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082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

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P(A ∩C) =P(C ∩A) =P(C)P(A|C)

P(A|C) = P(A∩C)

P(C)

28.6 Conditional probability: the multiplication law 923

NowP(A∩C)istheprobabilitythatthecomponentismanufacturedbymachineMand

isacceptable.Thisisknowntobe0.6175,usingEngineeringapplication28.12(b).Also

fromEngineering application 28.12(a) we seeP(C) = 0.908. Hence,

P(A|C) = 0.6175

0.908 = 0.68

Engineeringapplication28.13

Reliabilityofmanufacturedcomponents

A manufacturer studies the reliability of a certain component so that suitable guarantees

can be given: 83% of components remain reliable for at least 5 years; 92%

remain reliable for at least 3 years. What is the probability that a component which

has remainedreliable for3yearswill remainreliablefor5years?

Solution

We definethe events:

A: a component remainsreliableforatleast3years

B: a component remainsreliableforatleast5years

ThenP(A) = 0.92,P(B) = 0.83.Notethattheseareunconditionalprobabilities.We

requireP(B|A), a conditionalprobability.

P(A ∩B) =P(A)P(B|A)

A ∩Bisthecompoundeventacomponentremainsreliableforatleast3yearsandit

remainsreliableforatleast5years. Clearlythis isthe same asthe eventB. So

Hence

P(A ∩B) =P(B)

P(B) =P(A)P(B|A)

P(B|A) = P(B)

P(A) = 0.83

0.92 = 0.90

thatis,90%ofthecomponentswhichremainreliablefor3yearswillremainreliable

for atleast5years.

Alternatively, a tree diagram is shown in Figure 28.8. Starting with 100 components,92remainreliable3yearslater.Ofthese92,2yearslater,83arestillreliable.

So

P(component isreliable after 5 years, given itisreliable after 3 years) = 83

92

= 0.90 ➔

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