ECMT 3160: Exercise 3, Section 3.3
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ECMT 3160: Exercise 3, Section 3.3
2021
Question. Four children are picked at random (with no replacement) from a family which includes exactly two boys. The chance that neither boy is chosen is half the chance that both are chosen. How large is the family?
Answer. Let n denote the number of children in the family. From the question we deduce that there are (nn一!4)! permutations of 4 children from n. This information will be needed for sampling without replace-
ment. We also deduce from the question that we need to consider the following two events:
A = {no boy is chosen} and B = {2 boys are chosen},
as the information provided takes the following form: P (A）= P (B）/2. The sample space 。is the set of all
|
tuples of size 4, whose components we pick from a set of children of size n, which implies that |。| = Thus, we need to compute |A| and |B| to calculate P (A）and P (B）, as |
(n一4)! . |
P (A）= |A| and P (B）= |B|
and then use P (A）= P (B）/2 to find n.
Starting with the event A. No boy is chosen means that there are n - 2 children from which to select 4 without replacement. Hence, |A| = ╱n 4(一)2、.
For the event B, from selecting 2 boys, we have left n - 2 girls. Thus, there are ╱n 2(一)2、girls that can be chosen on the event B. This implies |B| = ╱n 2(一)2、.
Now we are in a position to use the equality P (A）= P (B）/2. Substituting the probabilities with their corresponding expressions, we obtain
╱
一4。
、 ÷e 2 ╱n4(-) 2、= ╱n2(-) 2、,
which is an equation in n. We can use it to solve for n. After some tedious algebra simplifying the above equation, we arrive to the quadratic equation: n2 - yn ＋ 14 = 〇. Finally, we can use the quadratic formula to compute the roots of this equation. We have n = 9士│82(1) 一4．14 , which simplifies to n ∈ {2, 7}. Since there
are two boys, and at least two girls, the answer must be n = 7. Hence, the family consists of 7 children, 2 of which are boys with 5 girls.
2022-04-06