Question

in a certain country, the true probability of a baby being a boy is 0.538. among the next five randomly selected births in the country, what is the probability that at least one of them is a girl?
the probability is
(round to three decimal places as needed.)

Explanation:

Step1: Find probability of all - boys

The probability of a baby being a boy is $p = 0.538$. The births are independent events. The probability that all 5 babies are boys is given by the binomial probability formula $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n = 5$, $k = 5$, $p=0.538$, and $C(n,k)=\frac{n!}{k!(n - k)!}$. When $k = 5$, $C(5,5)=\frac{5!}{5!(5 - 5)!}=1$. So $P(X = 5)=0.538^{5}$.
$P(X = 5)=0.538\times0.538\times0.538\times0.538\times0.538\approx0.042$.

Step2: Find probability of at - least one girl

The probability of at least one girl is the complement of the event of all - boys. Let $A$ be the event that at least one of the five births is a girl, and $B$ be the event that all five births are boys. Then $P(A)=1 - P(B)$.
Since $P(B)\approx0.042$, then $P(A)=1 - 0.042 = 0.958$.

Answer:

$0.958$