Seventy percent of all vehicles examined at a certain emissions inspection stati

Question

Seventy 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.
(a) P(all of the next three vehicles inspected pass)
(b) P(at least one of the next three inspected fail)
(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 proba-
bility that all three pass?

Explanation / Answer

Probability that any vehicle passes inspection = 70/100 = 7/10 a) P (all of the next three inspected vehicles pass) = 7/10 x 7/10 x 7/10 = 343/1000 = 0.343 b) P (atleast one of the next three inspected vehicles fails) = 1 - P (all of the next three inspected vehicles pass) = 1 - 0.343 = 0.657 c) P (exactly one of the next three inspected vehicles passes) = 7/10 x 3/10 x 3/10 x 3 = 0.189 d) P (at most of the next three inspected vehicles passes) = P (exactly one of the next three inspected vehicles passes) + P (none of the next three inspected vehicles passes) = 0.189 + 3/10 x 3/10 x 3/10 = 0.216 e) Let A - All three vehicles pass B - Atleast one of the three vehicles pass Thus, P (A/B) = P (AUB) / P (B) = P (A) / P (B) = 0.343 / [1 - P (all the next three inspected vehicles fail)] = 0.343 / (1 - 0.027) = 0.353

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