P(A|B)

P(A|B) denotes conditional probability. As far as any competitive exam is concerned, conditional probability P(A|B) has great importance. In this article, we learn the definition of conditional probability P(A|B), formula, and solved examples on conditional probability. Bayes’ theorem describes the probability of occurrence of an event related to any condition. This theorem is considered for the case of conditional probability.

P(A|B) Definition

Conditional probability is the probability of occurrence of any event A, when another event B in relation to A has already occurred. This also means the probability of event A depends on another event B. Venn diagram for conditional probability P(A|B) is given below.

P(A|B) Formula

Let P(A) be the probability of occurrence of event A, P(B) be the probability of occurrence of event B and P(A∩B) be the probability of happening of both A and B.

P(A|B) formula is given by

P(A|B) = P(A∩B)/P(B)

P(B|A) = P(A∩B)/P(A)

From these formulas, we can derive the product formulas of probability.

P(A∩B) = P(A|B) × P(B)

P(A∩B) = P(B|A) × P(A)

If A and B are independent events, then P(A|B) = P(A) or P(B|A) = P(B). If A and B are independent events, then P(A∩B) = P(A). P(B)

So P(A|B) = P(A). P(B)/P(B) = P(A)

Note: P(A|B)is not defined if P(B) = 0.

Let us discuss some special cases of conditional probability (P(A|B)).

Case 1: If A and B are disjoint.

Then A∩B = Ø

So P(A|B) = 0

When A and B are disjoint they cannot both occur at the same time. Thus, given that B has occurred, the probability of A must be zero.

Case 2: B is a subset of A

Then A∩B = B

So P(A|B) = P(A∩B)/P(B)

= P(B)/P(B)

= 1

Case 3: A is a subset of B

Then A∩B = A

So P(A|B) = P(A∩B)/P(B)

=> P(A|B) = P(A)/P(B)

Let us look at some solved examples.

Also Read

Properties of Probability

Probability JEE Main Previous Year Questions With Solutions

Solved Examples on P(A|B)

Example 1: A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared at least once is:

(a) 1/3

(b) 1/4

(c) 1/8

(d) 1/9

Solution:

Let A be the event that the sum obtained is a multiple of 4.

B be the event that the score of 4 has appeared at least once.

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

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

(A ∩ B) = (4, 4)

n(A ∩ B) = 1

Required probability = P(B|A)

= P(A ∩ B)/P(A)

= 1/9

Hence, option d is the answer.

Example 2: A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed, and this ball along with two additional balls of the same colour are returned to the bag. If a ball is now drawn at random from the bag, the probability that this drawn ball is red, is :

(a) 1/5

(b) 3/4

(c) 3/10

(d) 2/5

Solution:

There can be 2 possible cases: the first ball drawn is red or is black.

P(first ball black) = 6/10

P(first ball red) = 4/10

P(second ball red | first ball drawn is black) = 4/12

P(second ball red | first ball drawn is red) = 6/12 = 1/2

The probability of the second ball being red, P(R) = (4/10)×(1/2)+(6/10) ×(4/12)

= (1/5)+(1/5)

= 2/5

Hence, option (d) is the answer.

Example 3: There are three bags B₁, B₂, and B₃. Bag B₁ contains 5 red and 5 green balls, B₂ contains 3 red and 5 green balls, and B₃ contains 5 red and 3 green balls, Bags B₁, B₂ and B₃ have probabilities 3/10, 3/10, and 4/10, respectively, of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?

(a) Probability that the selected bag is B₃ and the chosen ball is green equals 3/10

(b) Probability that the chosen ball is green equals 39/80

(c) Probability that the chosen ball is green, given that the selected bag is B₃, equals 3/8

(d) Probability that the selected bag is B₃, given that the chosen balls is green, equals 5/13

Solution:

(i) P(B₃⋂G) = P(G|B₃).P(B₃)

= (⅜) × (4/10)

= 3/20

(ii) P(G) = (5/10)×(3/10) + (⅝)×(3/10) + (⅜)×(4/10)

= (60 + 75 + 60)/400

= 195/400

= 39/80

(iii) P(G|B₃) = 3/8

(iv) P(B₃|G) = P(G⋂B₃)/P(G)

= (3/20)/(39/80)

= 4/13

Hence, option b and c are correct.

Example 4: If A and B are two independent events such that P(A) = 1/2 and P(B) = 1/5, then P(A|B) equals

(a) ½

(b) ¼

(c) 1/5

(d) 1

Solution:

For independent events P(A⋂B) = P(A) ×P(B)

= (½)×(1/5)

= 1/10

P(A|B) = P(A⋂B)/P(B)

= (1/10)/(⅕)

= ½

Hence, option a is the answer.

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