Paper 4 · conditional probability / Bayes theorem

A manufacturing plant uses three machines, A, B and C, in its production process. The total daily output contributions of machines A, B and C are 40%, 45% and 15% respectively. It is known that 4% of the tins produced by A are defective, 3% of those produced by B are defective, and 1% of those produced by C are defective. What is the probability that one tin chosen at random from the day's production is defective?

0.027

0.031

0.032

0.034

Explanation: By the law of total probability,

P(\text{defective}) = 0.40(0.04) + 0.45(0.03) + 0.15(0.01) = 0.016 + 0.0135 + 0.0015 = 0.031

. Averaging the three defect rates without weighting by each machine's output share (i.e.

(0.04+0.03+0.01)/3

) gives

0.027

. Swapping the 40% and 45% output weights between A and B gives

0.032

. The remaining option is a plausible rounding slip in the addition.

Derived from ZIMSEC Maths Paper 1, June 2007, Q1 (Statistics)

More questions from this paper