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. (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)

(a) 0.343

(b) 0.973

(c) 0.189

Step-by-step explanation:

We are given that Probability of all vehicles examined at a certain emissions inspection station pass the inspection is 70% or 0.7.

(a) P (all of the next three vehicles inspected pass) = P (First vehicle passes the

inspection test) * P (Second vehicle passes the inspection test) * P (Third

vehicle passes the inspection test)

= 0.7 * 0.7 * 0.7 = 0.343.

(b) P (at least one of the next three inspected fails) = 1 - P (none of the next

three inspected fails)

= 1 - ((1 - 0.7) * (1 - 0.7) * (1 - 0.7)) = 1 - 0.027 = 0.973

{ Because if probability of passing the inspection is 0.7 then probability of not passing will be 1 - 0.7 }.

(c) P (exactly one of the next three inspected passes) = It means that one of the three vehicles get passed the inspection.

First vehicle passed the inspection, second and third does not pass. Second vehicle passed the inspection, first and third does not pass. Third vehicle passed the inspection, first and second does not pass.

Hence, P (exactly one of the next three inspected passes)

= 0.7 * (1-0.7) * (1-0.7) + (1-0.7) * 0.7 * (1-0.7) + (1-0.7) * (1-0.7) * 0.7

= 0.189.