Conditional Probability

Suppose we have two related questions:

(a) A family has 2 children. What is the probability of having 2 girls for the family?

(b) A family has 2 children. What is the probability of having 2 girls given that one of the children is girl?

Question (b) has a conditional probability element, while (a) doesn't. To answer (a), we need to list 4 possible cases of having 2 kids: gg, bg, gb, and bb. (b = boy, g = girl.) The probability is thus 1/4 for (a).

We only have 3 possible cases for (b): gg, bg, gb, so that the probability is 1/3.

The notion of conditional probability is the first hurdle for students taking the probability and statistics course I teach this term.

We write P(B | A) for the probability of B to occur, given that A occurs,

P(B | A) = P(B ∩ A)/P(A),

where B ∩ A means both events A and B occur.

If we want to apply this formula to answer part (b), then P(B ∩ A) = 1/4 since there is only one case (gg); P(A) = 3/4 since having 1 girl can happen with first child (gb), second child (bg), or both girls (gg). 

The formula P(B | A) = P(B ∩ A)/P(A) is counterintuitive though, so I would write it as

P(B ∩ A) = P(B | A) P(A).

That is, the probability of having A and B to occur is equal to (i) having A to occur, and (ii) having B to occur given A occurs. The last expression gives more clarity since if A and B are statistically independent, then

P(B | A) = P(B),

so that when A and B are independent, P (B ∩ A) = P(B) P(A).

Depending on available data, any P(B ∩ A), P(B | A), or P(A) can be computed, but clearly we need to know two in order to solve for the third.

Another useful probability formula is

P(A ∩ B) = P(A) + P(B) – P(A ∪ B),

where A ∪ B means either event A or B occurs.

The conditional probability can be used to answer an unconditional probability. For example, the probability of picking for the first time an odd number from the 5 numbers: 1, 2, 3, 4, 5 is 3/5 since there are 3 odd numbers out of 5. What is now the unconditional probability for picking an odd number for the second time (thus, regardless of the outcome of the first pick)?