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Question

Suppose you throw m=12 balls into n=8 bins. Let Xi be the number of balls that fall into bin i. Let Ti,j Be a random variable that is 1 if the j’th ball falls in the i’th bin. Clearly Ti,j,Ti,k are independent for 1≤j

What is Pr(Xi=0)? What is Pr(Xi=1)? What is E(Xi)?
What is var(Xi)?

I am stumped on these. Please explain how to do these?

What is k? And how does $T_{ij}$ have anything to do with the problem? — ndh, Oct 22 '13 at 05:16
@tsktsktsk: It should say that $T_{i,j}$ and $T_{i,k}$ are independent for $1\le j<k\le m$. See André’s and my answers for a way to use the $T_{i,j}$’s. — Brian M. Scott, Oct 22 '13 at 05:25

score 2 · Answer 1 · answered Oct 22 '13 at 05:23

HINTS: Presumably a uniform distribution is intended, so for each $j$, the probability that ball $j$ lands in bin $i$ is $\frac18$.

What is the probability that ball $j$ does not land in bin $i$?
There are $12$ balls. What is the probability that all of them land in bins other than bin $i$, i.e., that none of them lands in bin $i$? That figure is $\mbox{Pr}(X_i=0)$.

To get $X_i=1$ you must have one ball land in bin $i$ and the other $12$ in the other bins.

What is the probability that ball $1$ lands in bin $i$ and the remaining $11$ balls all land in other bins?
What is the probability that ball $2$ lands in bin $i$ and the remaining $11$ balls all land in other bins?
What is the total probability that one ball lands in bin $i$ the the remaining $11$ balls all land in other bins?

For the next part you can use linearity of expectation:

$$X_i=\sum_{j=1}^{12}T_{ij}\;,$$

so

$$\Bbb E[X_i]=\sum_{j=1}^{12}\Bbb E[T_{ij}]\;,$$

and you need only calculate $\Bbb E[T_{ij}]=\mbox{Pr}(T_{ij}=0)\cdot0+\mbox{Pr}(T_{ij}=1)\cdot1$, since $T_{ij}$ takes on only the values $0$ and $1$.

For the last part, $\mbox{Var}(X_i)=\Bbb E[X_i^2]-(\Bbb E[X_i])^2$.

André Nicolas · Answer 2 · 2013-10-22T05:53:39.747

We have $X_i=0$ precisely if all $12$ balls miss Bin $i$. The probability that any particular ball misses Bin $i$ is $\frac{7}{8}$. So the probability they all miss is $\left(\frac{7}{8}\right)^{12}$.

To find the probability that $X_i=1$, note that this means that one ball lands in Bin $i$, and the rest don't. The ball that hits can be chosen in $\binom{12}{1}$ ways. The probability that it hits is $\frac{1}{8}$. The probability that the rest miss is $\left(\frac{7}{8}\right)^{11}$. Thus the probability that $X_i=1$ is $\binom{12}{1}\left(\frac{1}{8}\right)\left(\frac{7}{8}\right)^{11}$.

The number $X_i$ of balls that land in Bin $i$ is given by $$X_i=T_{i,1}+T_{i,2}+\cdots+T_{i,12}.\tag{1}$$ Using the fact that the expectation of a sum is the sum of the expectations, we find from (1) that $$E(X_i)=\sum_{j=1}^{12}E(T_{i,j}).$$ But each $T_{i,j}$ has mean $\frac{1}{8}$. It follows that $E(X_i)=\frac{12}{8}$.

For the variance of $X_i$, use the fact that the $T_{i,j}$ are independent, and therefore the variance of their sum $X_i$ is the sum of the variances. The variance of any $T_{i,j}$ is not difficult to compute, since $T_{i,j}=1$ with probability $\frac{1}{8}$, and $0$ with probability $\frac{7}{8}$.

Remark: The random variable $X_i$ has binomial distribution. So you can find the mean and variance of $X_i$ by appealing to standard results. However, the fact that the $T_{i,j}$ were introduced probably means they were intended to be used to derive the formulas for mean and variance, at least for this particular choice of parameters $p=\frac{1}{8}$, $n=12$.

It is possible that you were not supposed to appeal to the result that the variance of an independent sum is the sum of the variances. In that case, you will have to find $E(X_i^2)$ by expanding $(T_{i,1}+T_{i,2}+\cdots +T_{i,12})^2$. Slightly messy, you get the $E(T_{i,j}^2)$, which are no problem, and cross-terms $E(T_{i,j}T_{i,k})=E(T_{i,j})E(T_{i,k})$, which are no problem either, except you have to keep track of how many there are.

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