Problem 2.1.5 from Basic Concepts of Probability and Statistics, 2nd Ed. by Hodges and Lehmann
Suppose you know that a family has 6 children, 2 boys and 4 girls, but you don't know their sexes according to age.
1. How many possible arrangements bbgggg, bgbggg,..., are there?
assume all arrangements of 1. to be equally likely, find probabilities of the following event:
2. the oldest child is a girl.
3. the two oldest children are both girls.
4. the oldest child is a boy and the youngest is a girl.
Need help on 2., 3., and 4.
My solutions:
1. I used 6! to find all the unique arrangements, but I see that using 6 choose 2 to find 15 combinations total is what they're looking for... *for 2.,3.,4. I used the 6! as the denominator..
2. I used [(4 choose 1)*5!]/6! = 2/3
3. I used [(4 choose 2)*4!]/6! = 1/5
4. I used [(2 choose 1)*4!*(4 choose 1)]/6! = 4/15
I have an "Unofficial" solutions manual (from someone who attempted all the problems in the book) that I looked at for the solutions for this problem (see attached)
Need help on:
1. someone to check my answers and the calculation processes I used to get to the answer.
2. someone to comment on the solutions for 2,3,4 in the solutions manual as found in the attached file. I don't quite understand the meaning behind the numerators used for the solutions to 2.,3.,4., for example I don't know what '5 choose 2' is actually choosing, and I am not sure if it is correct...
Thanks!,
Yong
Last edited by yl2j; 02-22-2015 at 02:43 PM.
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