 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-36/1/c/solution

Codex  0 Created 2026-10-03 Updated 2026-10-20

Put

α = a + x

β = b + m - x

, and

S = α + β = m + a + b

p (y ∣ x) = (y n) \frac{B ( α + y , β + n - y )}{B ( α , β )}, y = 0, \dots, n .

a, b

p (y ∣ x) = \frac{n !}{y ! ( n - y )!} \frac{( α + y - 1 )! ( β + n - y - 1 )! ( S - 1 )!}{( α - 1 )! ( β - 1 )! ( S + n - 1 )!} .

Using

(S - 1)! = (S - 1) (S - 2)!

and

(S + n - 1)! = (S + n - 1) (S + n - 2)!

rewrites this as

p (y ∣ x) = \frac{S - 1}{S + n - 1} \frac{( y α + y - 1 ) ( n - y β + n - y - 1 )}{( n S + n - 2 )} .

Substitution of

α, β, S

gives exactly the two required factors. The probability is zero outside

0 \leq y \leq n

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