_{i=0}^{9} \frac{9990-i}{10000-i}$

Given:

Total number of students = 100.

Number of students in Section 1 = 40.

Number of students in Section 2 = 60.

You and your friend are two specific students among the 100.

To Find:

(a) The probability that you and your friend are in the same section.

(b) The probability that you and your friend are in different sections.

Solution:

We can determine the probability by considering the sections you and your friend are assigned to. Imagine the sections are formed by randomly picking students without replacement.

(a) Probability that you both enter the same section.

This event can occur in two mutually exclusive ways:

Case 1: Both of you are in Section 1.

The probability that you are in Section 1 is $\frac{40}{100}$.

Given that you are in Section 1, there are 99 students remaining, and 39 spots left in Section 1.

The probability that your friend is also in Section 1 (given you are in Section 1) is $\frac{39}{99}$.

$P(\text{Both in Section 1}) = P(\text{You in S1}) \times P(\text{Friend in S1 | You in S1})$

$P(\text{Both in Section 1}) = \frac{40}{100} \times \frac{39}{99} = \frac{40 \times 39}{100 \times 99} = \frac{1560}{9900}$.

Case 2: Both of you are in Section 2.

The probability that you are in Section 2 is $\frac{60}{100}$.

Given that you are in Section 2, there are 99 students remaining, and 59 spots left in Section 2.

The probability that your friend is also in Section 2 (given you are in Section 2) is $\frac{59}{99}$.

$P(\text{Both in Section 2}) = P(\text{You in S2}) \times P(\text{Friend in S2 | You in S2})$

$P(\text{Both in Section 2}) = \frac{60}{100} \times \frac{59}{99} = \frac{60 \times 59}{100 \times 99} = \frac{3540}{9900}$.

The probability that you both enter the same section is the sum of the probabilities of these two mutually exclusive cases:

$P(\text{Both in same section}) = P(\text{Both in Section 1}) + P(\text{Both in Section 2})$

$P(\text{Both in same section}) = \frac{1560}{9900} + \frac{3540}{9900} = \frac{1560 + 3540}{9900} = \frac{5100}{9900}$.

Simplifying the fraction:

$P(\text{Both in same section}) = \frac{5100 \div 300}{9900 \div 300} = \frac{17}{33}$.

(b) Probability that you both enter different sections.

This event can occur in two mutually exclusive ways:

Case 1: You are in Section 1 and your friend is in Section 2.

$P(\text{You in S1}) = \frac{40}{100}$.

Given you are in Section 1, there are 99 students remaining, and 60 spots available in Section 2.

$P(\text{Friend in S2 | You in S1}) = \frac{60}{99}$.

$P(\text{You in S1 and Friend in S2}) = \frac{40}{100} \times \frac{60}{99} = \frac{2400}{9900}$.

Case 2: You are in Section 2 and your friend is in Section 1.

$P(\text{You in S2}) = \frac{60}{100}$.

Given you are in Section 2, there are 99 students remaining, and 40 spots available in Section 1.

$P(\text{Friend in S1 | You in S2}) = \frac{40}{99}$.

$P(\text{You in S2 and Friend in S1}) = \frac{60}{100} \times \frac{40}{99} = \frac{2400}{9900}$.

The probability that you both enter different sections is the sum of the probabilities of these two mutually exclusive cases:

$P(\text{Both in different sections}) = P(\text{You in S1 and Friend in S2}) + P(\text{You in S2 and Friend in S1})$

$P(\text{Both in different sections}) = \frac{2400}{9900} + \frac{2400}{9900} = \frac{4800}{9900}$.

Simplifying the fraction:

$P(\text{Both in different sections}) = \frac{4800 \div 300}{9900 \div 300} = \frac{16}{33}$.

Alternatively, the events "both in the same section" and "both in different sections" are complementary events. Therefore, the sum of their probabilities is 1.

$P(\text{Both in different sections}) = 1 - P(\text{Both in same section})$

$P(\text{Both in different sections}) = 1 - \frac{17}{33} = \frac{33}{33} - \frac{17}{33} = \frac{33 - 17}{33} = \frac{16}{33}$.

Given:

Three distinct letters are dictated to three persons.

Three distinct envelopes are addressed to each of the three persons.

The letters are inserted into the envelopes at random, one letter per envelope.

To Find:

The probability that at least one letter is in its proper envelope.

Solution:

Let the three letters be $L_1, L_2, L_3$ and the three corresponding addressed envelopes be $E_1, E_2, E_3$. A letter is in its proper envelope if letter $L_i$ is placed into envelope $E_i$.

The random insertion of letters into envelopes means we are considering all possible ways to arrange the three letters into the three envelopes, such that each envelope receives exactly one letter.

The total number of ways to arrange 3 distinct letters into 3 distinct envelopes is the number of permutations of 3 items, which is $3!$.

Total number of possible arrangements (outcomes) = $3! = 3 \times 2 \times 1 = 6$.

The sample space S consists of these 6 arrangements. Assuming the insertion is random, each arrangement is equally likely.

Let A be the event that at least one letter is in its proper envelope.

It is easier to calculate the probability of the complement event, A', which is the event that no letter is in its proper envelope.

The arrangements where no letter is in its proper envelope are called derangements. For 3 items, the derangements are the permutations where element $i$ is not in position $i$ for any $i$.

Let's list all 6 possible arrangements and identify the derangements:

Assume the envelopes are ordered $E_1, E_2, E_3$. An outcome is the sequence of letters placed in these envelopes.

1. ($L_1, L_2, L_3$): $L_1$ in $E_1$, $L_2$ in $E_2$, $L_3$ in $E_3$. (3 proper)

2. ($L_1, L_3, L_2$): $L_1$ in $E_1$, $L_3$ in $E_2$, $L_2$ in $E_3$. (1 proper: $L_1$)

3. ($L_2, L_1, L_3$): $L_2$ in $E_1$, $L_1$ in $E_2$, $L_3$ in $E_3$. (1 proper: $L_3$)

4. ($L_2, L_3, L_1$): $L_2$ in $E_1$, $L_3$ in $E_2$, $L_1$ in $E_3$. (0 proper) - Derangement

5. ($L_3, L_1, L_2$): $L_3$ in $E_1$, $L_1$ in $E_2$, $L_2$ in $E_3$. (0 proper) - Derangement

6. ($L_3, L_2, L_1$): $L_3$ in $E_1$, $L_2$ in $E_2$, $L_1$ in $E_3$. (1 proper: $L_2$)

The number of arrangements where no letter is in its proper envelope is 2.

Number of favorable outcomes for A' (no proper letter) = 2.

The probability of the complement event A' is:

$P(\text{no letter in proper envelope}) = \frac{\text{Number of derangements}}{\text{Total number of arrangements}} = \frac{2}{6} = \frac{1}{3}$

The probability of the event A (at least one letter in its proper envelope) is $1$ minus the probability of the complement event A'.

$P(\text{at least one proper}) = P(A) = 1 - P(A')$

$P(A) = 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{3 - 1}{3} = \frac{2}{3}$

The probability that at least one letter is in its proper envelope is $\frac{2}{3}$.

Alternate Solution using Principle of Inclusion-Exclusion:

Let $A_i$ be the event that letter $L_i$ is in envelope $E_i$, for $i=1, 2, 3$.

We want to find the probability of the union of these events: $P(A_1 \cup A_2 \cup A_3)$.

Using the Principle of Inclusion-Exclusion:

$P(A_1 \cup A_2 \cup A_3) = \sum P(A_i) - \sum P(A_i \cap A_j) + P(A_1 \cap A_2 \cap A_3)$

Calculate the individual probabilities:

$P(A_1)$: $L_1$ in $E_1$. The remaining 2 letters can be arranged in $2!$ ways in the remaining 2 envelopes. Total arrangements = $3!$.

$P(A_1) = \frac{2!}{3!} = \frac{2}{6} = \frac{1}{3}$. By symmetry, $P(A_1) = P(A_2) = P(A_3) = \frac{1}{3}$.

$\sum P(A_i) = P(A_1) + P(A_2) + P(A_3) = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$.

Calculate the probabilities of pairwise intersections:

$P(A_1 \cap A_2)$: $L_1$ in $E_1$ AND $L_2$ in $E_2$. The remaining letter $L_3$ must go into $E_3$ (1 way). Total arrangements = $3!$.

$P(A_1 \cap A_2) = \frac{1!}{3!} = \frac{1}{6}$. By symmetry, $P(A_1 \cap A_2) = P(A_1 \cap A_3) = P(A_2 \cap A_3) = \frac{1}{6}$.

$\sum P(A_i \cap A_j) = P(A_1 \cap A_2) + P(A_1 \cap A_3) + P(A_2 \cap A_3) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$.

Calculate the probability of the triple intersection:

$P(A_1 \cap A_2 \cap A_3)$: $L_1$ in $E_1$ AND $L_2$ in $E_2$ AND $L_3$ in $E_3$. There is only 1 way for this to happen. Total arrangements = $3!$.

$P(A_1 \cap A_2 \cap A_3) = \frac{1}{3!} = \frac{1}{6}$.

Substitute these values into the Inclusion-Exclusion formula:

$P(A_1 \cup A_2 \cup A_3) = 1 - \frac{1}{2} + \frac{1}{6}$

$P(A_1 \cup A_2 \cup A_3) = \frac{6}{6} - \frac{3}{6} + \frac{1}{6} = \frac{6 - 3 + 1}{6} = \frac{4}{6} = \frac{2}{3}$

Both methods yield the same result. The probability that at least one letter is in its proper envelope is $\frac{2}{3}$.

Given:

Probability of event A, $P(A) = 0.54$

Probability of event B, $P(B) = 0.69$

Probability of event A and B (intersection), $P(A \cap B) = 0.35$

To Find:

(i) $P(A \cup B)$

(ii) $P(A' \cap B')$

(iii) $P(A \cap B')$

(iv) $P(B \cap A')$

Solution:

(i) Find P(A $\cup$ B)

We use the Addition Rule for probabilities:

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

Substitute the given values into the formula:

$P(A \cup B) = 0.54 + 0.69 - 0.35$

$P(A \cup B) = 1.23 - 0.35$

$P(A \cup B) = 0.88$

(ii) Find P(A' $\cap$ B')

We use De Morgan's Law, which states that the complement of the union of two events is the intersection of their complements:

$A' \cap B' = (A \cup B)'$

The probability of the complement of an event is 1 minus the probability of the event:

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

From part (i), we found that $P(A \cup B) = 0.88$.

Substitute this value into the formula:

$P(A' \cap B') = 1 - 0.88$

$P(A' \cap B') = 0.12$

(iii) Find P(A $\cap$ B')

The event $A \cap B'$ represents the outcomes that are in A but not in B. This is the set difference $A \setminus B$.

The probability is given by:

$P(A \cap B') = P(A) - P(A \cap B)$

Substitute the given values:

$P(A \cap B') = 0.54 - 0.35$

$P(A \cap B') = 0.19$

(iv) Find P(B $\cap$ A')

The event $B \cap A'$ represents the outcomes that are in B but not in A. This is the set difference $B \setminus A$.

The probability is given by:

$P(B \cap A') = P(B) - P(A \cap B)$

Substitute the given values:

$P(B \cap A') = 0.69 - 0.35$

$P(B \cap A') = 0.34$

Given:

A group of 5 persons with the following particulars:

Total number of persons in the group = 5.

One person is selected at random to act as a spokesperson.

To Find:

The probability that the spokesperson will be either male or over 35 years.

Solution:

Let S be the sample space, which consists of the 5 persons in the group. The total number of outcomes when selecting one person is $|S| = 5$.

Let M be the event that the selected person is male.

Let A be the event that the selected person is over 35 years old.

From the table, the persons who are male are Harish, Rohan, and Salim.

Number of persons who are male = 3.

Probability of selecting a male: $P(M) = \frac{\text{Number of males}}{\text{Total number of persons}} = \frac{3}{5}$.

From the table, the persons who are over 35 years old are Sheetal (46) and Salim (41).

Number of persons who are over 35 years old = 2.

Probability of selecting a person over 35: $P(A) = \frac{\text{Number of persons over 35}}{\text{Total number of persons}} = \frac{2}{5}$.

The event "male and over 35 years" ($M \cap A$) consists of persons who are both male and over 35. From the table, only Salim is male and 41 years old (which is over 35).

Number of persons who are male and over 35 = 1.

Probability of selecting a person who is male and over 35: $P(M \cap A) = \frac{\text{Number of male persons over 35}}{\text{Total number of persons}} = \frac{1}{5}$.

We want to find the probability that the spokesperson will be either male or over 35 years, which is the probability of the union of events M and A, i.e., $P(M \cup A)$.

Using the Addition Rule for probabilities:

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

Substitute the calculated probabilities into the formula:

$P(M \cup A) = \frac{3}{5} + \frac{2}{5} - \frac{1}{5}$

$P(M \cup A) = \frac{3 + 2 - 1}{5}$

$P(M \cup A) = \frac{4}{5}$

Given:

Set of digits available: $\{0, 1, 3, 5, 7\}$.

We are forming 4-digit numbers.

The numbers must be greater than 5,000 (strictly > 5000).

A number is divisible by 5 if its units digit is 0 or 5.

To Find:

The probability of forming a number divisible by 5, for the cases where digits are repeated and where they are not repeated.

Solution:

Let the 4-digit number be represented as $d_1 d_2 d_3 d_4$, where $d_1$ is the thousands digit, $d_2$ is the hundreds digit, $d_3$ is the tens digit, and $d_4$ is the units digit.

For a 4-digit number, $d_1 \neq 0$. The available non-zero digits are $\{1, 3, 5, 7\}$.

For the number to be greater than 5,000, the thousands digit $d_1$ must be 5 or 7 (from the given digits). If $d_1=5$, the number must not be 5000.

For a number to be divisible by 5, the units digit $d_4$ must be 0 or 5.

Case (i): The digits are repeated.

First, find the total number of 4-digit numbers strictly greater than 5,000 that can be formed from the digits $\{0, 1, 3, 5, 7\}$ with repetition.

Total number of 4-digit numbers with repetition ($d_1 \neq 0$):

$d_1$ can be any of $\{1, 3, 5, 7\}$ (4 choices).

$d_2, d_3, d_4$ can be any of $\{0, 1, 3, 5, 7\}$ (5 choices each).

Total 4-digit numbers = $4 \times 5 \times 5 \times 5 = 500$.

Numbers $\leq 5000$ formed from these digits:

Numbers < 5000 must have $d_1 \in \{1, 3\}$ (2 choices).

$d_2, d_3, d_4$ can be any of the 5 digits (5 choices each).

Numbers < 5000 = $2 \times 5 \times 5 \times 5 = 250$.

The number 5000 can be formed with repetition (using 5 and 0). This number is not strictly greater than 5000.

Total numbers strictly > 5000 = (Total 4-digit numbers) - (Numbers < 5000) - (The number 5000 if formable)

Total numbers strictly > 5000 = $500 - 250 - 1 = 249$.

Total number of outcomes in the sample space for part (i) is 249.

Next, find the number of these numbers (strictly > 5000, repetition allowed) that are divisible by 5 ($d_4 \in \{0, 5\}$).

The number must be $d_1 d_2 d_3 d_4$ with $d_1 \in \{5, 7\}$, $d_4 \in \{0, 5\}$, and $d_i \in \{0, 1, 3, 5, 7\}$ for $i=2,3$, repetition allowed, and the number > 5000.

Consider numbers $\geq 5000$ that are divisible by 5:

$d_1 \in \{5, 7\}$ (2 choices).

$d_2 \in \{0, 1, 3, 5, 7\}$ (5 choices).

$d_3 \in \{0, 1, 3, 5, 7\}$ (5 choices).

$d_4 \in \{0, 5\}$ (2 choices).

Number of numbers $\geq 5000$ and divisible by 5 = $2 \times 5 \times 5 \times 2 = 100$.

From this count, we must exclude the number 5000 (if it is included and we need strictly > 5000). The number 5000 starts with 5 and ends with 0, so it is included in the 100 count.

Number of favorable outcomes (strictly > 5000 and divisible by 5) = $100 - 1 = 99$.

The probability is:

$P(\text{divisible by 5 | repetition allowed, > 5000}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

$P(\text{divisible by 5 | repetition allowed, > 5000}) = \frac{99}{249}$.

Simplifying the fraction:

$P = \frac{99 \div 3}{249 \div 3} = \frac{33}{83}$.

Case (ii): Repetition of digits is not allowed.

First, find the total number of 4-digit numbers strictly greater than 5,000 that can be formed from the digits $\{0, 1, 3, 5, 7\}$ without repetition.

$d_1$ must be 5 or 7 for the number to be > 5000.

If $d_1 = 5$ (1 choice): The number is $5 d_2 d_3 d_4$. Remaining digits are $\{0, 1, 3, 7\}$. We need to arrange 3 digits from these 4 in the remaining 3 positions ($d_2, d_3, d_4$). Number of ways = $P(4,3) = 4 \times 3 \times 2 = 24$. The smallest number formed is 5013, which is > 5000. All 24 numbers are > 5000.

If $d_1 = 7$ (1 choice): The number is $7 d_2 d_3 d_4$. Remaining digits are $\{0, 1, 3, 5\}$. We need to arrange 3 digits from these 4 in the remaining 3 positions ($d_2, d_3, d_4$). Number of ways = $P(4,3) = 4 \times 3 \times 2 = 24$. All these numbers are > 5000.

Total number of 4-digit numbers strictly > 5,000 without repetition = $24 + 24 = 48$.

Total number of outcomes in the sample space for part (ii) is 48.

Next, find the number of these numbers (strictly > 5000, no repetition) that are divisible by 5 ($d_4 \in \{0, 5\}$).

We consider cases based on the units digit $d_4$ and the thousands digit $d_1$ ($d_1 \in \{5, 7\}$, no repetition).

Case A: Units digit is 0 ($d_4 = 0$).

$d_1$: Must be > 5 and not 0. Can be 5 or 7 (2 choices).

If $d_1 = 5$: The number is $5 d_2 d_3 0$. Used digits are 5 and 0. Remaining digits for $d_2, d_3$ are from $\{1, 3, 7\}$ (3 digits). Number of ways to arrange 2 digits from these 3 is $P(3,2) = 3 \times 2 = 6$. These are all > 5000.

If $d_1 = 7$: The number is $7 d_2 d_3 0$. Used digits are 7 and 0. Remaining digits for $d_2, d_3$ are from $\{1, 3, 5\}$ (3 digits). Number of ways to arrange 2 digits from these 3 is $P(3,2) = 3 \times 2 = 6$. These are all > 5000.

Total numbers ending in 0 and strictly > 5000 = $6 + 6 = 12$.

Case B: Units digit is 5 ($d_4 = 5$).

$d_1$: Must be > 5 and not 5. Can be 7 (1 choice).

If $d_1 = 7$: The number is $7 d_2 d_3 5$. Used digits are 7 and 5. Remaining digits for $d_2, d_3$ are from $\{0, 1, 3\}$ (3 digits). Number of ways to arrange 2 digits from these 3 is $P(3,2) = 3 \times 2 = 6$. These are all > 5000.

Total numbers ending in 5 and strictly > 5000 = 6.

Total number of favorable outcomes (strictly > 5000 and divisible by 5 without repetition) = $12 + 6 = 18$.

The probability is:

$P(\text{divisible by 5 | no repetition, > 5000}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

$P(\text{divisible by 5 | no repetition, > 5000}) = \frac{18}{48}$.

Simplifying the fraction:

$P = \frac{18 \div 6}{48 \div 6} = \frac{3}{8}$.

Given:

A suitcase lock has 4 wheels.

Each wheel is labelled with digits from 0 to 9 (10 digits).

The lock opens with a sequence of four digits with no repeats.

To Find:

The probability of a person getting the right sequence to open the suitcase.

Solution:

The problem states that the lock opens with a sequence of four digits with no repeats. This means the order of the digits matters, and each digit used in the sequence must be unique.

This is a problem involving permutations, as we are arranging a selection of digits where order is important.

First, determine the total number of possible unique 4-digit sequences that can be formed from the 10 digits (0-9) without repetition.

For the first digit, there are 10 choices (0-9).

For the second digit (since repetition is not allowed), there are 9 remaining choices.

For the third digit, there are 8 remaining choices.

For the fourth digit, there are 7 remaining choices.

The total number of possible sequences is the number of permutations of 10 distinct items taken 4 at a time, denoted as $P(10, 4)$ or $_{10}P_4$.

Total number of outcomes = $10 \times 9 \times 8 \times 7$

$10 \times 9 = 90$

$90 \times 8 = 720$

$720 \times 7 = 5040$

Total number of possible unique 4-digit sequences = 5040.

Next, determine the number of favorable outcomes.

The problem states that the lock opens with "a sequence of four digits". This "right sequence" is unique.

Number of favorable outcomes (the right sequence) = 1.

The probability of getting the right sequence is the ratio of the number of favorable outcomes to the total number of possible outcomes.

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

Probability (right sequence) = $\frac{1}{5040}$

The probability of a person getting the right sequence to open the suitcase is $\frac{1}{5040}$.