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Question

In a bolt factory, machines A, B and C manufacture 25%, 35%, 40% respectively. Of the total of their output 5, 4 and 2% are defective. A bolt is drawn and is found to be defective. What are the probabilities that it was manufactured by the machines A, B and C?

A
2569,2869,1669.
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B
2569,2769,1769.
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C
2869,2569,1669.
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D
2769,2669,1669.
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Solution

The correct option is D 2569,2869,1669.
Here P(A)=25100,P(B)=35100,P(C)=40100

P(DA)=5100,P(DB)=4100,P(DC)=2100

where D denotes defective bolts
Now P(D)=P(A).P(DA)+P(B).P(DB)+P(C).P(DC)

=25100.5100+35100.4100+40100.2100=0.0345

P(AD)=P(A).P(DA)P(A).P(DA)+P(B).P(DB)+P(C).P(DC)

=25100.51000.0345=2569

P(BD)=P(B).P(DB)P(A).P(DA)+P(B).P(DB)+P(C).P(DC)

=35100.41000.0345=2869
P(CD)=P(C).P(DC)P(A).P(DA)+P(B).P(DB)+P(C).P(DC)

=40100.21000.0345=1669

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