Question

In: Math

It is known that of the articles produced by a factory, 20% come from Machine A,...

It is known that of the articles produced by a factory, 20% come from Machine A, 30% from Machine B, and 50% from Machine C. The percentages of satisfactory articles among those produced are 95% for A, 85% for B and 90% for C. An article is chosen at random. (a) What is the probability that it is satisfactory? (b) Assuming that the article is satisfactory, what is the probability that it was produced by Machine A? (c) Given that the article is satisfactory, what is the probability that it was produced by Machine C?

Solutions

Expert Solution

given data

probablity of product comes from machin A P(A)=0.2

probablity of satisfactory product comes from machin A P(sA)=0.95

probablity of product comes from machin B P(B)=0.3

probablity of satisfactory product comes from machin B P(sB)=0.85

probablity of product comes from machin C P(C)=0.5

probablity of satisfactory product comes from machin C P(sC)=0.9

(a)

now we have to find the probablity of getting a setisfactory part when one is chosen at random

so

Satisfactory part from machine A =P(A)*P(sA)=0.2*0.95=0.19 (of total)

Satisfactory part from machine B =P(B)*P(sB)=0.3*0.85=0.25 (of total)

Satisfactory part from machine C =P(C)*P(sC)=0.5*0.9=0.45 (of total)

So the probablity of getting a satisfactory part when one part is chosen at random=0.19+0.25+0.45=0.89

(b)

Satisfactory part from machine A =P(A)*P(sA)=0.2*0.95=0.19 (of total)

probablity of getting satisfactory part =0.89

probablity of getting part from mahin A given it is satisfactory =0.19/0.89=0.213

(c)

Satisfactory part from machine C =P(C)*P(sC)=0.5*0.9=0.45 (of total)

probablity of getting satisfactory part =0.89

probablity of getting part from machin C given it is satisfactory =0.45/0.89=0.505


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