 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/discrete-mathematics-continuous-mathematics-and-probability-resit/7/b/iv/solution

There is no such value of $p$ in the specified interval. Indeed,

P (Y = 1) = p e^{- p} > 0

P (X = 1 ∣ Y = 1) = 1 \neq = p = P (X = 1) .

This contradicts independence for every

0 < p < 0.8

. Equivalently,

(1, 2)

has joint probability zero but the product of its marginal probabilities is

p e^{- p} p^{2} /2 > 0

At the suggested value

r = e^{- r}

 New to topics? Read the docs here!