In a factory, machine A produces 60% and machine B produces 40% of the total output. Of their outputs, 2% and 5% respectively are defective. An item is chosen at random from the total output. The probability that it is defective is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.032
Let \(E_1\) and \(E_2\) be the events that the item is made by machine A and machine B, and let D be the event that it is defective.
\(P(E_1) = 0.6,\quad P(E_2) = 0.4\)
\(P(D|E_1) = 0.02,\quad P(D|E_2) = 0.05\)
By the theorem of total probability:
\(P(D) = P(E_1)P(D|E_1) + P(E_2)P(D|E_2)\)
\(P(D) = 0.6\times 0.02 + 0.4\times 0.05\)
\(P(D) = 0.012 + 0.020 = 0.032\)
Why the other options are incorrect
Option (1) 0.035 — Incorrect. This is the plain average of 2% and 5%, which ignores that machine A makes more of the items.
Option (2) 0.07 — Incorrect. This adds the two defect rates without weighting them by the machines’ shares.
Option (3) 0.038 — Incorrect. This pairs the shares the wrong way: \(0.4\times 0.02 + 0.6\times 0.05\).