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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ia/paper-2/10f/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 2 / 10F / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-14

Each waiting time has the geometric distribution on

1, 2, \dots

with success probability

1/2

: a wait of length

n

consists of

n - 1

failures followed by the prescribed success, so its probability is

2^{- n}

. Hence

G_{H_{j}} (s) = G_{T_{j}} (s) = n \geq 1 \sum (\frac{s}{2})^{n} = \frac{s}{2 - s}, E H_{j} = E T_{j} = G_{H_{j}}^{'} (1) = 2.

(1)

Disjoint successive blocks of independent tosses give independent waiting times. More explicitly, any specified list of block lengths determines exactly one allowed sequence of heads and tails and has probability

2^{- \sum_{j} (h_{j} + t_{j})}

, the product of the individual block probabilities.

There are

2 r

waiting times. Multiplication of their probability generating functions and addition of their expectations give

G_{Y_{r}} (s) = (\frac{s}{2 - s})^{2 r}, E Y_{r} = 4 r .

(2)

To recover its probability mass function, expand by the negative binomial series:

G_{Y_{r}} (s) = \frac{s ^{2 r}}{2 ^{2 r}} (1 - s /2)^{- 2 r} = \frac{s ^{2 r}}{2 ^{2 r}} \sum_{m \geq 0} (2 r - 1 m + 2 r - 1) (s /2)^{m} .

(3)

The coefficient of

s^{n}

is therefore

P (Y_{r} = n) = 2^{- n} (2 r - 1 n - 1), n \geq 2 r,

(4)

and it is zero below

2 r

. Equivalently there are

(2 r - 1 n - 1)

compositions of

n

into

2 r

positive block lengths, each with probability

2^{- n}

. This is the negative binomial distribution for a sum of geometric waiting times, even though the desired outcome alternates between blocks.

For

r = 1

P (Y_{1} \geq 5) = 1 - (\frac{1}{4} + \frac{2}{8} + \frac{3}{16}) = \frac{5}{16} .

(5)

Since

E Y_{1} = 4

, Markov's inequality gives

P (Y_{1} \geq 5) \leq 4/5

, and indeed

5/16 < 4/5

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