In a certain city, the true probability of a baby being a boy is 0.559. Among the next eight randomly selected births, find the probability that at least one of them is a boy. (round to three decimal places)

99.86%

Step-by-step explanation:

What we have here is a binomial distribution, let x be the baby number among 8 births.

Binomial (n = 8, p = 0.559)

We want to find, P (X ≥ 1)

P (X ≥ 1)    = 1 - P (X <1)

P (X ≥ 1)  = 1 - P (X = 0)

P (X = 0)   = 8C0 * (0.559) ^ 0 * (1 - 0.559) ^ (8-0)]

8C0 = 8! / (0! * (8-0)!) = 1

replacing we have:

P (X = 0)   = 1 * 1 * 0.0014

P (X = 0)  = 0.0014

P (X ≥ 1))  = 1 - 0.001 4

P (X ≥ 1) = 0.9986

Therefore the probability is 99.86%

i believe the answer is d.

hope it you! : d

77

step-by-step explanation: