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