Problems in Probability

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Fall 2011 Homework 1: Question 1
A certain family has 6 children, consisting of 3 boys and 3 girls. Assuming that all birth orders are equally likely, what is the probability that the 3 eldest children are the 3 girls?
Solution: Label the girls as 1, 2,3 and the boys as 4, 5, 6. Think of the birth order is a permutation of 1, 2, 3, 4, 5, 6, e.g., we can interpret 314265 as meaning that child 3 was born first, then child 1, etc. The number of possible permutations of the birth orders is 6!. Now we need to count how many of these have all of 1, 2, 3 appear before all of 4, 5, 6. This means that the sequence must be a permutation of 1, 2,3 followed by a permutation of 4, 5, 6. So with all birth orders equally likely, we have P(the 3 girls are the 3 eldest children) = (3!)^2/6! = 0.05. Alternatively, we can use the fact that there are 6C3 ways to choose where the girls appear in the birth order (without taking into account the ordering of the girls amongst themselves). These are all equally likely. Of these possibilities, there is only 1 where the 3 girls are the 3 eldest children. So again the probability is 1/(6C3)3 = 0.05.
"Mathematics is the logic of certainty, but statistics is the logic of uncertainty."
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