Question

In a certain​ country, the true probability of a baby being a boy is 0.511. Among the next eight randomly selected births in the​ country, what is the probability that at least one of them is a girl​?

Answer 1

Answer:

The probability that at least one of 8 babies born is a girl​  is 0.9954.

Step-by-step explanation:

Let X = a baby born is a girl.

The probability of a baby born being a girl is,

P (G) = 1 - P (B)

        = 1 - 0.511

     p  = 0.489

The number of births is, n = 8.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of the Binomial distribution is:

$P(X=x)={8\choose x}0.489^{x}(1-0.489)^{8-x};\ x=0,1,2,3...$

Compute the probability that out of 8 births at least one is a girl as follows:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             $=1-{8\choose 0}0.489^{0}(1-0.489)^{8-0}\\=1-(1\times1\times 0.00465)\\=1-0.00465\\=0.99535\\\approx 0.9954$

Thus, the probability that at least one of 8 babies born is a girl​  is 0.9954.