 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/7/solution

1 = c \sum_{n = 0}^{\infty} e^{- n} = \frac{c}{1 - e ^{- 1}}, c = 1 - e^{- 1} .

The tail is another geometric series:

P (X \geq N) = c n = N \sum \infty e^{- n} = \frac{c e ^{- N}}{1 - e ^{- 1}} = e^{- N} .

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