Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 2 / 4F (Probability)

Table of contents
- a 4F
  - Solution a
- b 4F
  - Solution b

a

Solution

The number of stops is

X \sim Bin (n, 1/ n)

. For each fixed

k \geq 0

,

P (X = k) = (k n) n^{- k} (1 - n^{- 1})^{n - k} ⟶ \frac{e ^{- 1}}{k !},

the mass function of a Poisson random variable of mean one.

b

Solution

At station

j

, the probability of stopping is

\frac{2}{n} - \frac{1}{n} (1 - \frac{1}{n})^{j - 1},

because the lower rate applies exactly when no earlier stop occurred. By linearity of expectation,

E X = 2 - \frac{1}{n} \sum_{j = 1}^{n} (1 - \frac{1}{n})^{j - 1} = 1 + (1 - \frac{1}{n})^{n} ⟶ 1 + e^{- 1} .

 Ancestors (9)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version