Question

in a certain country, the true probability of a baby being a girl is 0.476. among the next four randomly - selected births in the country, what is the probability that at least one of them is a boy? the probability is (round to three decimal places as needed.)

Explanation:

Step1: Find probability of all - girls

The probability of a baby being a girl is $p = 0.476$. Since the births are independent events, the probability that all 4 babies are girls is given by the binomial probability formula $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n = 4$, $k = 4$, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $C(4,4)=\frac{4!}{4!(4 - 4)!}=1$. So the probability that all 4 babies are girls is $P(X = 4)=0.476^{4}$.
$P(X = 4)=0.476^{4}=0.476\times0.476\times0.476\times0.476\approx0.051$.

Step2: Find probability of at - least one boy

The probability of at least one boy is the complement of the event of all - girls. Let $A$ be the event that at least one baby is a boy. Let $B$ be the event that all babies are girls. Then $P(A)=1 - P(B)$.
We know $P(B)\approx0.051$, so $P(A)=1 - 0.051 = 0.949$.

Answer:

$0.949$