Question

Suppose that 50% of all babies born in a particular hospital are girls. If 6 babies born in the hospital are randomly selected, what is the probability that more than 1 of them are girls?

Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places.

Answer 1

n = 6

p = 0.5

P(X = x) = 6Cx * 0.5^x * (1 - 0.5)^6-x

P(X > 1) = 1 - (P(X = 0) + P(X = 1))

              = 1 - (6C0 * 0.5⁰ * 0.5⁶ + 6C1 * 0.5¹ * 0.5⁵)

              = 1 - 0.1094

              = 0.8906

              = 0.89 (ans)