Paper 2 · probability

Using the same power-cut scenario (each generator equally likely to be switched on with probability 1/3; independent breakdown probabilities 0.2 for A, 0.3 for B, 0.25 for C), given that there was a generator breakdown, what is the probability that it was generator C?

A0.250

B0.300

C0.375

D0.333

Explanation: By Bayes' theorem, P(C | breakdown) = P(C and breakdown) / P(breakdown) = [(1/3)(0.25)] / 0.25 = (1/3), since the 0.25 factors cancel, giving approximately 0.333.

Derived from ZIMSEC Statistics Paper 2, November 2018, Q3

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Using the same power-cut scenario (each generator equally likely to be switched on with probability 1/3; independent breakdown probabilities 0.2 for A, 0.3 for B, 0.25 for C), given that there was a generator breakdown, what is the probability that it was generator C?

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