A packing company is inspecting a shipment of grapefruit by cutting 15 grapefruit selected at
random. The company will accept all grapefruit if 14 or all of these 15 grapefruit are of satisfactory
quality. Suppose that 95% of the grapefruit of the shipment are of satisfactory quality. What is the
probability that the shipment will be accepted?
Accepted Solution
A:
Probability that the shipment would be accepted is computed here as:
= P(X = 14) + P(X = 15)
= 15 * 0.95^14 * 0.05 + 0.9515 = 0.8290
Therefore 0.8290 is the required probability here.
Note that this was a case of a binomial distribution with n = 15 and p = 0.95