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A firm decided to have a sales event that they take name cards from 1000 of their customers, and randomly distribute the cards back to those customers. And the customers who receive their own card will get 50% sales for all of the firms products. When X is a random variable of number of people who gets back their own card, what is the variance of X

I tried by first let $A_i$ be the event that $i^{th}$ person getting back his own card. which so $P(A_i) = 1/1000$ and $P(A_i \bigcap A_j)=1/(1000)(999)$ from here I am kind of lost what to do

David Park
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    That is a good start. Now, recall that $\text{Var}[X] = E[X^2] - E[X]^2$ and (abusing notation a bit) that $X=A_1+A_2+\dots+A_{1000}$ where these $A_i$ are indicator random variables corresponding to each of their events whether or not they occurred (taking value of 1 if they did and 0 otherwise). Notice then that $X^2 = (A_1+A_2+\dots+A_{1000})(A_1+A_2+\dots+A_{1000})$ can be expanded and simplified and recall that expected value is linear. – JMoravitz Jan 04 '22 at 16:09
  • The main topic is 'derangements'. $E(X) = SD(X) = 1.$ For $n$ as large as yours, $X$ has approximately dist'n $\mathsf{Pois}(\lambda=1).$ But $P(X=n-1)=0$ and $P(X > n) = 0.$ – BruceET Jan 04 '22 at 16:36

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