Answer:
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.
[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.
In this problem we have that:
In a certain country, the true probability of a baby being a boy is 0.518. So [tex]p = 0.518[/tex]
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
[tex]P(X = 4) + P(X < 4) = 1[/tex]
We want to find [tex]P(X < 4)[/tex] when [tex]n = 4[/tex]. So
[tex]P(X < 4) = 1 - P(X = 4)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 4) = C_{4,4}.(0.518)^{4}.(0.482)^{0} = 0.072[/tex]
[tex]P(X < 4) = 1 - P(X = 4) = 1 - 0.072 = 0.928[/tex]
0.928 = 92.8% probability that at least one of them is a girl