Solutions:
- 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).
Of the eight scenarios above, 6/8 (or 3/4) would have one boy and one girl.
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.
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.
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). Back to Puzzling Adventures main page
Back to Puzzling Adventures main page
