 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/7/b/solution

For

0 < p < 1

, independence of the remaining indicators gives

P (E_{1} \cap {S_{n} = k}) = (k - 1 n - 1) p^{k} (1 - p)^{n - k} = \frac{k}{n} P (S_{n} = k) .

Because

0 < k < n

, the latter probability is positive for interior

p

. The two events are independent events exactly when their joint probability equals

p P (S_{n} = k)

, which is equivalent to

p = k / n

p = 0

p = 1

{S_{n} = k}

is null, and a null event is independent of every event. Therefore the full answer, including the degenerate endpoints, is

p \in {0, \frac{k}{n}, 1} .

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