Skip to content

CMA Foundation · Fundamentals of Business Mathematics and Statistics · Probability

In a Pune factory, Machine A makes 60% of the output and Machine B makes the remaining 40%. Of A's items 5% are defective, and of B's items 10% are defective. An item picked at random is found defective. What is the probability that it came from Machine A?

The probability is 3/7. Machine A contributes 0.03 of defective items (0.6 x 0.05) and Machine B contributes 0.04 (0.4 x 0.10), so the total defective proportion is 0.07. Dividing A's share by the total gives 0.03/0.07 = 3/7 by Bayes' theorem.

  1. A3/7Correct
  2. B4/7
  3. C3/5
  4. D1/2

Explanation

P(A and defective) = 0.6 x 0.05 = 0.03 and P(B and defective) = 0.4 x 0.10 = 0.04. Total P(defective) = 0.07. P(A | defective) = 0.03/0.07 = 3/7. The option 4/7 is the probability for Machine B, and 3/5 is just the prior share of A.

Did you get it right without looking?

One question tells you little. A timed set on Probability shows your real accuracy, how long you take and where you lose marks.

Practise ProbabilityFree diagnostic mock

More Probability questions