If you have 10 balls and 5 boxes what is the expected number of boxes with no balls. The probability that each ball goes independently into box $i$ is $p_i$ with the $\sum_{i=1}^5 p_i =1$. Also, what is the expected number of boxes that have exactly one ball. For part 1, isn't the answer related to the number of solutions to the equation $x_1+x_2+x_3+x_4+x_5 = 10$ where all the $x$s can take on nonnegative integers? And for the second part, isn't it the number of only positive solutions?
Probability – How to Find the Expected Number of Empty Boxes
probability
Related Solutions
Let $\chi_i$ be the binary random variable representing the event that ball $i$ is placed into box $i$. That is to say, $\chi_i=\begin{cases}1&\text{if ball}~i~\text{is matched with box}~i\\0&\text{otherwise}\end{cases}$
Question (1) can be rephrased as calculating $Pr(\chi_k=1)$. As alluded to in the comments we recognize that the arrangement of balls in boxes is precisely a permutation and each position in the permutation is equally likely to be any of the numbers (or equivalently each box is equally likely to have any of the balls). For a more formal explanation, one can approach this with multiplication principle noting that there are $(n-1)!$ ways to arrange balls into boxes satisfying that box $k$ gets ball $k$. As a result, the probability we are looking for is $Pr(\chi_k=1)=\frac{1}{n}$.
Letting $X$ represent the total number of boxes with their correct balls (i.e. the number of fixed points in the permutation) we recognize that $X=\chi_1+\chi_2+\dots+\chi_n$
Question (2) asks us to find $E[X]$ which using the above and linearity of expectation gives us $E[X]=E[\chi_1+\chi_2+\dots+\chi_n]=E[\chi_1]+E[\chi_2]+\dots+E[\chi_n]=n\cdot\frac{1}{n}=1$
Question (3) asks us to find $Var[X]$ which we have the option to expand as $E[X^2]-E[X]^2$. We already found $E[X]=1$ earlier, so all that remains apart from arithmetic is to calculate $E[X^2]$. Using that $X=\chi_1+\chi_2+\dots+\chi_n$ again we find:
$$\begin{array}{rl}X^2 &= (\chi_1+\chi_2+\dots+\chi_n)(\chi_1+\chi_2+\dots+\chi_n)\\ &= \chi_1\chi_1+\chi_1\chi_2+\chi_1\chi_3+\dots+\chi_n\chi_{n-1}+\chi_n\chi_n\\ &=\sum\limits_{k=1}^n\chi_k^2 + 2\sum\limits_{k=1}^n\sum\limits_{j=k+1}^n\chi_k\chi_j\end{array}$$
We have then $E[X^2]=E\left[\sum\limits_{k=1}^n\chi_k^2 + 2\sum\limits_{k=1}^n\sum\limits_{j=k+1}^n\chi_k\chi_j\right]$ which simplifies quite nicely due to symmetry and properties of binomial random variables.
$E[X^2]=nE[\chi_1] + 2\binom{n}{2}E[\chi_1\chi_2]$
From here there is one small probability calculation left and then a bit of arithmetic to finish.
You need to calculate $E[\chi_1\chi_2]$. Remember that the product of two binary random variables is again a binary random variable and the expected value of a binary random variable is just the probability that the random variable takes the value 1. That is to say $\chi_1\chi_2=\begin{cases}1&\text{if ball}~1~\text{is in box}~1~\text{and ball}~2~\text{is in box}~2\\0&\text{otherwise}\end{cases}$ and $E[\chi_1\chi_2]=Pr(\chi_1\chi_2=1)$
Continued:
$Pr(\chi_1\chi_2=1)=Pr(\chi_1=1)\cdot Pr(\chi_2=1\mid \chi_1=1)=\frac{1}{n}\cdot\frac{1}{n-1}$ (note, these random variables are dependent on one another). We get then $$Var[X]=E[X^2]-E[X]^2=n\cdot\frac{1}{n}+2\binom{n}{2}\frac{1}{n(n-1)}-1=1+1-1=1$$
Best Answer
We first describe informally the probability model. We grab the first ball, and choose at random a box to put this ball into, with all choices equally likely. We then independently go through the same procedure with the second ball, the third ball, and so on.
Expected Number of Empty Boxes: For $i=1, 2, \dots, 5$, define the random variable $X_i$ by $X_i=1$ if Box $i$ ends up with zero balls, and by $X_i=0$ otherwise. Let $$X=X_1+X_2+X_3+X_4+X_5.$$ Then $X$ is the total number of boxes that end up with zero balls in them. Note that $$E(X)=E(X_1+X_2+\cdots+X_5)=E(X_1)+E(X_2)+\cdots+E(X_5).$$ Next we calculate $E(X_i)$. For any $i$, $X_i=1$ if $10$ times in a row we chose one of the other boxes. Thus $P(X_i=1)=(4/5)^{10}$. It follows that $$E(X_i)=\left(\frac{4}{5}\right)^{10}.$$ Now the calculation of $E(X)$ is easy: $$E(X)=5\left(\frac{4}{5}\right)^{10}.$$
Expected Number with $1$ Ball: The same idea works. Let random variable $Y_i$ have value $1$ if Box $i$ ends up with $1$ ball, and value $0$ otherwise. Let $Y=Y_1+Y_2+\cdots+Y_5$. Then $Y$ is the number of boxes with precisely $1$ ball. We want $E(Y)$.
The probability that Box $i$ has precisely one ball is given by $$P(Y_i=1)=\binom{10}{1}\left(\frac{1}{5}\right)^1\left(\frac{4}{5}\right)^9.$$ Then $E(Y_i)=P(Y_i=1)$, and $E(Y)=5E(Y_i)$.
Comment: We could in principle deal with the first question by finding the probability distribution function of the random variable $X$, and then using the ordinary expression for expectation. Similarly, we could find the probability distribution function of the random variable $Y$. But the probability distribution functions are a little bit tricky to compute. The (standard) procedure that we used bypasses the problem of finding these distributions. It is a very powerful technique, with many applications.