7. + -15 points DevoreStat9 2..50TXP. My Notes Ask Your Teacher Seventy percent

Question

7. + -15 points DevoreStat9 2..50TXP. My Notes Ask Your Teacher 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) Pall of the next three vehicles inspected pass) (b) P(at least one of the next three inspected fails) (c) Prexactly one of the next three inspected pases) (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)? (Round your answer to three decimal places.) Need Help?Read It Talk to a Tutor

Explanation / Answer

(a) P(all of the next three vehicles inspected pass) = (Probability of a vehicle passing the inspection)3

We are given, Probability of a vehicle passing the inspection = 0.70

P(all of the next three vehicles inspected pass) = 0.703 = 0.343

(b) P(atleast one of the next three vehicles fails) = 1 - P(none of the next three vehicles fails)=

= 1 - P(All of the next three vehicles pass) = 1 - 0.343 = 0.657

(c) We will use binomial distribution in this case,

P(exactly one of the next three inspected passses) = nC1 * p1 * ( 1 - p )n - 1

We have n = 3 and p = 0.70

P(exactly one of the next three inspected passses) = 3C1 * 0.701 * ( 1 - 0.70 )3 - 1

= ( 3! / ( 2! * 1! ) ) * 0.70 * 0.302

= 0.189

(d) P(at most one of the next three vehicles inspected passes) = P( no vehicles passes) + P(exactly one vehicle passes)

P( no vehicle passes) = 0.303 = 0.027

P(exactly one vehicle passes) = 0.189 [ Calculated in part(c) ]

P(at most one of the next three vehicles inspected passes) = 0.027 + 0.189 = 0.216

(e) P(all three pass / atleast one of the next three pass) =

= P(all three pass and atleast one of the next three pass) / P(atleast one of the next three pass)

= P(all three pass) / P(atleast one of the next three pass)

P(all three pass) = 0.343 [ Calculated in part (a) ]

P(atleast one of the next three pass) = 1 - P(none of the next three pass) = 1 - 0.33 = 0.973

P(all three pass / atleast one of the next three pass) = 0.343 / 0.973 = 0.353

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