 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/probability/9/b/i/solution

Write

M_{n} = max_{i} E_{i}

1 - e^{- x}

for

x \geq 0

P (M_{n} \leq x) = {0, (1 - e^{- x})^{n}, x < 0, x \geq 0.

For any fixed

y

, the threshold

y + ln n

is nonnegative for all sufficiently large

n

. Then

P (M_{n} - ln n \leq y) = (1 - \frac{e ^{- y}}{n})^{n} ⟶ exp (- e^{- y}) .

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