Respuesta :
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 [tex]\bigcap[/tex] 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)
= [tex]\frac{0.8*0.8*0.8}{1-(0.2*0.2*0.2)}[/tex] = 0.516 .
Therefore, Probability that all three pass given that at least one of the next three vehicles passes inspection = 0.516 .
