 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/discrete-mathematics-continuous-mathematics-and-probability-resit/8/b/ii/solution

For

0 \leq k \leq n

{X = k, X + Y = n}

{X = k, Y = n - k}

P (X = k ∣ X + Y = n) = \frac{e ^{- λ} λ ^{k} / k ! e ^{- μ} μ ^{n - k} / ( n - k )!}{e ^{- (λ + μ)} ( λ + μ ) ^{n} / n !} = (k n) (\frac{λ}{λ + μ})^{k} (\frac{μ}{λ + μ})^{n - k} .

Outside

0 \leq k \leq n

the conditional probability is zero. Consequently

X ∣ (X + Y = n) \sim Bin (n, \frac{λ}{λ + μ}) .

n = 0

, when it is concentrated at zero. Conditioning on the total converts the independent counts into an allocation of that total between the two rates.

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