**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)****<em> = </em>****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) = **

<em> 1 - P(none of the next three inspected fails) = 1 - P(all next three passes)</em>

= 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 ;**

- <em>First vehicle pass the inspection, second and third vehicle doesn't pass</em>
- <em>Second vehicle pass the inspection, first and third vehicle doesn't pass</em>
- <em>Third vehicle pass the inspection, first and second vehicle doesn't pass</em>

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) <em>P(at most one of the next three vehicles inspected passes) = P(that none</em>

<em> of the next three vehicle passes) + P(Only one of the next three passes)</em>

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;**

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

= P( All next three passes 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)

= = 0.516 .

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