✔ 最佳答案
(a) P(Contains 3 or more bacteria) = 1 - P(No bacteria) - P(Contains 1 bacteria) - P(Contain 2 bacteria) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
= 1 - e^(-3)[1 + 3 + 9/2] = 1 - e^(-3)(8.5) = 0.5768
(b) This is a binomial dist. with n = 5 and p = 0.5768
P(2 out of 5 samples containing 3 or more bacteria) = (5C2)(0.5768^2)(1 - 0.5768)^3 = 0.25217.