A family has two children. Assume that each child is equally likely to be a boy or a girl. Let E be the event “both children are of the same sex”, F the event “at least one child is a boy” and G the event “the elder child is a girl”. Consider the following statements:
(A) \(P(E|F) = \displaystyle\frac{1}{3}\)
(B) E and G are independent.
(C) F and G are mutually exclusive.
(D) \(P(G|F) = \displaystyle\frac{1}{2}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (B) only
Writing the elder child first, the four equally likely outcomes are BB, BG, GB, GG. So
E = {BB, GG}, F = {BB, BG, GB}, G = {GB, GG}
The common parts are \(E\cap F = \{BB\}\), \(E\cap G = \{GG\}\) and \(F\cap G = \{GB\}\).
Given F, the reduced sample space has 3 outcomes; two events are independent when the probability of both equals the product of their probabilities.
So \(P(E|F) = \frac{1}{3}\), \(P(E)P(G) = \frac{1}{4} = P(E\cap G)\), \(F\cap G\) is not empty and \(P(G|F) = \frac{1}{3}\).
Checking the statements
(A) \(P(E|F) = \frac{1}{3}\) — Correct. Of the 3 outcomes BB, BG, GB, only BB has both children of the same sex.
(B) E and G are independent. — Correct. \(P(E)P(G) = \frac{1}{2}\times\frac{1}{2} = \frac{1}{4} = P(E\cap G)\).
(C) F and G are mutually exclusive. — Incorrect. GB (elder girl, younger boy) lies in both F and G.
(D) \(P(G|F) = \frac{1}{2}\) — Incorrect. Of BB, BG, GB, only GB has an elder girl, so \(P(G|F) = \frac{1}{3}\).