Some couples planning a new family would prefer at least one child of each sex. The probability that a couple’s first child is a boy is 0.512. In the absence of technological intervention, the probability that their second child is a boy is independent of the sex of their first child, and so remains 0.512. Imagine that you are helping a new couple with their planning. If the couple plans to have only two children: (a) What is the probability of getting one child of each sex?

For each children, there are only two possible outcomes. Either they are a boy, or they are a girl. The sex of a children is independent of other children, so we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

The probability that a couple’s first child is a boy is 0.512.

This means that [tex]p = 0.512[/tex]

The will have two children:

This means that [tex]n = 2[/tex]

(a) What is the probability of getting one child of each sex?

This is P(X = 1).

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 1) = C_{2,1}.(0.512)^{1}.(0.488)^{1} = 0.4997[/tex]

49.97% probability of getting one child of each sex