Conditional Probability

Two events A & B are expressed to be dependent when B can occur only when A is acknowledge to have occurred or vice-versa. Probability attached to such type of event is known as the conditional probability and is represented by P(A/B), and in another words the probability of A given B has occurred.

If two events A & B are dependent, so the conditional probability of B given A is:

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

Proof: Assume a1 is the number of cases for the simultaneous happening of A & B out of a1 + a2 situation in which A can happen with or without happening of B.

∴ P(B/A) = a₁/(a₁ + a₂) = (a₁/n)/(a₁ + a₂)/n = P(AB)/P(A)

Similarly it can be shown that
P(A/B) = P[(AB)/P(B)]

The general rule of multiplication in its modified form in terms of conditional probability becomes:

P(A and B) = P(B) × P(A/B)
Or, P(A and B) = P(A) × P(B/A)

For three events A, B and C we have
P(ABC) = P(A) × P(B/A) × P(C/AB)

i.e. the probability of occurrence of A, B & C is equal to the probability of A, times the probability of B given that A has occurred, and times the probability of C given that both A & B have occurred.

Illustration: a bag contains 5 white & 3 black balls. Two balls are drawn at random one after other without replacement. Find the probability that both balls drawn are black.

Solution: the probability of drawing a black ball in the first attempt is

P(A) = 3/(5 + 3) = 3/8

the Probability of drawing the second black ball given that the first ball drawn is black

P(B/A) = 2/(5 + 2) = 2/7

∴ The probability that both balls drawn are black is given by:

P(AB) = P(A) × P(B/A) = 3/8 × 2/7 = 3/28