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.
Best Answer
That's wrong. More precisely, that's correct only if $C_i >0$. If $C_i=0$ then the number of items in box $i$ can be 0 or 1.