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Puzzling Adventures: A Kingdom for My Child -- August 2008 Solution




Cloe Liane Shasha

Solutions:

1. Let's consider again the three children that a family could have and associate the number at which they would stop. (As a thought experiment, we imagine that they have three children in a certain order, but "undo," or send away, the extras once they have a boy.):

000—3
001—3
010—2
011—2
100—1
101—1
110—1
111—1

So, the average number is under two (14/8).

2. Of the eight scenarios above, 6/8 (or 3/4) would have one boy and one girl.

3. We write down the number of children needed to achieve completeness in each case:

000—3
001—3
010—2
011—2
100—2
101—2
110—3
111—3

This gave us an average of 20/8 children per family, or 2.5.

4. If the youngest happened to be a girl, then the others would either both be boys or both girls (otherwise their parents would have stopped already). Therefore, the probability of at least one boy is 1/2.

5. Once again we are, with equal likelihood, in one of these eight family situations. There are 10 girls available (because some families stop before three) and in 7/10 of the cases, there is a boy, so the likelihood is 7/10.

000—3
001—3
01x—2
01x—2
10x—2
10x—2
110—3
111—3

For more on this subject, see Statistics is Easy! by Dennis Shasha and Manda Wilson (Morgan & Claypool, July 2008).

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