## Chapter 1 Probability

**Conditional Probability And Probability Of Independent Events**

The concept of probability and various results on it were discussed in class 9. In this chapter, we shall be dealing with problems based on conditional probability and probability of independent events.

**Conditional Probability**

Let A and B be two events associated with the same random experiment. Then, the probability of occurrence of A under the condition B has already occurred and P(B) ≠ 0, is called conditional probability, denoted by P(A/B).

We define:

P(A/B) = \(\frac{P(A \cap B)}{P(B)}\), where P(B) ≠ 0

and P(B/A) = \(\frac{P(B \cap A)}{P(A)}=\frac{P(A \cap B)}{P(A)}\), where P(A) ≠ 0.

**Read and Learn More Class 12 Math Solutions**

## Solved Examples

**Example 1 A die is rolled. If the outcome is an odd number, what is the probability that it is prime?**

**Solution**

**Given**

**A die is rolled. If the outcome is an odd number,**

When a die is rolled, the sample space is given by

S = {1,2,3,4,5,6}.

Let A = event of getting a prime number, and

B = event of getting an odd number.

Then, A = {2,3,5}, B = {1,3,5} and A ∩ B = {3,5}.

∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{3}{6}=\frac{1}{2}\), P(B) = \(\frac{n(B)}{n(S)}=\frac{3}{6}=\frac{1}{2}\)

and P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{2}{6}=\frac{1}{3}\)

Suppose B has already occured and then A occurs.

So, we have to find P(A/B).

Now, P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{(1 / 3)}{(1 / 2)}=\left(\frac{1}{3} \times \frac{2}{1}\right)=\frac{2}{3} \text {. }\)

P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{(1 / 3)}{(1 / 2)}=\left(\frac{1}{3} \times \frac{2}{1}\right)=\frac{2}{3} .\)

Hence, the required probability is \(\frac{2}{3}\).

**Example 2 Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?**

**Solution**

**Given**

**Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3**

Clearly, the sample is S = {1,2,3,4,5,6,7,8,9,10}.

Let A = event that the number on the drawn card is even, and

B = event that number on the drawn card is more than 3.

Then A = {2,4,6,8,10}, B = {4,5,6,7,8,9,10}

and A ∩ B = {4,6,8,10}.

∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{5}{10}=\frac{1}{2}\), P(B) = \(\frac{n(B)}{n(S)}=\frac{7}{10}\) and

P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{4}{10}=\frac{2}{5} .\)

Suppose B has already occured and then A occurs.

So, we have to find P(A/B).

Now, P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{(2 / 5)}{(7 / 10)}=\left(\frac{2}{5} \times \frac{10}{7}\right)=\frac{4}{7} .\)

Hence, the required probability is \(\frac{4}{7}\).

**Example 3 A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?**

**Solution**

**Given**

**A die is thrown twice and the sum of the numbers appearing is observed to be 6. **

We know that when a die is thrown twice, then the sample space has 36 possible outcomes.

Let A = event that 4 appears at least once, and

B = event that the sum of the numbers appearing is 6.

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

and B = {(1,5), (2,4), (3,3), (4,2), (5,1)}.

∴ A ∩ B = {(2,4), (4,2)}.

So, P(A) = \(\frac{n(A)}{n(S)}=\frac{11}{36}, P(B) = \frac{n(B)}{n(S)}=\frac{5}{36}\)

and P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{2}{36}=\frac{1}{18}\)

Suppose B has already occured and then A occurs.

So, we have to find P(A/B).

Now, P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{(1 / 18)}{(5 / 36)}=\left(\frac{1}{18} \times \frac{36}{5}\right)=\frac{2}{5}\)

Hence, the required probability is \(\frac{2}{5}\).

**Example 4 Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (1) the youngest child is a girl, (2) at least one of the children is a girl?**

**Solution**

**Given**

**Assume that each born child is equally likely to be a boy or a girl. If a family has two children**

We may write the sample space as

S = {G_{1}G_{2}, G_{1}B_{2}, B_{1}G_{2}, B_{1}B_{2}}, where the youngest child appears later.

(1) Let A = event that both the children are girls, and

B = event that the youngest child is a girl.

Then, A = {G_{1},G_{2}}, B = {G_{1}G_{2}, B_{1}G_{2}} and A ∩ B = {G_{1},G_{2}}.

∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{1}{4}, P(B) = \frac{n(B)}{n(S)}=\frac{2}{4}=\frac{1}{2}\)

and P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{1}{4}\)

Suppose B has already occured and then A occurs.

So, we have to find P(A/B).

Now, P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{(1 / 4)}{(1 / 2)}=\left(\frac{1}{4} \times \frac{2}{1}\right)=\frac{1}{2} .\)

Hence, the required probability is \(\frac{1}{2}\).

(2) Let A = event that both the children are girls, and

E = event that at least one of the children is a girl.

Then, A = {G_{1}G_{2}}, E = {G_{1}B_{2}, B_{1}G_{2}, G_{1}G_{2}} and A ∩ E = {G_{1}G_{2}}.

∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{1}{4}, P(E) = \frac{n(E)}{n(S)}=\frac{3}{4}\)

and P(A ∩ B) = \(\frac{n(A \cap E)}{n(S)}=\frac{1}{4}\)

Suppose E has already occured and then A occurs.

So, we have to find P(A/E).

Now, P(A/E) = \(\frac{P(A \cap E)}{P(E)}=\frac{(1 / 4)}{(3 / 4)}=\left(\frac{1}{4} \times \frac{4}{3}\right)=\frac{1}{3} \text {. }\)

Hence, the required probability is \(\frac{1}{3}\).

**Example 5 An instructor has a question bank consisting of 300 easy true/false questions; 200 difficult true/false questions; 500 easy multiple-choice questions and 400 difficult multiple-choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple-choice question?**

**Solution**

**Given**

An instructor has a question bank consisting of 300 easy true/false questions; 200 difficult true/false questions; 500 easy multiple-choice questions and 400 difficult multiple-choice questions. If a question is selected at random from the question bank

Clearly, the sample space consists of 1400 questions.

∴ n(S) = 1400.

Let A = event of selecting an easy question, and

B = event of selecting a multiple-choice question.

Then, A ∩ B = event of selecting an easy multiple-choice question.

∴ n(A) = (300 + 500) = 800, n(B) = (500 + 400) = 900

and n(A ∩ B) = 500.

So, P(A) = \(\frac{n(A)}{n(S)}=\frac{800}{1400}=\frac{4}{7}, P(B) = \frac{n(B)}{n(S)}=\frac{900}{1400}=\frac{9}{14}\)

and P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{500}{1400}=\frac{5}{14}\)

Suppose B has already occured and then A occurs.

Thus we have to find P(A/B).

Now, P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{(5 / 14)}{(9 / 14)}=\left(\frac{5}{14} \times \frac{14}{9}\right)=\frac{5}{9}\)

Hence, the required probability is \(\frac{5}{9}\).

**Example 6 Two numbers are selected at random from the integers 1 through 9. If the sum is even, find the probability that both the numbers are odd.**

**Solution**

**Given**

Two numbers are selected at random from the integers 1 through 9. If the sum is eve

Out of the numbers from 1 to 9, there are 5 odd numbers and 4 even numbers.

Let A = event of choosing two odd numbers. and

B = event of choosing two numbers whose sum is even.

Then, n(A) = number of ways of choosing 2 odd numbers out of 5 = ^{5}C_{2}.

n(B) = number of ways of choosing 2 numbers whose sum is even

= (^{4}C_{2} +^{5}C_{2}) [ 2out of 4 even and 2 out of 5 odd].

n(A ∩ B) = number of ways of choosing 2 odd numbers out of 5 = ^{5}C_{2}.

Suppose B has already occurred and then A occurs.

Then, we have to find P(A/B).

Now, P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{n(A \cap B)}{n(B)}=\frac{{ }^5 C_2}{{ }^4 C_2+{ }^5 C_2}\)

= \(\frac{5 \times 4}{2} \times \frac{2}{(4 \times 3+5 \times 4)}=\frac{20}{32}=\frac{5}{8} .\)

Hence, the required probability is \(\frac{5}{8}\).

**Properties of Conditional Probability**

**Theorem 1 Let A and B be events of a sample space S of an experiment. Then, prove that P(S/B) = P(B/B) = 1.**

**Proof**

We know that:

P(S/B) = \(\frac{P(S \cap B)}{P(B)}=\frac{P(B)}{P(B)}=1\) [∵ S ∩ B = B].

And, P(B/B) = \(\frac{P(B \cap B)}{P(B)}=\frac{P(B)}{P(B)}=1 .\)

Hence, P(S/B) = P(B/B) = 1.

Theorem 2 Let A and B be two events of a sample space S and let E be an event such that P(E) ≠ 0. Then, prove that

P[(A ∪ B)/E] = P(A/E) + P(B/E) – P[(A ∩ B)/E].

Proof

We have

P[(A ∪ B)/E] = \(\frac{P[(A \cup B) \cap E]}{P(E)}\)

= \(\frac{P[(A \cap E) \cup(B \cap E)]}{P(E)}\)

= \(\frac{P(A \cap E)+P(B \cap E)-P[(A \cap E) \cap(B \cap E)]}{P(E)}\)

= \(\frac{P(A \cap E)+P(B \cap E)-P(A \cap B \cap E)}{P(E)}\)

= \(\frac{P(A \cap E)}{P(E)}+\frac{P(B \cap E)}{P(E)}-\frac{P[(A \cap B) \cap E]}{P(E)}\)

= P(A/E) + P(B/E) – P[(A ∩ B)/E].

Hence, P[(A ∪ B)/E] = P(A/E) + P(B/E) – P[(A ∩ B)/E].

**Corollary** If A and B are disjoint events, prove that

P[(A ∪ B)/E] = P(A/E) + P(B/E).

**Proof**

For any event A and B, we have

P[(A ∪ B)/E] = P(A/E) + P(B/E) – P[(A ∩ B)/E]

If A and B are disjoint, the P[(A ∩ B)/E] = 0.

Hence, in this case, we have

P[(A ∪ B)/E] = P(A/E) + P(B/E).

**Theorem 3 For any events A and B of a sample space S, prove that** \(P(\bar{A} / B)=1-P(A / B)\)**, where** \(\bar{A}\) **denotes ‘not A’.**

**Proof**

We know that

P(S/B) = 1 ⇒ \(P[(A \cup \bar{A}) / B]=1\) [∵ S = A ∪ \(\bar{A}\)]

⇒ \(P(A / B)+P(\bar{A} / B)=1\) [∵ \(A \cap \bar{A}=\varphi\)]

⇒ \(P(\bar{A} / B)=1-P(A / B)\).

**Example 7 If A and B are two events such that P(A) =** \(\frac{3}{5}\)**, P(B) =** \(\frac{7}{10}\) **and P(A ∪ B) =** \(\frac{9}{10}\)**, then find (1) P(A ∩ B) (2) P(A/B) (3) P(B/A).**

**Solution**

(1) We know that

P(A ∩ B) = P(A) + P(B) – P(A ∪ B)

= \(\left(\frac{3}{5}+\frac{7}{10}-\frac{9}{10}\right)=\frac{(6+7-9)}{10}=\frac{4}{10}=\frac{2}{5}\)

(2) P(A/B) = \(\frac{(A \cap B)}{P(B)}=\frac{(2 / 5)}{(7 / 10)}=\left(\frac{2}{5} \times \frac{10}{7}\right)=\frac{4}{7}\)

(3) P(B/A) = \(\frac{(B \cap A)}{P(A)}=\frac{P(A \cap B)}{P(A)}=\left(\frac{2 / 5}{3 / 5}\right)=\left(\frac{2}{5} \times \frac{5}{3}\right)=\frac{2}{3} .\)

**Example 8 Evaluate P(A ∪ B), if 2P(A) = P(B) =** \(\frac{6}{13}\)** and P(A/B) =** \(\frac{1}{3}\).

**Solution**

2P(A) = P(B) = \(\frac{6}{13}\) ⇒ P(A) = \(\frac{3}{13}\) and P(B) = \(\frac{6}{13}\).

∴ P(A/B) = \(\frac{P(A \cap B)}{P(B)} \Rightarrow P(A \cap B)=P(A / B) \cdot P(B)=\left(\frac{1}{3} \times \frac{6}{13}\right)=\frac{2}{13}\)

So, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= \(\left(\frac{3}{13}+\frac{6}{13}-\frac{2}{13}\right)=\frac{(3+6-2)}{13}=\frac{7}{13}\)

Hence, P(A ∪ B) = \(\frac{7}{13}\).

**Example 9 Let A and B be events such that P(A) =** \(\frac{1}{3}\)**, P(B) =** \(\frac{1}{4}\) **and P(A ∩ B) =** \(\frac{1}{5}\)**. Find: (1) P(A/B) (2) P(B/A) (3) P(A ∪ B) (4)** P(\(\bar{B} / \bar{A}\))

**Solution**

We have:

(1) P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{(1 / 5)}{(1 / 4)}=\left(\frac{1}{5} \times \frac{4}{1}\right)=\frac{4}{5} .\)

(2) P(B/A) = \(\frac{P(B \cap A)}{P(A)}=\frac{P(A \cap B)}{P(A)}=\frac{(1 / 5)}{(1 / 3)}=\left(\frac{1}{5} \times \frac{3}{1}\right)=\frac{3}{5}\)

(3) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= \(\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)=\frac{(20+15-12)}{60}=\frac{23}{60}\)

(4) \(P(\bar{B} / \bar{A})=\frac{P(\bar{B} \cap \bar{A})}{P(\bar{A})}=\frac{P(\bar{A} \cap \bar{B})}{P(\bar{A})}=\frac{P(\overline{A \cup B})}{P(\bar{A})}\)

= \(\frac{1-P(A \cup B)}{1-P(A)}=\frac{\left(1-\frac{23}{60}\right)}{\left(1-\frac{1}{3}\right)}=\frac{(37 / 60)}{(2 / 3)}=\left(\frac{37}{60} \times \frac{3}{2}\right)=\frac{37}{40}\)

**Multiplication Theorem on Probability**

Let A and B be two events associated with a sample space S. Then, the simultaneous occurrence of two events A and B is denoted by (A ∩ B) and also written as AB.

**Multiplication Theorem** Let A and B be two events associated with a sample space S. Then, prove that

P(AB) = P(A ∩ B) = P(A).P(B/A) = P(B).P(A/B), provided P(A) ≠ 0 and P(B) ≠ 0.

**Proof**

For any events A and B, we have

P(A/B) = \(\frac{P(A \cap B)}{P(B)}\), where P(A) ≠ 0.

∴ P(A ∩ B) = P(B).P(A/B). …(1)

Again, P(B/A) = \(\frac{P(B \cap A)}{P(A)}=\frac{P(A \cap B)}{P(A)}\), where P(B) ≠ 0.

∴ P(A ∩ B) = P(A).P(B/A). …(2)

From (1) and (2), we get

P(A ∩ B) = P(B).P(A/B) = P(A).P(A/B), where P(A) ≠ 0 and P(B) ≠ 0.

**Multiplication Rule for Three Events**

For any three events A, B, C of the same sample space, we have

P(A ∩ B ∩ C) = P(A).P(B/A).P[C/(A ∩ B)]

= P(A).P(B/A).(C/AB).

This rule can be extended for four or more events.

## Solved Examples

**Example 1 An urn contains 8 white and 4 red balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are white?**

**Solution**

**Given**

An urn contains 8 white and 4 red balls. Two balls are drawn from the urn one after the other without replacement.

Let A and B denote respectively the events that first and second balls both drawn are white.

Then, we have to find P(A ∩ B).

Now, P(A) = P(white ball in the first draw) = \(\frac{8}{12}\).

After the occurrence of event A, we are left with 7 white and 4 red balls.

The probability of drawing second white ball, given that the first ball drawn is white, is clearly the conditional probability of occurrence of B, given that A has occurred.

∴ P(B/A) = \(\frac{7}{11}\).

By multiplication rule o probability, we have

P(A ∩ B) = P(A).P(B/A) = \(\left(\frac{8}{12} \times \frac{7}{11}\right)=\frac{14}{33}\)

Hence, the required probability is \(\frac{14}{33}\).

The probability that both drawn balls are white is \(\frac{14}{33}\).

**Example 2 Three cards are drawn successively without replacement from a pack of 52 well-shuffled cards. What is the probability that first two cards are queens and the third card drawn is a king?**

**Solution**

**Given**

Three cards are drawn successively without replacement from a pack of 52 well-shuffled cards.

Let Q denote the event that the card drawn is a queen and K be the event that the card drawn is a king. Then, we have to find P(QQK).

Probability of drawing first queen is P(Q) = \(\frac{4}{52}\).

Now, there are 3 queens in remaining 51 cards.

Let P(Q/Q) be the probability of getting the second queen with the condition that one queen has already been drawn.

∴ P(Q/Q) = \(\frac{3}{51}\).

Lastly, P(K/QQ) is probability of third drawn card to be a king, with the condition that two queens have already been drawn.

Now, there are 4 kings in remaining 50 cards.

∴ P(K/QQ) = \(\frac{4}{50}\).

By multiplication law of probability, we have

P(QQK) = P(Q ∩ Q ∩ K)

= P(Q).P(Q/Q).P(K/QQ)

= \(\left(\frac{4}{52} \times \frac{3}{51} \times \frac{4}{50}\right)=\left(\frac{1}{13} \times \frac{1}{17} \times \frac{2}{25}\right)=\frac{2}{5525} .\)

Hence, the required probability is \(\frac{2}{5525}\).

The probability that first two cards are queens and the third card drawn is a king is \(\frac{2}{5525}\).

**Independent Events**

Two events A and B are said to be independent if

P(A/B) = P(A), where P(B) ≠ 0

and P(B/A) = P(B), where P(A) ≠ 0.

i.e., the event A does not depend on the occurrence of event B and vice versa.

**Condition for Independent of Two Events**

For any two events A and B, we have

P(A ∩ B) = P(A).P(B/A). …(1)

If A and B are independent, we have P(B/A) = P(B).

∴ (1) becomes **P(A ∩ B) = P(A) x P(B).**

Thus, two events A and B associated with the same random experiment are said to be independent if P(A ∩ B) = P(A) x P(B).

**Note:** Two events A and B are said to be dependent if they are not independent, i.e., if P(A ∩ B) ≠ P(A) x P(B).

**Difference between Two Mutually Exclusive and Independent Events**

Two events A and B are said to be mutually exclusive if A ∩ B = φ and in this case P(A ∩ B) = P(φ) = 0.

Also we know that two events A and B are independent if P(A ∩ B) = P(A) x P(B).

Clearly, two independent events with nonzero probabilities cannot be mutually exclusive.

Also, two mutually exclusive events with nonzero probabilities cannot be mutually independent.

Three events A, B and C are said to be mutually independent, if

P(A ∩ B) = P(A) x P(B), P(A ∩ C) = P(A) x P(C), P(B ∩ C) = P(B) x P(C) and P(A ∩ B ∩ C) = P(A) x P(B) x P(C).

If at least one of the above is not true for true for three given events A, B and C, then we say that these events are not independent.

**Example 3 Let E _{1} and E_{2} be two events such that P(E_{1}) = 0.3, P(E_{1} ∪ E_{2}) = 0.4 and P(E_{2}) = x. Find the value of x such that**

**(1) E _{1} and E_{2} are mutually exclusive,**

**(2) E _{1} and E_{2} are independent.**

**Solution**

(1) Let E_{1} and E_{2} be mutually exclusive. Then, E_{1} ∩ E_{2} = φ.

∴ P(E_{1} ∪ E_{2}) = P(E_{1}) + P(E_{2})

⇒ 0.4 = 0.3 + x

⇒ x = 0.1

Thus, when E_{1} and E_{2} mutually exclusive, then x = 0.1.

(2) Let E_{1} and E_{2} be two independent events. Then,

P(E_{1} ∩ E_{2}) = P(E_{1}) x P(E_{2}) = 0.3 x x = 0.3x.

∴ P(E_{1} ∪ E_{2}) = P(E_{1}) + P(E_{2}) – P(E_{1} ∩ E_{2})

⇒ 0.4 = 0.3 + x – 0.3x

⇒ 0.7x = 0.1

⇒ x = \(\frac{0.1}{0.7}=\frac{1}{7}\)

Thus, when E_{1} and E_{2} are independent, then x = \(\frac{1}{7}\).

**Example 4 Let E _{1} and E_{2} are two independent events such that P(E_{1}) = 0.35 and P(E_{1} ∪ E_{2}) = 0.60, find P(E_{2}).**

**Solution**

**Given:**

Let E_{1} and E_{2} are two independent events such that P(E_{1}) = 0.35 and P(E_{1} ∪ E_{2}) = 0.60

Let P(E_{2}) = x.

Then, E_{1} and E_{2} being independent events, we have

P(E_{1} ∩ E_{2}) = P(E_{1}) x P(E_{2}) = 0.35 x x = 0.35x.

Now, P(E_{1} ∪ E_{2}) = P(E_{1}) + P(E_{2}) – P(E_{1} ∩ E_{2})

⇒ 0.60 = 0.35 + x – 0.35x

⇒ 0.65x = 0.25

⇒ x = \(\frac{0.25}{0.65}=\frac{25}{65}=\frac{5}{13}\).

Hence, P(E_{2}) = \(\frac{5}{13}\).

**Example 5 A coin is tossed thrice. Let the event e be ‘the first throw results in a head’, and the event F be ‘the last throw results in a tail’. Find whether the events E and F are independent.**

**Solution**

**Given:**

A coin is tossed thrice. Let the event e be ‘the first throw results in a head’, and the event F be ‘the last throw results in a tail’

When a coin is tossed three times, the sample space is given by

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

Now, E = event that the first throw results in a head.

∴ E = {HHH, HHT, HTH, HTT}.

And, F = event that the last throw results in a tail.

∴ F = {HHT, THT, HTT, TTT}.

So, (E ∩ F) = {HHT, HTT}.

Clearly, n(E) = 4, n(F) = 4, n(E ∩ F) = 2 and n(S) = 8.

∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{4}{8}=\frac{1}{2}, P(F) = \frac{n(F)}{n(S)}=\frac{4}{8}=\frac{1}{2}\)

and P(E ∩ F) = \(\frac{n(E \cap F)}{n(S)}=\frac{2}{8}=\frac{1}{4}\)

Thus, P(E ∩ F) = P(E) x P(F).

Hence, E and F are independent events.

**Example 6 An unbiased die is tossed twice. Find the probability of getting a 4, 5 or 6 on the first toss and a 1, 2, 3 or 4 on the second toss.**

**Solution**

**Given**

An unbiased die is tossed twice.

In each case, the sample space is given by S = {1,2,3,4,5,6}.

Let E = event of getting a 4,5 or 6 on the first toss.

And, F = event of getting a 1,2,3 or 4 on the second toss.

Then, P(E) = \(\frac{3}{6}\) = \(\frac{1}{2}\) and P(F) = \(\frac{4}{6}\) = \(\frac{2}{3}\).

Clearly, E and F are independent events.

∴ required probability = P(E ∩ F) = P(E) x P(F) [∵ E and F are independent]

= \(\left(\frac{1}{2} \times \frac{2}{3}\right)=\frac{1}{3}\)

**Example 7 Ramesh appears for an interview for two posts, A and B, for which the selection is independent. The probability for his selection for Post A is (1/6) and for Post B, it is (1/7). Find the probability that Ramesh is selected for at least one post.**

**Solution**

**Given**

Ramesh appears for an interview for two posts, A and B, for which the selection is independent. The probability for his selection for Post A is (1/6) and for Post B, it is (1/7).

Let E_{1} = event that Ramesh is selected for the post A,

and E_{2} = event that Ramesh is selected for the post B.

Then, P(E_{1}) = \(\frac{1}{6}\) and P(E_{2}) = \(\frac{1}{7}\).

Clearly E_{1} and E_{2} are independent events.

∴ P(E_{1} ∩ E_{2}) = P(E_{1}) x P(E_{2}) = \(\left(\frac{1}{6} \times \frac{1}{7}\right)=\frac{1}{42}\).

∴ P(Ramesh is selected for at least one post)

= P(E_{1} ∪ E_{2})

= P(E_{1}) + P(E_{2}) – P(E_{1} ∩ E_{2})

= \(\left(\frac{1}{6}+\frac{1}{7}-\frac{1}{42}\right)=\frac{12}{42}=\frac{2}{7}\)

Hence, the requried probability is \(\frac{2}{7}\).

**Example 8 A can solve 90% of the problems given in a book, and B can solve 70%. What is the probability that at least one of them will solve a problem selected at random from the book?**

**Solution**

**Given**

A can solve 90% of the problems given in a book, and B can solve 70%.

Let E_{1} = event that A solves the problem,

and E_{2} = event that B solves the problem.

Then, P(E_{1}) = \(\frac{90}{100}\) = \(\frac{9}{10}\) and P(E_{2}) = \(\frac{70}{100}\) = \(\frac{7}{10}\).

Clearly, E_{1} and E_{2} are independent events.

∴ P(E_{1} ∩ E_{2}) = P(E_{1}) x P(E_{2}) = \(\left(\frac{9}{10} \times \frac{7}{10}\right)=\frac{63}{100}\)

∴ P(at least one of them will solve the problem)

= P(E_{1} ∪ E_{2})

= P(E_{1}) + P(E_{2}) – P(E_{1} ∩ E_{2})

= \(\left(\frac{9}{10}+\frac{7}{10}-\frac{63}{100}\right)=\frac{(90+70-63)}{100}=\frac{97}{100}\)

Hence, the required probability is 0.97.

**Example 9 The probability that A hits a target is (1/3) and the probability that B hits it is (2/5). What is the probability that the target will be hit if both A and B shoot at it?**

**Solution**

**Given:**

The probability that A hits a target is (1/3) and the probability that B hits it is (2/5).

Let E_{1} = event that A hits the target,

and E_{2} = event that B hits the target.

Then, P(E_{1}) = \(\frac{1}{3}\) and P(E_{2}) = \(\frac{2}{5}\).

Clearly, E_{1} and E_{2} are independent events.

∴ P(E_{1} ∩ E_{2}) = P(E_{1}) x P(E_{2}) = \(\left(\frac{1}{3} \times \frac{2}{5}\right)=\frac{2}{15}\)

∴ P(target is hit) = P(A hits or B hits)

= P(E_{1} ∪ E_{2})

= P(E_{1}) + P(E_{2}) – P(E_{1} ∩ E_{2})

= \(\left(\frac{1}{3}+\frac{2}{5}-\frac{2}{15}\right)=\frac{9}{15}=\frac{3}{5}.\)

Hence, the required probability is \(\frac{3}{5}\).

**Example 10 A and B appear for an interview for two posts. The probability of A’s selection is (1/3) and that of B’s selection is (2/5). Find the probability that only one of them will be selected.**

**Solution**

**Given:**

A and B appear for an interview for two posts. The probability of A’s selection is (1/3) and that of B’s selection is (2/5)

Let E_{1 }= event that A is selected,

and E_{2} = event that B is selected.

Then, P(E_{1}) = \(\frac{1}{3}\) and P(E_{2}) = \(\frac{2}{5}\)

⇒ \(P\left(\bar{E}_1\right)=\left(1-\frac{1}{3}\right)=\frac{2}{3} \text { and } P\left(\bar{E}_2\right)=\left(1-\frac{2}{5}\right)=\frac{3}{5} \text {. }\)

∴ P(event that only one of them is selected)

= P[(E_{1} and not E_{2}) or (E_{2} and not E_{1})]

= \(P\left[\left(E_1 \cap \bar{E}_2\right) \text { or }\left(E_2 \cap \bar{E}_1\right)\right]\)

= \(P\left(E_1 \cap \bar{E}_2\right)+P\left(E_2 \cap \bar{E}_1\right)\) [∵ \(\left(E_1 \cap \bar{E}_2\right) \cap\left(E_2 \cap \bar{E}_1\right)=\phi\)]

= \(P\left(E_1\right) \cdot P\left(\bar{E}_2\right)+P\left(E_2\right) \cdot P\left(\bar{E}_1\right)\)

[∵ E_{1} and \(\bar{E}_2\) are independent, and E_{2} and \(\bar{E}_1\) are independent]

= \(\left(\frac{1}{3} \times \frac{3}{5}\right)+\left(\frac{2}{5} \times \frac{2}{3}\right)\)

= \(\left(\frac{1}{5}+\frac{4}{15}\right)=\frac{7}{15}\)

**Example 11 A speaks the truth in 60% of the cases, and B in 90% of the cases. In what percentage of cases are they likely to contradict each other in stating the same fact?**

**Solution**

**Given:**

A speaks the truth in 60% of the cases, and B in 90% of the cases.

Let E_{1} = event that A speaks the truth,

and E_{2 }= event that B speaks the truth.

Then, \(\bar{E}_1\) = event that A tells a lie,

and \(\bar{E}_2\) = event that B tells a lie.

Clearly, E_{1} and E_{2} are independent events.

Also, (E_{1} and \(\bar{E}_2\)) as well as (\(\bar{E}_1\) and E_{2}) are independent.

Now, P(E_{1}) = \(\frac{60}{100}\) = \(\frac{3}{5}\); P(E_{2}) = \(\frac{90}{100}\) = \(\frac{9}{10}\);

∴ P(A and B contradict each other) = P[(A speaks the truth and B tells a lie) or (A tells a lie and B speaks the truth)]

= \(P\left[\left(E_1 \cap \bar{E}_2\right) \cup\left(\bar{E}_1 \cap E_2\right)\right]\)

= \(P\left(E_1 \cap \bar{E}_2\right)+P\left(\bar{E}_1 \cap E_2\right) [∵ \left(E_1 \cap \bar{E}_2\right) \cap\left(E_2 \cap \bar{E}_1\right)=\phi]\)

= \(\left\{P\left(E_1\right) \times P\left(\bar{E}_2\right)\right\}+\left\{P\left(\bar{E}_1\right) \times P\left(E_2\right)\right\}\)

= \(\left(\frac{3}{5} \times \frac{1}{10}\right)+\left(\frac{2}{5} \times \frac{9}{10}\right)=\left(\frac{3}{50}+\frac{18}{50}\right)=\frac{21}{50}\)

Percentage of cases in which A and B contradict each other

= \(\left(\frac{21}{50} \times 100\right) \%=42 \%\)

**Example 12 The probabilities of a specific problem being solved independently by A and B are 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that**

**(1) the problem is solved**

**(2) exactly one of them solves the problem.**

**Solution**

Let E_{1} = event that A solves the problem,

and E_{2} = event that B solves the problem.

Then, P(E_{1}) = \(\frac{1}{2}\) and P(E_{2}) = \(\frac{1}{3}\)

⇒ \(P\left(\bar{E}_1\right)=\left(1-\frac{1}{2}\right)=\frac{1}{2} \text { and } P\left(\bar{E}_2\right)=\left(1-\frac{1}{3}\right)=\frac{2}{3} \text {. }\)

Clearly, E_{1} and E_{2} are independent events.

∴ P(E_{1} ∩ E_{2}) = P(E_{1}) x P(E_{2}) = \(\left(\frac{1}{2} \times \frac{1}{3}\right)=\frac{1}{6}\)

(1) P(the problem is solved)

= P(at least one of A and B solves the problem)

= P(E_{1} or E_{2}) = P(E_{1} ∪ E_{2})

= P(E_{1}) + P(E_{2}) – P(E_{1} ∩ E_{2})

= \(\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\right)=\frac{4}{6}=\frac{2}{3} .\)

(2) P(exactly one of them solves the problem)

= P[(E_{1} and not E_{2}) or (E_{2} and not E_{1})]

= \(P\left(E_1 \cap \bar{E}_2\right)+P\left(E_2 \cap \bar{E}_1\right)\)

= \(P\left(E_1\right) \times P\left(\bar{E}_2\right)+P\left(E_2\right) \times P\left(\bar{E}_1\right)\)

= \(\left(\frac{1}{2} \times \frac{2}{3}\right)+\left(\frac{1}{3} \times \frac{1}{2}\right)=\left(\frac{1}{3}+\frac{1}{6}\right)=\frac{3}{6}=\frac{1}{2}\)

**Example 13 Amit and Nisha appear for an interview for two vacancies in a company. The probability of Amit’s selection is 1/5 and that of Nisha’s selection is 1/6. What is the probability that**

**(1) both of them are selected?**

**(2) only one of them is selected?**

**(3) none of them is selected?**

**Solution**

Let E_{1} = event that Amit is selected,

and E_{2} = event that Nisha is selected.

Then, P(E_{1}) = \(\frac{1}{5}\) and P(E_{2}) = \(\frac{1}{6}\).

Clearly, E_{1} and E_{2} are independent events.

(1) P(both are selected) = P(E_{1} ∩ E_{2})

= P(E_{1}) x P(E_{2})

[∵ E_{1} and E_{2} are independent]

= \(\left(\frac{1}{5} \times \frac{1}{6}\right)=\frac{1}{30} .\)

(2) P(only one of them is selected)

= P[(E_{1} and not E_{2}) or (E_{2} and not E_{1})]

= P(E_{1} and not E_{2}) + P(E_{2} and not E_{1})

= P(E_{1}).P(not E_{2}) + P(E_{2}).P(not E_{1})

= P(E_{1}).[1 – P(E_{2})] + P(E_{2}).[1 – P(E_{1})]

= \(\frac{1}{5} \cdot\left(1-\frac{1}{6}\right)+\frac{1}{6} \cdot\left(1-\frac{1}{5}\right)=\left(\frac{1}{5} \times \frac{5}{6}\right)+\left(\frac{1}{6} \times \frac{4}{5}\right)\)

= \(\left(\frac{1}{6}+\frac{2}{15}\right)=\frac{9}{30}=\frac{3}{10}\)

(3) P(none of them is selected)

= P(not E_{1} and not E_{2})

= P(not E_{1}) and P(not E_{2})

= [1 – P(E_{1})].[1 – P(E_{2})]

= \(\left(1-\frac{1}{5}\right) \cdot\left(1-\frac{1}{6}\right)=\left(\frac{4}{5} \times \frac{5}{6}\right)=\frac{2}{3}\)

**Example 14 Three groups of children contain 3 girls and 1 boy; 2 girls and 2 boys; and 1 girl and 3 boys. One child is selected at random from each group. Find the chance that the three children selected comprise 1 girl and 2 boys.**

**Solution**

**Given:**

Three groups of children contain 3 girls and 1 boy; 2 girls and 2 boys; and 1 girl and 3 boys. One child is selected at random from each group.

Let G_{1}, G_{2}, G_{3} be the events of selecting a girl from the first, second and third group respectively, and let B_{1}, B_{2}, B_{3} be the events of selecting a boy from the first, second and third group respectively.

Then,

P(G_{1}) = \(\frac{3}{4}\); P(G_{2}) = \(\frac{2}{4}\) = \(\frac{1}{2}\); P(G_{3}) = \(\frac{1}{4}\).

P(B_{1}) = \(\frac{1}{4}\); P(B_{2}) = \(\frac{2}{4}\) = \(\frac{1}{2}\) and P(B_{3}) = \(\frac{3}{4}\).

∴ P(selecting 1 girl and 2 boys)

= P[(G_{1}B_{2}B_{3}) or (B_{1}G_{2}B_{3}) or (B_{1}B_{2}G_{3})]

= P(G_{1}B_{2}B_{3}) + P(B_{1}G_{2}B_{3}) + P(B_{1}B_{2}G_{3})

= {P(G_{1})xP(B_{2})xP(B_{3})} + {P(B_{1})xP(G_{2})xP(B_{3})} + {P(B_{1})xP(B_{2})xP(G_{3})}

= \(\left(\frac{3}{4} \times \frac{1}{2} \times \frac{3}{4}\right)+\left(\frac{1}{4} \times \frac{1}{2} \times \frac{3}{4}\right)+\left(\frac{1}{4} \times \frac{1}{2} \times \frac{1}{4}\right)=\left(\frac{9}{32}+\frac{3}{32}+\frac{1}{32}\right)=\frac{13}{32}\)

Hence, the chances of selecting 1 girl and 2 boys are \(\frac{13}{32}\).

**Example 15 A problem is given to three students whose chances of solving it are 1/3, 2/7 and 3/8. What is the probability that the problem will be solved?**

**Solution**

**Given:**

A problem is given to three students whose chances of solving it are 1/3, 2/7 and 3/8.

Let the three students be named A, B, and C respectively. Let E_{1}, E_{2}, E_{3} be the events that the problem is solved by A, B, C respectively. Then,

P(E_{1}) = \(\frac{1}{3}\), P(E_{2}) = \(\frac{2}{7}\), P(E_{3}) = \(\frac{3}{8}\).

∴ \(P\left(\bar{E}_1\right)=\left(1-\frac{1}{3}\right)=\frac{2}{3} ; P\left(\bar{E}_2\right)=\left(1-\frac{2}{7}\right)=\frac{5}{7} \text { and } P\left(\bar{E}_3\right)=\left(1-\frac{3}{8}\right)=\frac{5}{8} \text {. }\)

∴ P(none solves the problem)

= P[(not E_{1}) and (not E_{2}) and (not E_{3})]

= \(P\left(\bar{E}_1 \cap \bar{E}_2 \cap \bar{E}_3\right)\)

= \(P\left(\bar{E}_1\right) \times P\left(\bar{E}_2\right) \times P\left(\bar{E}_3\right)\) [∵ \(\bar{E}_1, \bar{E}_2, \bar{E}_3\) are independent]

= \(\left(\frac{2}{3} \times \frac{5}{7} \times \frac{5}{8}\right)=\frac{25}{84}\)

∴ P(that the problem is solved)

= 1 – P(none solves the problem)

= \(\left(1-\frac{25}{84}\right)=\frac{59}{84}\)

Hence, the required probability is \(\frac{59}{84}\).

**Example 16 A problem in mathematics is given to three students whose chances of solving it correctly are 1/2, 1/3 and 1/4 respectively. What is the probability that only one of them solves it correctly?**

**Solution**

**Given**

A problem in mathematics is given to three students whose chances of solving it correctly are 1/2, 1/3 and 1/4 respectively.

Let A, B, C be the given students and let E_{1}, E_{2} and E_{3} be the events that the problem is solved by A, B, C respectively. Then, \(\bar{E}_1, \bar{E}_2, and \bar{E}_3\) are the events that the given problem is not solved by A, B, C respectively. Then,

P(E_{1}) = \(\frac{1}{2}\); P(E_{2}) = \(\frac{1}{3}\); P(E_{3}) = \(\frac{1}{4}\);

P(exactly one of them solves the problem)

= \(P\left[\left(E_1 \cap \bar{E}_2 \cap \bar{E}_3\right) \text { or }\left(\bar{E}_1 \cap E_2 \cap \bar{E}_3\right) \text { or }\left(\bar{E}_1 \cap \bar{E}_2 \cap E_3\right)\right]\)

= \(P\left(E_1 \cap \bar{E}_2 \cap \bar{E}_3\right)+P\left(\bar{E}_1 \cap E_2 \cap \bar{E}_3\right)+P\left(\bar{E}_1 \cap \bar{E}_2 \cap E_3\right)\)

= \(\left\{P\left(E_1\right) \times P\left(\bar{E}_2\right) \times P\left(\bar{E}_3\right)\right\}+\left\{P\left(\bar{E}_1\right) \times P\left(E_2\right) \times P\left(\bar{E}_3\right)\right\}+\left\{P\left(\bar{E}_1\right) \times P\left(\bar{E}_2\right) \times P\left(E_3\right)\right\}\)

= \(\left(\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}\right)+\left(\frac{1}{2} \times \frac{1}{3} \times \frac{3}{4}\right)+\left(\frac{1}{2} \times \frac{2}{3} \times \frac{1}{4}\right)\)

= \(\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{12}\right)=\frac{11}{24}\)

Hence, the required probability is \(\frac{11}{24}\).

**Example 17 Three critics review a book. For the three critics, the odds in favour of the book are (5:2), (4:3) and (3:4) respectively. Find the probability that the majority is in favour of the book.**

**Solution**

**Given**

Three critics review a book. For the three critics, the odds in favour of the book are (5:2), (4:3) and (3:4) respectively.

Let A, B, C denote the events that the book be favoured by the first, second and third critic respectively. Then,

P(A) = \(\frac{5}{7}\); P(B) = \(\frac{4}{7}\); P(C) = \(\frac{3}{7}\);

\(P(\bar{A})=\left(1-\frac{5}{7}\right)=\frac{2}{7} ; P(\bar{B})=\left(1-\frac{4}{7}\right)=\frac{3}{7} \text { and } P(\bar{C})=\left(1-\frac{3}{7}\right)=\frac{4}{7} \text {. }\)Required probability

= P(2 critics favour the book or 3 critics favour the book)

= P(2 critics favour the book) + P(3 critics favour the book)

= P[{A and B and not C} or {A and C and not B} or {B and C and not A}] + P(A and B and C)

= \(P(A \cap B \cap \bar{C})+P(A \cap \bar{B} \cap C)+P(\bar{A} \cap B \cap C)+P(A \cap B \cap C)\)

= \(\{P(A) \times P(B) \times P(\bar{C})\}+\{P(A) \times P(\bar{B}) \times P(C)\}+\{P(\bar{A}) \times P(B) \times P(C)\}+\)

\(\{P(A) \times P(B) \times P(C)\}\)= \(\left(\frac{5}{7} \times \frac{4}{7} \times \frac{4}{7}\right)+\left(\frac{5}{7} \times \frac{3}{7} \times \frac{3}{7}\right)+\left(\frac{2}{7} \times \frac{4}{7} \times \frac{3}{7}\right)+\left(\frac{5}{7} \times \frac{4}{7} \times \frac{3}{7}\right)\)

= \(\left(\frac{80}{343}+\frac{45}{343}+\frac{24}{343}+\frac{60}{343}\right)=\frac{209}{343} .\)

Hence, the required probability is \(\frac{209}{343}\).

The probability that the majority is in favour of the book is \(\frac{209}{343}\).

**Example 18 The odds against a man who is 45 years old, living till he is 70 are 7:5, and the odds against his wife who is now 36, living till she is 61 are 5:3. Find the probability that**

**(1) the couple will be alive 25 years hence**

**(2) at least one of them will be alive 25 years hence.**

**Solution**

Let E_{1} = event that the husband will be alive 25 years hence, and

E_{2} = event that the wife will be alive 25 years hence.

Then, P(E_{1}) = \(\frac{5}{12}\) and P(E_{2}) = \(\frac{3}{8}\).

∴ \(P\left(\bar{E}_1\right)=\left(1-\frac{5}{12}\right)=\frac{7}{12} \text { and } P\left(\bar{E}_2\right)=\left(1-\frac{3}{8}\right)=\frac{5}{8} \text {. }\)

Clearly, E_{1} and E_{2} are independent events.

(1) P(the couple will be alive 25 years hence)

= P(E_{1} and E_{2}) = P(E_{1} ∩ E_{2})

= \(P\left(E_1\right) \cdot P\left(E_2\right)=\left(\frac{5}{12} \times \frac{3}{8}\right)=\frac{5}{32} .\)

(2) P(at least one of them will be alive 25 years hence)

= 1 – P(none will be alive 25 years hence)

= 1 – P[(not E_{1}) and (not E_{2})]

= 1 – \(P\left(\bar{E}_1 \cap \bar{E}_2\right)\)

= 1 – \(\left[P\left(\bar{E}_1\right) \cdot P\left(\bar{E}_2\right)\right]\) [∵ \(\bar{E}_1\) and \(\bar{E}_2\) are independent]

= \(1-\left(\frac{7}{12} \times \frac{5}{8}\right)=\left(1-\frac{35}{96}\right)=\frac{61}{96} .\)

**Example 19 A, B and C shoot to hit a target. Iff A hits the target 4 times in 5 trails; B hits it 3 times in 4 trails and C hits it 2 times in 3 trails, what is the probability that the target is hit by at least 2 persons?**

**Solution**

**Given**

A, B and C shoot to hit a target. Iff A hits the target 4 times in 5 trails; B hits it 3 times in 4 trails and C hits it 2 times in 3 trails,

Let E_{1}, E_{2} and E_{3} be the events that A hits the target, B hits the target and C hits the target respectively. Then,

P(E_{1}) = \(\frac{4}{5}\), P(E_{2}) = \(\frac{3}{4}\), P(E_{3}) = \(\frac{2}{3}\);

**Case 1** A, B, C all hit the target

In this case, P(A, B and C all hit the target)

= P(E_{1} and E_{2} and E_{3})

= P(E_{1}).P(E_{2}).P(E_{3}) [∵ E_{1}, E_{2}, E_{3} are independent]

= \(\left(\frac{4}{5} \times \frac{3}{4} \times \frac{2}{3}\right)=\frac{2}{5}\)

**Case 2** A and B hit but not C

In this case, P(A and B hit but not C)

= P(E_{1} and E_{2} and not E_{3})

= P(E_{1} ∩ E_{2} ∩ \(\bar{E}_3\))

= P(E_{1}).P(E_{2}).P(\(\bar{E}_3\)) [∵ E_{1}, E_{2}, \(\bar{E}_3\) are independent]

= \(\left(\frac{4}{5} \times \frac{3}{4} \times \frac{1}{3}\right)=\frac{1}{5}\)

**Case 3** A and C both hit but not B

In this case, P(A and C hit but not B)

= P(E_{1 }and E_{3} and \(\bar{E}_2\))

= P(E_{2}).P(E_{3}).P(\(\bar{E}_1\)) [∵ E_{2}, E_{3}, \(\bar{E}_1\) are independent]

= \(\left(\frac{3}{4} \times \frac{2}{3} \times \frac{1}{5}\right)=\frac{1}{10}\)

Clearly, all these are mutually exclusive.

Hence, required probability = \(\left(\frac{2}{5}+\frac{1}{5}+\frac{2}{15}+\frac{1}{10}\right)=\frac{5}{6}\)

## Chapter 2 Bayes’ Theorem And Its Applications

**Theorem of Total Probability**

**Theorem** **Let E _{1}, E_{2},….,E_{n} be mutually exclusive and exhaustive events associated with a random experiment and let E be an event that occurs with some E_{i}. then, prove that**

**P(E) =** \(\sum_{i=1}^n P\left(E / E_i\right) \cdot P\left(E_j\right)\)

**Proof**

Let S be the sample space. Then,

S = E_{1} ∪ E_{2} ∪ … E_{n} and E_{i} ∩ E_{j }= Φ for i ≠ j.

∴ E = E ∩ S = E ∩(E_{1} ∪ E_{2} ∪ … ∪ E_{n})

= (E ∩ E_{1}) ∪ (E ∩ E_{2})∪ … ∪(E ∩ E_{n})

⇒ P(E) = P{(E ∩ E_{1}) ∪ (E ∩ E_{2}) ∪ … ∪(E ∩ E_{n})}

= P(E ∩ E_{1}) + P(E ∩ E_{2}) + … + P(E ∩ E_{n})

{∵ (E ∩ E_{1}), (E ∩ E_{2}), …,(E ∩ E_{n}) are pairwise disjoint}

= P(E/E_{1}).P(E_{1}) + P(E/E_{2}).P(E_{2}) + … + P(E/E_{n}).P(E_{n}) [by multiplication theorem]

P(E)= \(\sum_{i=1}^n P\left(E / E_i\right) \cdot P\left(E_i\right)\)

**Example There are three urns containing 3 white and 2 black balls; 2 white and 3 black balls; 1 black and 4 white balls respectively.**

**Solution**

Let E_{1}, E_{2} and E_{3} be the events of choosing the first, second and third urn respectively. Then,

P(E_{1}) = P(E_{2}) = P(E_{3}) = \(\frac{1}{3}\).

Let E be the event that a white ball is drawn. Then,

P(E/E_{1}) = \(\frac{3}{5}\), P(E/E_{2}) = \(\frac{2}{5}\) and P(E/E_{3}) = \(\frac{4}{5}\).

By the theorem of total probability, we have

P(E) = P(E/E_{1}).P(E_{1}) + P(E/E_{2}).P(E_{2}) + P(E/E_{3}).P(E_{3})

= \(\left(\frac{3}{5} \times \frac{1}{3}+\frac{2}{5} \times \frac{1}{3}+\frac{4}{5} \times \frac{1}{3}\right)=\left(\frac{1}{5}+\frac{2}{15}+\frac{4}{15}\right)=\frac{9}{15}=\frac{3}{5} .\)

**Bayes’ Theorem** Let E_{1}, E_{2}, …, E_{n} be mutually exclusive and exhaustive events, associated with a random experiment, and let E be any event that to occurs with some E_{i}. Then,

**Proof**

By the theorem of total probability, we have

\(P(E)=\sum_{i=1}^n P\left(E / E_i\right) \cdot P\left(E_i\right)\) …(1)

∴ \(P\left(E_i / E\right)=\frac{P\left(E \cap E_i\right)}{P(E)}\) [by multiplication theorem]

= \(\frac{P\left(E / E_i\right) \cdot P\left(E_i\right)}{P(E)}\left[\quad P\left(E / E_i\right)=\frac{P\left(E \cap E_i\right)}{P\left(E_i\right)}\right]\)

= \(\frac{P\left(E / E_i\right) \cdot P\left(E_i\right)}{\sum_{i=1}^n P\left(E / E_i\right) \cdot P\left(E_i\right)}\) [using(1)].

Hence, P(E_{i}/E) = \(\frac{P\left(E / E_i\right) \cdot P\left(E_i\right)}{\sum_{i=1}^n P\left(E / E_i\right) \cdot P\left(E_i\right)}\)

## Solved Examples

**Example 1 A factory has three machines, X, Y and Z, producing 1000, 2000 and 3000 bolts per day respectively. The machine X produces 1% defective bolts, Y produces 1.5% defective bolts and Z produces 2% defective bolts. At the end of the day, a bolt is drawn at random and it is found to be defective. What is the probability that this defective bolt has been produced by the machine X?**

**Solution**

**Given:**

A factory has three machines, X, Y and Z, producing 1000, 2000 and 3000 bolts per day respectively. The machine X produces 1% defective bolts, Y produces 1.5% defective bolts and Z produces 2% defective bolts. At the end of the day, a bolt is drawn at random and it is found to be defective.

Total number of bolts produced in a day = (1000 + 2000 + 3000) = 6000.

Let E_{1}, E_{2} and E_{3} be the events of drawing a bolt produced by machines X, Y and Z respectively. Then,

P(E_{1}) = \(\frac{1000}{6000}\) = \(\frac{1}{6}\); P(E_{2}) = \(\frac{2000}{6000}\) = \(\frac{1}{3}\) and P(E_{3}) = \(\frac{3000}{6000}\) = \(\frac{1}{2}\).

Let E be the event of drawing a defective bolt. Then,

P(E/E_{1}) = probability of drawing a defective bolt, given that it is produced by the machine X

= \(\frac{1}{100}\).

P(E/E_{2}) = probability of drawing a defective bolt, given that it is produced by the machine Y

= \(\frac{1.5}{100}\) = \(\frac{15}{1000}\) = \(\frac{3}{200}\).

P(E/E_{3}) = probability of drawing a defective bolt, given that it is produced by the machine Z

= \(\frac{2}{100}\) = \(\frac{1}{50}\).

Required probability

= P(E_{1}/E)

= probability that the bolt drawn is produced by X, given that it is defective

= \(\frac{P\left(E_1\right) \cdot P\left(E / E_1\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)+P\left(E_3\right) \cdot P\left(E / E_3\right)}\)

= \(\frac{\left(\frac{1}{6} \times \frac{1}{100}\right)}{\left(\frac{1}{6} \times \frac{1}{100}\right)+\left(\frac{1}{3} \times \frac{3}{200}\right)+\left(\frac{1}{2} \times \frac{1}{50}\right)}\)

= \(\left(\frac{1}{600} \times \frac{600}{10}\right)=\frac{1}{10}=0.1\)

Hence, the required probability is 0.1.

**Example 2 In a bolt factory, three machines, A, B, C, manufacture 5%, 35% and 40% of the total production respectively. Of thier respective outputs, 5%, % and 2% are defective. Find the probability that it was manufactured by the machine C.**

**Solution**

**Given**

In a bolt factory, three machines, A, B, C, manufacture 5%, 35% and 40% of the total production respectively. Of thier respective outputs, 5%, % and 2% are defective.

Let E_{1}, E_{2} and E_{3} be the events of drawing a bolt produced by machine A, B and C respectively. Then,

P(E_{1}) = \(\frac{25}{100}\) = \(\frac{1}{4}\), P(E_{2}) = \(\frac{35}{100}\) = \(\frac{7}{20}\), and P(E_{3}) = \(\frac{40}{100}\) = \(\frac{2}{5}\).

Let E be the event of drawing a defective bolt. Then,

P(E/E_{1}) = probability of drawing a defective bolt, given that it is produced by the machine A

= \(\frac{5}{100}\) = \(\frac{1}{20}\).

P(E/E_{2}) = probability of drawing a defective bolt, given that it is produced by the machine B

= \(\frac{2}{100}\) = \(\frac{1}{50}\).

Probability that the bolt drawn is manufactured by C, given that it is defective

= P(E_{3}/3)

= \(\frac{P\left(E / E_3\right) \cdot P\left(E_3\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)+P\left(E / E_3\right) \cdot P\left(E_3\right)}\) [by Bayes’ theorem]

= \(\frac{\left(\frac{1}{50} \times \frac{2}{5}\right)}{\left(\frac{1}{20} \times \frac{1}{4}\right)+\left(\frac{1}{25} \times \frac{7}{20}\right)+\left(\frac{1}{50} \times \frac{2}{5}\right)}=\left(\frac{1}{125} \times \frac{2000}{69}\right)=\frac{16}{69} \text {. }\)

Hence, the required probability is \(\frac{16}{69}\).

**Example 3 A company has two plants to manufacture bicycles. The first plant manufactures 60% of the bicycles and the second plant, 40%. Also, 80% of the bicycles are rated of standard quality at the first plant and 90% of standard quality at the second plant. A bicycle is picked up at random and found to be of standard quality. Find the probability that it comes from the second plant.**

**Solution**

**Given:**

A company has two plants to manufacture bicycles. The first plant manufactures 60% of the bicycles and the second plant, 40%. Also, 80% of the bicycles are rated of standard quality at the first plant and 90% of standard quality at the second plant. A bicycle is picked up at random and found to be of standard quality.

Let E_{1} and E_{2} be the events of choosing a bicycle from the first plant and the second plant respectively. Then,

P(E_{1}) = \(\frac{60}{10}\) = \(\frac{3}{5}\), and P(E_{2}) = \(\frac{40}{100}\) = \(\frac{2}{5}\).

Let E be the event of choosing a bicycle of standard quality. then,

P(E/E_{1}) = probability of choosing a bicycle of standardd quality, given that it is produced by the first plant

= \(\frac{80}{100}\) = \(\frac{4}{5}\).

P(E/E_{2}) = probability of choosing a bicycle of standard quality, given that it is produced by the second plant

= \(\frac{90}{100}\) = \(\frac{9}{10}\).

The required probability

P(E_{2}/E) = probability of choosing a bicycle from the second plant, given that it is of standard quality

= \(\frac{P\left(E_2\right) \cdot P\left(E / E_2\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)}\) [by Bayes’ theorem]

= \(\frac{\left(\frac{2}{5} \times \frac{9}{10}\right)}{\left(\frac{3}{5} \times \frac{4}{5}\right)+\left(\frac{2}{5} \times \frac{9}{10}\right)}=\frac{3}{7} .\)

**Example 4 An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident involving a scooter, a car and a truck is** \(\frac{1}{100}\), \(\frac{3}{100}\) **and** \(\frac{3}{20}\) **respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?**

**Solution**

**Given:**

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident involving a scooter, a car and a truck is \(\frac{1}{100}\), \(\frac{3}{100}\) and \(\frac{3}{20}\) respectively. One of the insured persons meets with an accident.

Total number of persons insured = (2000 + 4000 + 6000) = 12000.

Let E_{1}, E_{2} and E_{3} be the events of choosing a scooter driver, a car driver and a truck driver respectively. Then,

P(E_{1}) = \(\frac{2000}{12000}\) = \(\frac{1}{6}\), P(E_{2}) = \(\frac{4000}{12000}\) = \(\frac{1}{3}\), and P(E_{3}) = \(\frac{6000}{12000}\) = \(\frac{1}{2}\).

Let E be the event of an insured person meeting with an accident. Then,

P(E/E_{1}) = probability that an insured person meets with an accident, given that he is a scooter driver

= \(\frac{1}{100}\)

Similarly, P(E/E_{2}) = \(\frac{3}{100}\) and P(E/E_{3}) = \(\frac{3}{20}\).

Required Probability

= P(E_{1}/E) [by Bayes’ theorem]

= probability of choosing a scooter driver, given that he meets with an accident

= \(\frac{P\left(E / E_1\right) \cdot P\left(E_1\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)+P\left(E / E_3\right) \cdot P\left(E_3\right)}\)

= \(\frac{\left(\frac{1}{100} \times \frac{1}{6}\right)}{\left(\frac{1}{100} \times \frac{1}{6}\right)+\left(\frac{3}{100} \times \frac{1}{3}\right)+\left(\frac{3}{20} \times \frac{1}{2}\right)}=\frac{1}{52} .\)

Hence, the required probability is \(\frac{1}{52}\).

**Example 5 A doctor is to visit a patient. From past experience, it is known that the probabilities that he will come by train, bus, scooter or by car are respectively** \(\frac{3}{10}\)**,** \(\frac{1}{5}\), \(\frac{1}{10}\) **and** \(\frac{2}{5}\)**. The probabilities that he will be late are** \(\frac{1}{4}\), \(\frac{1}{3}\) **and** \(\frac{1}{12}\)**, if he comes by train, bus and scooter respectively; but if he comes by car, he will not be late. When he arrives, he is late. What is the probability that he come by train?**

**Solution**

**Given:**

A doctor is to visit a patient. From past experience, it is known that the probabilities that he will come by train, bus, scooter or by car are respectively \(\frac{3}{10}\), \(\frac{1}{5}\), \(\frac{1}{10}\) and \(\frac{2}{5}\). The probabilities that he will be late are \(\frac{1}{4}\), \(\frac{1}{3}\) and \(\frac{1}{12}\), if he comes by train, bus and scooter respectively; but if he comes by car, he will not be late.

Let E_{1}, E_{2}, E_{3} and E_{4} be the events that the doctor comes by train, bus, scooter and car respectively. Then,

P(E_{1}) = \(\frac{3}{10}\), P(E_{2}) = \(\frac{1}{5}\), P(E_{3}) = \(\frac{1}{10}\) and P(E_{4}) = \(\frac{2}{5}\).

Let E be the event that the doctor is late. Then,

P(E/E_{1}) = probability that the doctor is late, given that he comes by train

= \(\frac{1}{4}\).

P(E/E_{2}) = probability that the doctor is late, given that he comes by b us

= \(\frac{1}{3}\).

P(E/E_{3}) = probability that the doctor is late, given that he comes by scooter

= \(\frac{1}{12}\).

P(E/E_{4}) = probability that the doctor is late, given that he comes by car

= 0.

Probability that he comes by train, given that he is late

= P(E_{1}/E)

= \(\frac{P\left(E_1\right) \cdot P\left(E / E_1\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)+P\left(E_3\right) \cdot P\left(E / E_3\right)+P\left(E_4\right) \cdot P\left(E / E_4\right)}\)

[by Bayes’ theorem]

= \(\frac{\left(\frac{3}{10} \times \frac{1}{4}\right)}{\left(\frac{3}{10} \times \frac{1}{4}\right)+\left(\frac{1}{5} \times \frac{1}{3}\right)+\left(\frac{1}{10} \times \frac{1}{12}\right)+\left(\frac{2}{5} \times 0\right)}=\left(\frac{3}{40} \times \frac{120}{18}\right)=\frac{1}{2} \text {. }\)

Hence, the required probability is \(\frac{1}{2}\).

**Example 6 A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.**

**Solution**

**Given**

A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six.

In a throw of a die, let

E_{1} = event of getting a six,

E_{2} = event of not getting a six, and

E = event that the man reports that it is a six.

Then, P(E_{1}) = \(\frac{1}{6}\), and \(P\left(E_2\right)=\left(1-\frac{1}{6}\right)=\frac{5}{6}\).

P(E/E_{1}) = [probability that the man reports that six occurs, when six has actually occurred

= probability that the man speaks the truth

= \(\frac{3}{4}\).

P(E/E_{2}) = probability that the man reports that six occurs, when six has not actually occurred

= probability that the man does not speak the truth

= \(\left(1-\frac{3}{4}\right)=\frac{1}{4}\).

Probability of getting a six, given that the man reports it to be six

= P(E_{1}/E)

= \(\frac{P\left(E / E_1\right) \cdot P\left(E_1\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)}\) [by Bayes’ theorem]

= \(\frac{\left(\frac{3}{4} \times \frac{1}{6}\right)}{\left(\frac{3}{4} \times \frac{1}{6}\right)+\left(\frac{1}{4} \times \frac{5}{6}\right)}=\left(\frac{1}{8} \times 3\right)=\frac{3}{8} .\)

Hence, the required probability is \(\frac{3}{8}\).

**Example 7 In an examination, an examine either guesses or copies or knows the answer to a multiple-choice question with four choices. The probability that he makes a guess is (1/3) and the probability that he copies the answer is (1/6). The probability that his answer is correct, given that he copied it, is (1/8). The probability that his answer is correct, given that he guessed it, is (1/4). Find the probability that he knew the answer to the question, given that he correctly answered it.**

**Solution**

**Given**

In an examination, an examine either guesses or copies or knows the answer to a multiple-choice question with four choices. The probability that he makes a guess is (1/3) and the probability that he copies the answer is (1/6). The probability that his answer is correct, given that he copied it, is (1/8). The probability that his answer is correct, given that he guessed it, is (1/4).

Let E_{1} = event that the examinee guesses the answer,

E_{2} = event that he copies the answer,

E_{3} = event that he knows the answer, and

E = event that he answers correctly.

Then, P(E_{1}) = \(\frac{1}{3}\), P(E_{2}) = \(\frac{1}{6}\), and P(E_{3}) = \(1-\left(\frac{1}{3}+\frac{1}{6}\right)=\frac{1}{2}\)

[∵ E_{1}, E_{2}, E_{3} are mutually exclusive and exhaustive].

∴ P(E/E_{1}) = probability that he answers correctly, given that he guesses

= \(\frac{1}{4}\).

P(E/E_{2}) = probability that he answers correctly, given that he copies

= \(\frac{1}{8}\).

P(E/E_{3}) = probability that he answers correctly, given that he knew the answer

= 1.

Required probability

= P(E_{3}/E)

= \(\frac{P\left(E / E_3\right) \cdot P\left(E_3\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)+P\left(E / E_3\right) \cdot P\left(E_3\right)}\)

[by Bayes’ theorem]

= \(\frac{\left(1 \times \frac{1}{2}\right)}{\left(\frac{1}{4} \times \frac{1}{3}\right)+\left(\frac{1}{8} \times \frac{1}{6}\right)+\left(1 \times \frac{1}{2}\right)}=\frac{24}{29} .\)

Hence, the required probability is \(\frac{24}{29}\).

**Example 8 By examining the chest X-ray, the probability that a person is diagnosed with TB when he is actually suffering from it, is 0.99. The probability that the doctor incorrectly diagnoses a person to be having TB, on the basis of X-ray reports, is 0.001. In a certain city, 1 in 1000 persons suffers from TB. A person is selected at random and is diagnosed to have TB. What is the chance that he actually has TB?**

**Solution**

**Given:**

By examining the chest X-ray, the probability that a person is diagnosed with TB when he is actually suffering from it, is 0.99. The probability that the doctor incorrectly diagnoses a person to be having TB, on the basis of X-ray reports, is 0.001. In a certain city, 1 in 1000 persons suffers from TB. A person is selected at random and is diagnosed to have TB.

Let E = event that the doctor diagnoses TB,

E_{1} = event that the person selected is suffering from TB, and

E_{2} = event that the person selected is not suffering from TB.

Then, P(E_{1}) = \(\frac{1}{1000}\) and P(E_{2}) = \(\left(1-\frac{1}{1000}\right)=\frac{999}{1000}\)

P(E/E_{1}) = probability that TB is diagnosed, when the person actually has TB

= \(\frac{99}{100}\).

P(E\E_{2}) = probability that Tb is diagnosed, when the person has no TB

= \(\frac{1}{1000}\).

Using Bayes’ theorem, we have

P(E_{1}/E) = probability of a person actually having TB, if it is known that he is diagnosed to have TB

= \(\frac{P\left(E / E_1\right) \cdot P\left(E_1\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)}\)

= \(\frac{\left(\frac{99}{100} \times \frac{1}{1000}\right)}{\left(\frac{99}{100} \times \frac{1}{1000}\right)+\left(\frac{1}{1000} \times \frac{999}{1000}\right)}=\frac{110}{221} .\)

Hence, the required probability is \(\frac{110}{221}\).

**Example 9 Bag A contains 2 white and 3 red balls, and bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bag and it is found to be red. Find the probability that it was drawn from bag B.**

**Solution**

**Given**

Bag A contains 2 white and 3 red balls, and bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bag and it is found to be red.

Let E_{1} = event of choosing bag A,

E_{2} = event of choosing bag B, and

E = event of drawing a red ball.

Then, P(E_{1}) = \(\frac{1}{2}\) and P(E_{2}) = \(\frac{1}{2}\).

Also, P(E/E_{1}) = event of drawing a red ball from bag A = \(\frac{3}{5}\), and

P(E/E_{2}) = event of drawing a red ball from bag B = \(\frac{5}{9}\).

Probability of drawing a ball from B, it being given that it is red

= P(E_{2}/E)

= \(\frac{P\left(E / E_2\right) \cdot P\left(E_2\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)}\)

[by Bayes’ theorem]

= \(\frac{\left(\frac{5}{9} \times \frac{1}{2}\right)}{\left(\frac{3}{5} \times \frac{1}{2}\right)+\left(\frac{5}{9} \times \frac{1}{2}\right)}=\frac{25}{52} .\)

Hence, the required probability is \(\frac{25}{52}\).

**Example 10 There are 5 bags, each containing 5 white balls and 3 black balls. Also, there are 6 bags, each containing 2 white balls and 4 black balls. A white ball is drawn at random. Find the probability that this white ball is from a bag of the first group.**

**Solution**

**Given**

There are 5 bags, each containing 5 white balls and 3 black balls. Also, there are 6 bags, each containing 2 white balls and 4 black balls. A white ball is drawn at random.

Let E_{1} = event of selecting a bag from the first group,

E_{2} = event of selecting a bag from the second group, and

E = event of drawing a white ball.

Then, P(E_{1}) = \(\frac{5}{11}\) and P(E_{2}) = \(\frac{6}{11}\).

P(E/E_{1}) = probability of getting a white ball, given that it is from a bag of the first group

= \(\frac{5}{8}\).

P(E/E_{2}) = probability of getting a white ball, given that it is from a bag of the second group

= \(\frac{2}{6}\) = \(\frac{1}{3}\).

Probability of getting the ball from a bag of the first group, given that it is white

= P(E_{1}/E)

= \(\frac{P\left(E / E_1\right) \cdot P\left(E_1\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)}\) [by Bayes’ theorem]

= \(\frac{\left(\frac{5}{8} \times \frac{5}{11}\right)}{\left(\frac{5}{8} \times \frac{5}{11}\right)+\left(\frac{1}{3} \times \frac{6}{11}\right)}=\frac{75}{123}\)

**Example 11 Urn A contains 1 white, 2 black and 3 red balls; urn B contains 2 white, 1 black and 1 red ball; and urn C contains 4 white, 5 black and 3 red balls. One urn is chosen at random and two balls are drawn. These happen to be one white and one red. What is the probability that they come from urn A?**

**Solution**

**Given**

Urn A contains 1 white, 2 black and 3 red balls; urn B contains 2 white, 1 black and 1 red ball; and urn C contains 4 white, 5 black and 3 red balls. One urn is chosen at random and two balls are drawn. These happen to be one white and one red.

Let E_{1}, E_{2}, E_{3} be the events that the balls are drawn from urn A, urn B and urn C respectively, and let E be the event that the balls drawn are one white and one red. Then, P(E_{1}) = P(E_{2}) = P(E_{3}) = \(\frac{1}{3}\).

P(E/E_{1}) = probability that the balls drawn are one white and one red, given that the balls are from urn A

= \(\frac{{ }^1 C_1 \times{ }^3 C_1}{{ }^6 C_2}=\frac{3}{15}=\frac{1}{5}\)

P(E/E_{2}) = probability that the balls drawn are one white and one red, given that the balls are from urn B

= \(\frac{{ }^2 C_1 \times{ }^1 C_1}{{ }^4 C_2}=\frac{2}{6}=\frac{1}{3} .\)

P(E/E_{3}) = probability that the balls drawn are one white and one red, given that the balls are from urn C

= \(\frac{{ }^4 C_1 \times{ }^3 C_1}{{ }^{12} C_2}=\frac{12}{66}=\frac{2}{11}\)

Probability that the balls drawn are from urn A, it being given that the balls drawn are one white and one red

= P(E_{1}/E)

= \(\frac{P\left(E / E_1\right) \cdot P\left(E_1\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)+P\left(E / E_3\right) \cdot P\left(E_3\right)}\)

[by Bayes’ theorem]

= \(\frac{\left(\frac{1}{5} \times \frac{1}{3}\right)}{\left(\frac{1}{5} \times \frac{1}{3}\right)+\left(\frac{1}{3} \times \frac{1}{3}\right)+\left(\frac{2}{11} \times \frac{1}{3}\right)}\)

= \(\left(\frac{1}{15} \times \frac{495}{118}\right)=\frac{33}{118}\)

Hence, the required probability is \(\frac{33}{118}\).

**Example 12 A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both spades. Find the probability of the lost card being a spade.**

**Solution**

**Given:**

A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both spades.

Let E_{1}, E_{2}, E_{3} and E_{4} be the events of losing a card of spades, clubs, hearts and diamonds respectively.

Then, P(E_{1}) = P(E_{2}) = P(E_{3}) = P(E_{4}) = \(\frac{13}{52}\) = \(\frac{1}{4}\).

Let E be the event of drawing 2 spades from the remaining 51 cards. Then,

P(E/E_{1}) = probability of drawing 2 spades, given that a card of spades is missing

= \(\frac{{ }^{12} C_2}{{ }^{51} C_2}=\frac{(12 \times 11)}{2 !} \times \frac{2 !}{(51 \times 50)}=\frac{22}{425} .\)

P(E/E_{2}) = probability of drawing 2 spades, given that a card of clubs is missing

= \(\frac{{ }^{13} C_2}{{ }^{51} C_2}=\frac{(13 \times 12)}{2 !} \times \frac{2 !}{(51 \times 50)}=\frac{26}{425}\)

P(E/E_{3}) = probability of drawing 2 spades, given that a card of hearts is missing

= \(\frac{{ }^{13} C_2}{{ }^{51} C_2}=\frac{26}{425}\)

P(E/E_{4}) = probability of drawing 2 spades, given that a card of diamonds is missing

= \(\frac{{ }^{13} C_2}{{ }^{51} C_2}=\frac{26}{425}\)

∴ P(E_{1}/E) = probability of the lost card being a spade, given that 2 spades are drawn from the remaining 51 cards

= \(\frac{P\left(E_1\right) \cdot P\left(E / E_1\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)+P\left(E_3\right) \cdot P\left(E / E_3\right)+P\left(E_4\right) \cdot P\left(E / E_4\right)}\)

= \(\frac{\left(\frac{1}{4} \times \frac{22}{425}\right)}{\left(\frac{1}{4} \times \frac{22}{425}\right)+\left(\frac{1}{4} \times \frac{26}{425}\right)+\left(\frac{1}{4} \times \frac{26}{425}\right)+\left(\frac{1}{4} \times \frac{26}{425}\right)}\)

= \(\frac{22}{100}\) = 0.22.

Hence, the required probability is 0.22.