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10 kids each place a jacket on a hangar. Later, each of them randomly grabs a jacket. What's the variance for the number of kids who end up grabbing their own jacket?

I know we should expect just a single kid to grab his/her own jacket, but how do we use that to find the variance?

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I'm surprised TSA would let kids be running around an airport like that these days... –  Arkamis Nov 8 '12 at 21:15
    
Updated for clarity. Thanks! –  David Faux Nov 8 '12 at 21:40
    
If they are randomly distribute i.i.d then wouldn't the expectation be just $\frac{1}{10}$ and you can proceed in finding the variance? –  diimension Nov 8 '12 at 21:40
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up vote 5 down vote accepted

Let $X_i=1$ if the $i$-th kid grabs her/his jacket, and $0$ otherwise. Then the number $Y$ of kids who get their own jacket is given by $$Y=X_1+X_2+X_3+\cdots +X_n,$$ where $n$ is the number of kids, here $10$. One common way to compute the variance is to use the fact that the variance is $E(Y^2)-(E(Y))^2$. You know the second part, so we concentrate on $E(Y^2)$.

We have $E(Y^2)=E((X_1+X_2+\cdots+X_n)^2)$. Expand the square, and use the linearity of expectation. We get $$E(Y^2)=\sum_{i=1}^n E(X_i^2) +2\sum_{1\le i\lt j\le n}E(X_iX_j).$$

Calculate. Since $X_i$ is $0$ or $1$. $E(X_i^2)=\Pr(X_i=1)=\dfrac{1}{n}$.

For the other part, to find $E(X_iX_j)$, it is enough to find $\Pr(X_iX_j=1)$. The probability that $X_i$ is $1$ is $\dfrac{1}{n}$. Given that $X_i=1$, the probability that $X_j=1$ is $\dfrac{1}{n-1}$. Thus $E(X_iX_j)=\dfrac{1}{n(n-1)}$.

Finally, put things together. Things simplify very considerably.

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