Machines A, B and C in a factory produce 20%, 30% and 50% of the total output respectively. Of their outputs, 4%, 3% and 1% respectively are defective. An item chosen at random from the total output is found to be defective. The probability that it was produced by machine B is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{9}{22}\)
Let \(E_1, E_2, E_3\) be the events that the item is made by A, B and C, and let D be the event that it is defective.
\(P(E_1)P(D|E_1) = 0.20\times 0.04 = 0.008\)
\(P(E_2)P(D|E_2) = 0.30\times 0.03 = 0.009\)
\(P(E_3)P(D|E_3) = 0.50\times 0.01 = 0.005\)
By the theorem of total probability, these add up to P(D):
\(P(D) = 0.008 + 0.009 + 0.005 = 0.022\)
By Bayes’ theorem, the share of machine B in the defective items is
\( \displaystyle P(E_2|D) = \frac{P(E_2)P(D|E_2)}{P(D)} = \frac{0.009}{0.022} = \frac{9}{22} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{4}{11}\) — Incorrect. This is \(\frac{0.008}{0.022}\), the chance that the defective item came from machine A.
Option (3) \(\displaystyle\frac{5}{22}\) — Incorrect. This is \(\frac{0.005}{0.022}\), the chance that it came from machine C.
Option (4) \(\displaystyle\frac{3}{10}\) — Incorrect. This is machine B’s share of the whole output, before we know the item is defective.