The event coordinator asks you to determine how many students participated in the track-and-field day. The total number of stude
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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 .