 Solution

ID: past-exam-of-the-computer-science-course-of-the-university-of-oxford/2015/probability-and-computing/1/d/solution

Let

D_{i}

be the result of roll

i

, so

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

. Each

D_{i}

{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

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

All the required moments are finite. Taking

a = 50

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.

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