 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-2/16/b/ii/solution

The sole green ball comes from the first bag exactly on the event

(X, Y) = (1, 0)

. Divide this joint probability by the conditioning event's probability:

P (X = 1, Y = 0 ∣ X + Y = 1) = \frac{( 3/5 ) ( 2/3 )}{13/30} = \frac{12}{13} .

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