Question
In a factory which manufactures calculators, machines A,
B and C manufacture respectively 25%, 35% and 40% of the calculators. Of their outputs, 5, 4 and 2 percent are respectively defective calculators. A calculator is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine B?Solution
Let events C1, C2, C3 be the following : C1 : the calculator is manufactured by machine A C2 : the calculator is manufactured by machine B C3 : the calculator is manufactured by machine C Clearly, C1, C2, C3 are mutually exclusive and exhaustive events and hence, they represent a partition of the sample space. Let the event E be ‘the calculator is defective’. The event E occurs with C1 or with C2 or with C3. Given that, P(C1) = 25% = 0.25, P (C2) = 0.35 and P(C3) = 0.40 Again P(E|C1) = Probability that the calculator is defective given that it is manufactured by machine A = 5% = 0.05 Similarly, P(E|C2) = 0.04, P(E|C3) = 0.02. Hence, by Bayes' Theorem, we have P(C2|E) = [P(C2)P(E|C2)]/[P(C1)P(E|C1)+P(C2)P(E|C2)+P(C3)P(E|C3)] => [0.35 × 0.04]/[0.25 × 0.05 + 0.35 × 0.04 + 0.40 × 0.02] => 0.0140/0.0345 = 28/69
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