In a certain country, the true probability of a baby being a boy is 0.518. Among the next four randomly selected births in the country, what is the probability that at least one of them is a girl? (Round to three decimal places as needed.)

0.928 = 92.8% probability that at least one of them is a girl

Step-by-step explanation:

For each baby, there are only two possible outcomes. Either they are boys, or they are not. The probabilities of each baby being a boy is independent from other babies. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

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

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.

In this problem we have that:

In a certain country, the true probability of a baby being a boy is 0.518. So

Among the next four randomly selected births in the country, what is the probability that at least one of them is a girl?

Either all four babies are boys, or at least one is girl. The sum of these events is decimal 1. So

We want to find when . So

In which

0.928 = 92.8% probability that at least one of them is a girl