 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/16/c/solution

Of the ten outcomes in

F

, all except

(1, 1, 1)

have exactly one repeated pair. Hence

P (T ∣ F) = 9/10

. Applying Bayes' theorem in the requested direction,

P (F ∣ T) = \frac{P ( T ∣ F ) P ( F )}{P ( T )} = \frac{( 9/10 ) ( 10/216 )}{90/216} = \frac{1}{10} .

Equivalently there are nine favorable outcomes among the ninety outcomes in

T

. This direct count confirms the conditional probability calculation.

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