 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/discrete-mathematics-continuous-mathematics-and-probability/7/a/iii/solution

E (S_{n}) = \sum_{i = 1}^{n} p = n p

. Part (ii) gives

Var (X_{i}) = E (X_{i}^{2}) - [E (X_{i})]^{2} = p - p^{2}

. Independent random variables have zero pairwise covariance, so the variance of their sum is the sum of their variances. Therefore

E (S_{n}) = n p, Var (S_{n}) = n p (1 - p) .

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