Question

In a bulb factory, three machines, A, B, C, manufacture 60%, 25% and 15% of the total production respectively. Of their respective outputs, 1%, 2% and 1% are defective. A bulb is drawn at random from the total product, and it is found to be defective. Find the probability that it was manufactured by machine C. Select the correct answer from above options

Answer 1

Let D : Bulb is defective We want to find P(C|D), i.e. probability that the selected defective bulb is manufactured by C Where, P(A) = probability that bulb is made by machine A = \(\frac{60}{100}\) P(B) = probability that bulb is made by machine B = \(\frac{25}{100}\) P(C) = probability that bulb is made by machine C = \(\frac{15}{100}\) P(D|A) = probability of defective bulb from machine A = \(\frac{1}{100}\) P(D|B) = probability of defective bulb from machine B = \(\frac{2}{100}\) P(D|C) = probability of defective bulb from machine C = \(\frac{1}{100}\) P(C|D) = \(\frac{15}{60+50+15}\) = \(\frac{15}{125}\) = \(\frac{3}{25}\) Conclusion: Therefore, the probability of selected defective bulb is from machine C is \(\frac{3}{25}\)