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

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954 Chapter 29 Statistics and probability distributions

Engineeringapplication29.3

Samplingcomponentsmadebyamachine

Amachinemakescomponents.Theprobabilitythatacomponentisacceptableis0.9.

(a) If three components are sampled find the probability that the first is acceptable,

the second isacceptable and the thirdisnot acceptable.

(b) If three components are sampled what is the probability that exactly two are

acceptable?

Solution

Let the eventsAandBbe defined thus:

A:thecomponentis acceptable,P(A) = 0.9

B:thecomponentis notacceptable,P(B) = 0.1

(a) We denote byAAB the compound event that the first is acceptable, the second

is acceptable, the third is not acceptable. Since the three events are independent

the multiplication lawinSection 28.7 gives

P(AAB) =P(A)P(A)P(B) = (0.9)(0.9)(0.1) = (0.9) 2 (0.1) = 0.081

(b) Weareinterestedinthecompoundeventinwhichtwocomponentsareacceptable

and one is not acceptable. We denote byAAB the compound event that the first

is acceptable, the second is acceptable, the third is not acceptable. Compound

eventsABA andBAA have obvious interpretations.

If exactly two components are acceptable then either AAB or ABA or BAA

occurs.Thesecompoundeventsaremutuallyexclusiveandwecanthereforeuse

the addition law (see Section 28.3). Hence

P(exactly two acceptable components) =P(AAB) +P(ABA) +P(BAA)

From (a) P(AAB) = 0.081 and by similar reasoning P(ABA) = P(BAA) =

0.081. Hence,

P(exactly two acceptable components) = 3(0.081) = 0.243

29.10.1 Probabilityofksuccessesfromntrials

Let us now return to the general problem posed earlier. We define the compound

eventC:

C:Aoccursktimes andBoccursn −k times

( n

The k occurrences of event A can be distributed amongst the n trials in different

k)

ways (see Example 29.17). The probability of a particular distribution

(

ofkoccurrences

n

ofAandn−koccurrencesofBisp k (1−p) n−k .Sincethereare distinctdistributions

k)

possible then

P(C) =

( n

k)

p k (1−p) n−k

k=0,1,2,...,n

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