Question

in a certain country, the true probability of a baby being a girl is 0.474. among the next six 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 $p = 0.474$. The births are independent events. The probability that all 6 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 $n = 6$, $k = 6$, and $C(n,k)=\frac{n!}{k!(n - k)!}$. When $n = 6$, $k = 6$, $C(6,6)=\frac{6!}{6!(6 - 6)!}=1$. So $P(X = 6)=0.474^{6}$.
$P(X = 6)=0.474^{6}\approx0.016$

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 of the 6 babies is a boy, and $B$ be the event that all 6 babies are girls. Then $P(A)=1 - P(B)$.
$P(A)=1-0.474^{6}\approx1 - 0.016 = 0.984$

Answer:

$0.984$