Question

in a certain country, the true probability of a baby being a boy is 0.515. among the next six 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 the probability of all - boys

The probability of a baby being a boy is $p = 0.515$. The births are independent events. The probability that all 6 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 $n = 6$, $k = 6$, and $C(n,k)=\frac{n!}{k!(n - k)!}$. When $k = 6$, $C(6,6)=\frac{6!}{6!(6 - 6)!}=1$. So $P(X = 6)=0.515^{6}$.
$P(X = 6)=0.515^{6}\approx0.018$

Step2: Find the 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 6 babies is a girl, and $B$ be the event that all 6 babies are boys. Then $P(A)=1 - P(B)$.
$P(A)=1 - 0.515^{6}\approx1 - 0.018 = 0.982$

Answer:

$0.982$