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This week's Challenge Question:
You have three colleagues: Arthur, Bill and Charlie. Each has two (and only two) children. Bill mentioned that today is his daughter's birthday and Charlie brought his daughter to work.
Assuming 50:50 male:female distribution and no further information about your colleagues or their families, what are the probabilities that each has a son?
Thanks to ExPat for providing us with this Challenge.
And the Answer is....
Arthur: 3/4
Bill: 2/3
Charlie: 1/2
Most won't have difficulty with Arthur and Charlie; Bill's case is less intuitive, at least until the problem is broken down.
Explanation:
Designate the two children and Child A and Child B. Each child can be a boy or a girl (B or G). For Arthur, there are four permutations, each with equal probability; three include at least one boy, so the probability that Arthur has a son is 3/4. Child A Child B Has Son?
| Child A | Child B | Has son? |
|---|
| B | B | Y |
| B | G | Y |
| G | B | Y |
| G | G | N |
For Bill, we know that B-B is not an available solution, leaving three permutations, each with equal probability; two include a boy, so the probability that Bill has a son is 2/3. Child A Child B Has Son?
| Child A | Child B | Has son? |
|---|
| B | G | Y |
| G | B | Y |
| G | G | N |
For Charlie, we have already met one child (let's designate her Child A), so solutions B-B and B-G are unavailable, leaving two permutations, each with equal probability; one includes a boy, so the probability that Charlie has a son is 1/2. Child A Child B Has Son?
| Child A | Child B | Has son? |
|---|
| G | B | Y |
| G | G | N |
(We could of course designate the known daughter as Child B, resulting in the same probabilities.)
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