Question

Eighty 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)? (Round your answer to three decimal places.)

Answer 1

Answer:

(a) P(all of the next three vehicles inspected pass) = 0.512 .

(b) P(at least one of the next three inspected fails) = 0.488 .

(c) P(exactly one of the next three inspected passes) = 0.096 .

(d) P(at most one of the next three vehicles inspected passes) = 0.104 .

(e) Probability that all three pass given that at least one of the next three vehicles passes inspection = 0.516 .

Step-by-step explanation:

We are given the Probability of all vehicles examined at a certain emissions inspection station pass the inspection to be 80%.

So, Probability that the next vehicle examined fails the inspection is 20%.

Also, it is given that successive vehicles pass or fail independently of one another.

(a) P(all of the next three vehicles inspected pass) = Probability that first vehicle, second vehicle and third vehicle also pass the inspection

              = 0.8 * 0.8 * 0.8 = 0.512

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

        1 - P(none of the next three inspected fails) = 1 - P(all next three passes)

    = 1 - (0.8 * 0.8 * 0.8) = 1 - 0.512 = 0.488 .

(c) P(exactly one of the next three inspected passes) is given by ;

• First vehicle pass the inspection, second and third vehicle doesn't pass
• Second vehicle pass the inspection, first and third vehicle doesn't pass
• Third vehicle pass the inspection, first and second vehicle doesn't pass

Hence, P(exactly one of the next three inspected passes) = Add all above cases ;

             (0.8 * 0.2 * 0.2) + (0.2 * 0.8 * 0.2) + (0.2 * 0.2 * 0.8) = 0.096 .

(d) P(at most one of the next three vehicles inspected passes) = P(that none

     of the next three vehicle passes) + P(Only one of the next three passes)

We have calculated the P(Only one of the next three passes) in the above part of this question;

 Hence, P(at most one of the next three vehicles inspected passes) =

              (0.2 * 0.2 * 0.2) + (0.8 * 0.2 * 0.2) + (0.2 * 0.8 * 0.2) + (0.2 * 0.2 * 0.8)

             = 0.008 + 0.096 = 0.104 .

(e) Probability that all three pass given that at least one of the next three vehicles passes inspection is given by;

P(All three passes/At least one of the next three vehicles passes inspection)

= P( All next three passes $\bigcap$ At least one of next three passes) /

      P(At least one of next three passes)

=  P( All next three passes ) / P(At least one of next three passes)

= P( All next three passes ) / 1 - P(none of the next three passes)

 =  $\frac{0.8*0.8*0.8}{1-(0.2*0.2*0.2)}$ = 0.516 .

Therefore, Probability that all three pass given that at least one of the next three vehicles passes inspection = 0.516 .