 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/18/b/iii/solution

The joint probability of supplier

C

and a non-defective bulb is

P (C \cap D^{c}) = (2/5) (1 - α /4)

. Normalize by the positive non-defective probability:

P (C ∣ D^{c}) = \frac{( 2/5 ) ( 1 - α /4 )}{1 - α /2} = \frac{4 - α}{5 ( 2 - α )} .

This is defined even at

α = 0

, where it reduces to the original supplier fraction

2/5

. At

α = 1

it is

3/5

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