Past exam of the computer science course of the University of Oxford / 2015 / Probability and computing / 1 / d

Table of contents
- Solution d

Solution

Let

D_{i}

be the result of roll

i

, so

X = \sum_{i = 1}^{100} D_{i}

. Each

D_{i}

has the uniform distribution on a finite set

{1, \dots, 6}

E D_{i} = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{7}{2}, E D_{i}^{2} = \frac{1 + 4 + 9 + 16 + 25 + 36}{6} = \frac{91}{6},

and hence variance

Var (D_{i}) = 91/6 - (7/2)^{2} = 35/12

. The rolls are independent, so linearity of expectation and variance additivity for independent random variables give

E X = 350, Var (X) = 100 \frac{35}{12} = \frac{875}{3} .

All the required moments are finite. Taking

a = 50

in Chebyshev's inequality now yields

P (∣ X - 350∣ \geq 50) \leq \frac{875/3}{2500} = \frac{7}{60} \approx 0.1167.

This is an upper bound, rather than an exact tail probability.

 Ancestors (10)

 View article source

 Discussion (0) New discussion

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