Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 2 / 4F (Probability)

Table of contents
- i 4F
  - Solution i
- ii 4F
  - Solution ii
- iii 4F
  - Solution iii

i

Solution

S_{n} = \sum I_{i}

, each

E I_{i} = 1/ n

. Distinct indicators have joint probability

1/ [n (n - 1)]

and zero covariance, so

E S_{n} = 1

and

var S_{n} = n (1/ n) (1 - 1/ n) = 1 - 1/ n

.

ii

Solution

Cauchy-Schwarz inequality applied to

X 1_{X > 0}

gives

(E X)^{2} \leq E (X^{2}) P (X > 0)

.

iii

Solution

Apply part (ii):

P (S_{n} > 0) \geq 1/ E (S_{n}^{2}) = 1/ (2 - 1/ n) \geq 1/2

.

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