In a certain country, the true probability of a baby being a girl is 0.479. A mound the next four randomly selected births in the country, what is the probability that at least one of them is a boy ? Round to 3 decimal places as needed

0.947

Step-by-step explanation:

We are told that the true probability of a baby being a girl is 0.479.

This means that probability of a baby being a boy is;

p = 1 - 0.479

p = 0.521

Now,using binomial probability distribution formula we can solve this problem.

P(X=x) = C(n, x) × p^(x) × (1 - p)^(n - x)

We are told that a mound the next four randomly selected births in the country, what is the probability that at least one of them is a boy.

This is;

P(X ≥ 1) = P(1) + P(2) + P(3) + P(4)

P(1) = C(4, 1) × 0.521^(1) × (1 - 0.521)^(4 - 1)

P(1) = 0.229

P(2) = C(4, 2) × 0.521^(2) × (1 - 0.521)^(4 - 2)

P(2) = 0.3737

P(3) = C(4, 3) × 0.521^(3) × (1 - 0.521)^(4 - 3)

P(3) = 0.27096

P(4) = C(4, 4) × 0.521^(4) × (1 - 0.521)^(4 - 4)

P(4) = 0.0737

Thus;

P(X ≥ 1) = 0.229 + 0.3737 + 0.27096 + 0.0737

P(X ≥ 1) = 0.94736

P(X ≥ 1) ≈ 0.947

the answers c

what are you trying to solve here

step-by-step explanation: