 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/discrete-mathematics-continuous-mathematics-and-probability-resit/8/b/i/solution

For

X = m

m - 1

digits must all differ from

3

and the next digit must be

3

P (X = m) = (\frac{2}{3})^{m - 1} \frac{1}{3}, m = 1, 2, \dots .

Thus

X

has the geometric distribution counting trials up to and including the first success, with success probability

1/3

E [X] = \sum_{n = 0}^{\infty} P (X > n) = \sum_{n = 0}^{\infty} (\frac{2}{3})^{n} = 3 .

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