# We have $100$ balls (numbered $1,2,\cdots,100$) and $50$ boxes (numbered $1,2,\cdots,50$).

We put the balls, independently and randomly, into the boxes.

Let $X$ be the number of boxes that receive no balls.

Find $E(X) =$ ?

Notes:

I am a little confused as to what type of probability distribution this is. Any help is greatly appreciated!

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For $i=1$ to $50$, define random variable $X_i$ by $X_i=1$ if box $i$ has no balls, and $X_i=0$ otherwise. Then $X=X_1+\cdots +X_{50}$, so we want $E(X_1+\cdots+X_{50}$).

By the linearity of expectation this is $E(X_1)+\cdots+E(X_{50})$.

But $\Pr(X_i=1)$ is the probability all the balls miss box $i$. This is $\left(\frac{49}{50}\right)^{100}$, and is $E(X_i)$.

So our required expectation $E(X)$ is $50\left(\frac{49}{50}\right)^{100}$.

Remarks: The technique of indicator random variables that was used in the solution is quite powerful. We used it to bypass the task of finding the exact distribution of the sum, and then doing the painful summation $\sum k\Pr(X=k)$.

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$\Pr(X_i)=1$ should say $\Pr(X_i=1)$. – user2357112 Jul 29 '14 at 2:37
Thank you! Parenthesis fixed. – André Nicolas Jul 29 '14 at 2:40
@AndréNicolas Awesome!! Thank you for you thorough explanation! I'm starting to really enjoy probability! – user161154 Jul 29 '14 at 2:53
You are welcome. Probability is a nice subject, fun to teach. – André Nicolas Jul 29 '14 at 2:55
The $X_i$'s are not quite iid since they can't all simultaneously be zero, so they're not independent. Doesn't change the answer, but I thought it was worth pointing out. – DavidButlerUofA Jul 29 '14 at 3:05

Prob$(X=i)$=Prob(All 100 balls go into $(50-i)$ boxes)=${50\choose{50-i}}*\frac{(50-i)^{50+i}}{50^{100}}$

The denominator is the total number of ways 50balls can be put in 100 bxes, each ball can be put on any of the 50 boxes, i.e in 50 ways, so 100 balls can be put in $50^{100}$ ways.

The numerator is the product of number of ways to choose the (50-i) boxes, times the number of ways the balls can go to those chosen boxes, such that none of the the (50-i) boxes is empty. One way to do it would be to put one ball in each of the (50-i) boxes to start with. Each of the rest of the $(100-(50-i))$ can be put in any of the (50-i) boxes.

So, E(X)=$\sum_{i}$i.Prob$(X=i)$

** I have assumed that the balls are symmetric.

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Wow, the other answer is much nicer! You learn something new everyday. – Juanito Jul 29 '14 at 2:45