Question

A restaurant owner rejects his produce shipment if he finds more than 3 crates with any bruised

or spoiled food in it. The probability that a crate has bruised or spoiled food in it is 0.11. What
is the probability that he will reject a shipment of 15 crates?

Answer 1

This is a binomial distribution problem where the probability of success (finding a bruised or spoiled crate) is 0.11 and the number of trials is 15. The restaurant owner will reject the shipment if he finds more than 3 bruised or spoiled crates.

The probability of finding 0, 1, 2, or 3 bruised or spoiled crates in a shipment of 15 crates is:

P(X = 0) = (15 choose 0) * (0.11)^0 * (0.89)^15 = 0.27
P(X = 1) = (15 choose 1) * (0.11)^1 * (0.89)^14 = 0.39
P(X = 2) = (15 choose 2) * (0.11)^2 * (0.89)^13 = 0.25
P(X = 3) = (15 choose 3) * (0.11)^3 * (0.89)^12 = 0.09

The probability of finding more than 3 bruised or spoiled crates is:

P(X > 3) = 1 - P(X ≤ 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3))
= 1 - (0.27 + 0.39 + 0.25 + 0.09) = 0.01

Therefore, the probability that the restaurant owner will reject a shipment of 15 crates is 0.01 or 1%.