 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability-resit/7/b/i/solution

To obtain the

(k + 1)

st head on toss

n + 1

n

tosses must contain exactly

k

heads, and the last toss must be a head. There are

(k n)

k

head positions. By independence, each such arrangement followed by a head has probability

r^{k} (1 - r)^{n - k} r

. The arrangements are disjoint, so

h_{n, k} (r) = (k n) (1 - r)^{n - k} r^{k + 1}, n \geq k \geq 0.

n + 1

counting all tosses up to and including the

(k + 1)

st success. It is not merely the probability of

k + 1

heads somewhere among

n + 1

tosses: the last toss is prescribed to be a head.

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