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Exercise 3.1

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If the $k$ -th person’s arrival results in the first birthday match, the first $k - 1$ people have $365 * 364 * \dots * (365 - k + 2)$ choices of birthday assignments such that no two people have the same birthday. The $k$ -th person has $k - 1$ choices of birthdays, since their birthday must match that of one of the first $k - 1$ people.

Thus,

P (X = k) = \frac{365 * 364 * \dots * (365 - k + 2)}{36 5^{k - 1}} \frac{k - 1}{365}

fifthist

2021-12-05 00:00

Exercise 3.1

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