Sixty percent of all vehicles examined at a certain emissions inspection station

Question

Sixty percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehicles pass or fail independently of one another, calculate the following probabilities. (Enter your answers to three decimal places.) (a) P(all of the next three vehicles inspected pass) (b) P(at least one of the next three inspected fails) (c) P(exactly one of the next three inspected passes) (d) P(at most one of the next three vehicles inspected passes) (e) Given that at least one of the next three vehicles passes inspection, what is the probability that all three pass (a conditional probability)?

Explanation / Answer

P(pass) = 0.6

P(fail) = 0.4

A) P(all of the next three vehicles inspected pass)

= 0.6 * 0.6 * 0.6 = 0.216

B) P(at least one of the next three inspected fails)

= FPP + FFP + FFF

= 0.4 * 0.6 * 0.6 + 0.4 * 0.4 * 0.6 + 0.4 * 0.4 * 0.4 = 0.304

C) P(exactly one of the next three inspected passes)

= PFF + FPF + FFP

= 0.6 * 0.4 * 0.4 + 0.4 * 0.6 * 0.4 + 0.4 * 0.4 * 0.6 = 0.288

D) P(at most one of the next three vehicles inspected passes)

= PFF + FPF + FFP + FFF

= 0.6 * 0.4 * 0.4 + 0.4 * 0.6 * 0.4 + 0.4 * 0.4 * 0.6 + 0.4 * 0.4 * 0.4 = 0.352

E) at least one of the passes = PFF + PFP + PPP

. = 0.6 * 0.4 * 0.4 + 0.6 * 0.4 * 0.6 + 0.6 * 0.6 * 0.6 = 0.456

P(all three passes | at least one of the next three vehicles passes inspection) = P(all three passes and at least one of the passes) / at least one of the passes = 0.216 / 0.456 = 0.474

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