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I have the following problem:

Let $X$ be the number of distinct birthdays in a group of 110 people (i.e., the number of days in a year such that at least one person in the group has that birthday). Under the usual assumptions (no February 29, all the other 365 days of the year are equally likely, and the day when one person is born is independent of the days when the other people are born), find the mean and variance of $X$.

The solution proceeds as follows:

Let $I_j$ be the indicator r.v. for the event that at least one of the people was born on the $j$th day of the year, so $X = \sum_{j = 1}^{365}$ with $I_j \sim \text{Bern}(p)$, where $p = 1−(364/365)^{110}$.

I'm wondering how the author got that $p = 1−(364/365)^{110}$?

I would greatly appreciate it if someone could please take the time to clarify this.

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For every fixed day $j$ there is a probability of $\left(\frac{364}{365}\right)^{110}$ that all $110$ persons have their birthday on one of the $364$ other days.

So the probability that this is not the case (i.e. day $j$ is the birthday of at least one of the persons) equals: $$1-\left(\frac{364}{365}\right)^{110}$$

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  • $\begingroup$ Interesting. Thanks for the answer. So what is the combinatorial reasoning that gets us $\left(\frac{364}{365}\right)^{110}$ in the first place? $\endgroup$ – The Pointer Oct 27 '19 at 14:57
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    $\begingroup$ There are $110$ (independent) persons. Number them. The probability that the one with number $1$ has his/her birthday on one of the other days is $\frac{364}{365}$ on base of equiprobability of the days. Also the probability that the one with number $2$ has his/her birthday on one of the other days is $\frac{364}{365}$. This for all $110$ persons. The independence then tells us that the product of these factors is the probability that all $110$ have their birthday on one of the other $364$ days. $\endgroup$ – drhab Oct 27 '19 at 15:16

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