Question

# A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that was produced by A?

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Solution

## Let P( A ), P( B ) P( C ) and P( D ) be the probabilities that are defined below, The probability that the item produced by operation A is P( A ). The probability that the item produced by operation B is P( B ). The probability that the item produced by operation C is P( C ) The probability that the item is defective is P( D ). Find the probability that the item is produced by operation A if it is defective that is P( A D ), P( A D )= P( A )P( D A ) P( A )P( D A )+P( B )P( D B )+P( C )⋅P( D C ) (1) Given that, P( A )=50% = 1 2 The probability of a defective item produced by operation A is, P( D A )=1% =0.01 P( B )=30% =0.3 The probability of a defective item produced by operation B is, P( D B )=5% =0.05 P( C )=20% =0.2 The probability of a defective item produced by operation C is, P( D C )=7% =0.07 Put these values in equation (1), P( A D )= 0.5×0.01 0.5×0.01+0.3×0.05+0.2×0.07 = 0.005 0.034 = 5 34 Thus, the required probability is 5 34 .

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