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

Consider D as the defective bulb Here we should determine P (C/D) – probability that the selected defective bulb is manufactured by C We know that Probability that bulb is made by machine A P (A) = 60/100 Probability that bulb is made by machine B P (B) = 25/100 Probability that bulb is made by machine C P (C) = 15/100 Probability of defective bulb from machine A P (D/A) = 1/100 Probability of defective bulb from machine B P (D/B) = 2/100 Probability of defective bulb from machine C P (D/C) = 1/100 So we get On further calculation = 15/125 = 3/25 Hence, the probability of selected defective bulb is from machine C is 3/25.