Customers who purchase a certain make of automobile can order an engine in any o

Question

Customers who purchase a certain make of automobile can order an engine in any one of three capacities. Of all the cars sold, 45% have the smallest engine, 35% have the medium sized one and the remaining 20% have the largest engine. Of vehicles with the smallest engine, 10% fail an emission test within 2 years of purchase, while 12% of those with the medium-sized engine fail and 15% of those with the largest engine fail.
(a) What is the probability that a randomly chosen vehicle will fail an emission test within two years?
(b) If a vehicle chosen at random fails an emission test, what is the probability that it has a small engine?
(c) If you choose 20 vehicles at random from the total population of vehicles, what is the probability that you choose at most 2 with the largest engine?
(d) If you choose 5 vehicles at random from the total population of vehicles every day for 4 days, what is the probability that you choose one vehicle exactly each day with the largest engine?

Explanation / Answer

probability that the cars sold have the smallest engine, P(S) =0.45 probability that the cars sold have the medium engine, P(M) =0.35 probability that the cars sold have the largest engine, P(L) =0.20 given that vehicles have the smallest engine, probability of failing an emission test within 2 years of purchase, P(F/S) =0.1 given that vehicles have the medium engine, probability of failing an emission test within 2 years of purchase, P(F/M) = 0.12 given that vehicles have the largest engine, probability of failing an emission test within 2 years of purchase, P(F/L) =0.15
(a) Using Bayes' theorem, the probability that a randomly chosen vehicle will fail an emission test within two years =P(F) = P(F/S) * F(S) +P(F/M)*P(M) +P(F/L)*P(L) =(0.1*0.45) + (0.12*0.35) + (0.15*0.2) =0.117
(b) As per Bayes' theorem, If a vehicle chosen at random fails an emission test, the probability that it has a small engine =P(S/F) =[P(F/S)*P(S)]/P(F) = (0.1*0.45) /0.117 =0.385
(c) Using binomial theorem, If you choose 20 vehicles at random from the total population of vehicles, the probability that you choose at most 2 with the largest engine = P(0)+P(1)+P(2) = [20!/(0!20!)]*0.20 * 0.820 + [20!/(1!19!)]*0.21 * 0.819 + [20!/(2!18!)]*0.22 * 0.818 =0.2061
(d) Again using binomial probability, If you choose 5 vehicles at random from the total population of vehicles every day for 4 days, the probability that you choose one vehicle exactly each day with the largest engine =P(1) = [5!/(1!4!)]*0.21 * 0.84 =0.4096 =0.4096

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.