4.-110 points My Notes Ask Your Teach Ninety percent of all vehicles examined at

Question

4.-110 points My Notes Ask Your Teach Ninety percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehidles 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 vehidles inspected pass) (b) P(at least one of the next three inspected fails) (c) Pexactly one of the next three inspected passes) (d) 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? Round your answer to three decimal places. Hint: This may be written as P(all three pass at least one passes)

Explanation / Answer

P(pass) = 0.9

P(fail) = 0.1

n = 3

A) P(X = 3) = 3C3 * (0.9)^3 * (0.1)^0 = 0.729

B) P(X > 1) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - 3C0 * (0.1)^0 * (0.9)^3  

= 1 - 0.729 = 0.271

C) P(X = 1) = 3C1 * (0.9)^1 * (0.1)^2 = 0.027

D) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= 3C0 * (0.9)^0 * (0.1)^3 + 3C1 * (0.9)^1 * (0.1)^2 + 3C2 * (0.9)^2 * (0.1)^1 + 3C3 * (0.9)^3 * (0.1)^0 = 1

e) P(all three pass | at least one passes) = P(X = 3 | X > 3) = P(X = 3)/(P(X > 1) = P(X = 3)/(1 - P(X < 1))

= P(X = 3)/(1 - (P(X = 0))

= (3C3 * (0.9)^3 * (0.1)^0)/(1 - (3C0 * (0.9)^0 * (0.1)^3) = 0.729/(1 - 0.001)

= 0.729/0.999 = 0.7297

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