Question

11. According to the National Institute of Health 32% of all women fracture a hip by the age of 90. If 8 women aged 90 are selected at random,

a) What is the probability that exactly 5 suffered from a hip fracture?

b) What is the probability that at most 3 suffered from a hip fracture?

c) What is the probability that none suffered from a hip fracture?

d) What is the probability that at least one had a hip fracture?

Answer 1

p = 0.32

n = 8

a) P(X = 5) = 8C5 * 0.32⁵ * (1 - 0.32)^8-5 = 0.0591

b) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

                  = 8C0 * 0.32⁰ * 0.68⁸ + 8C1 * 0.32¹ * 0.68⁷ + 8C2 * 0.32² * 0.68⁶ + 8C3 * 0.32³ * 0.68⁵

                  = 0.7681

c) P(X = 0) = 8C0 * 0.32⁰ * 0.68⁸ = 0.0457

d) P(X > 1) = 1 - P(X = 0) = 1 - 0.0457 = 0.9543