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.

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.