 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ia/paper-2/4f/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 2 / 4F / Solution by

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

Let

X

be one score. It has the discrete uniform distribution on

1, \dots, 6

, so its expected value and second moment are

E X = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{7}{2}, E X^{2} = \frac{1 ^{2} + 2 ^{2} + 3 ^{2} + 4 ^{2} + 5 ^{2} + 6 ^{2}}{6} = \frac{91}{6} .

Therefore

E X = \frac{7}{2}, Var (X) = \frac{91}{6} - \frac{49}{4} = \frac{35}{12} .

For independent throws, linearity of expectation and additivity of variance give

E Y_{n} = 7 n /2

and

Var (Y_{n}) = 35 n /12

. The sample average consequently has expectation

7/2

35/ (12 n)

.

The Chebyshev inequality follows by applying the Markov inequality to the nonnegative square of a centered random variable. It gives

P (\frac{Y _{n}}{n} - \frac{7}{2} > \frac{3}{2}) \leq \frac{35/ ( 12 n )}{( 3/2 ) ^{2}} = \frac{35}{27 n} .

This is at most

1/10

if

n \geq 350/27

. Thus $n = 13$ suffices, as does every larger integer. This is the smallest integer guaranteed by this particular Chebyshev inequality bound; it need not be the smallest integer satisfying the actual probability requirement.

 Read the full article

 New to topics? Read the docs here!