Question
A family has 2 children. Given that at least one is a boy, what is the probability that both are boys? (Assume boys and girls are equally likely.)
Solution — Step by Step
For two children, the equally likely outcomes are: BB, BG, GB, GG (where order matters: first child, second child).
Each has probability 1/4.
“At least one is a boy” rules out GG. The reduced sample space is {BB, BG, GB}, three equally likely outcomes.
Numerator: .
Denominator: .
Final answer:
Why This Works
Conditioning shrinks the sample space. Once we know at least one child is a boy, GG is impossible — we should only count outcomes consistent with the given information.
The non-intuitive answer (1/3, not 1/2) trips up most students. The instinct is “the other child is independent, so 1/2”. But the conditioning event isn’t “the first is a boy” — it’s “at least one is a boy”, which includes BG and GB. The asymmetry produces 1/3.
Alternative Method
Think of it as directly:
- = “both boys”, = “at least one boy”
- (because BB satisfies both), so
Common Mistake
The classic wrong answer: “We know one is a boy, so the other is independent → 1/2”. This implicitly conditions on a specific child being a boy (e.g., “the older is a boy”), which is a different event. Conditioning on “at least one” includes the BG and GB cases, not just one of them.
If the question said “the elder child is a boy”, the answer would be 1/2.
Always write out the sample space when in doubt about conditional probability. The mechanical method (count favourable / count conditioning) never lies.