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Given in the question that, Suppose that $10$balls are put into $5$boxes, with each ball independently being put in box $i$with probability $p_i,\underset{i=1}{\overset{5}{鈭�}}{p}_i=1$.

Given:

$10$balls are put into $5$boxes independently

Probability of a ball being placed in box $i$is $p_i$.

$\underset{i=1}{\overset{5}{鈭�}}{p}_i=1$

The number of successes among a selected number of independent trials with a persistent probability of success observes a binomial distribution. Definition binomial probability:

$P(X=k)=\left(\begin{array}{l}n\\ k\end{array}\right)路p^k路(1-p{)}^{n-k}=\frac{n!}{k!(n-k)!}路p^k路(1-p{)}^{n-k}$

Let us evaluate the definition of binomial probability at $n=10,p=p_i$and $k=0$:

$P($box $i$is empty $)=P(X=0)$

$=\left(\begin{array}{c}10\\ 0\end{array}\right)路p_i^0路{\left(1-p_i\right)}^{10-0}$

$=1路1路{\left(1-p_i\right)}^{10}$

$={\left(1-p_i\right)}^{10}$

The expected value (or mean) is the sum of the product of each possibility $x$(number of boxes) with its probability $P\left(x\right)$.

$E($empty boxes $)=鈭�xP(x)$

$=\underset{i=1}{\overset{5}{鈭�}}1脳{\left(1-p_i\right)}^{10}$

$=\underset{i=1}{\overset{5}{鈭�}}{\left(1-p_i\right)}^{10}$

The expected number of boxes that do not have any balls is$\underset{i=1}{\overset{5}{鈭�}}鈥�{\left(1鈭�p_i\right)}^{10}$

Suppose that$10$balls are put into $5$ boxes, with each ball independently being put in box$i$ with probability$p_i,\underset{i=1}{\overset{5}{鈭�}}{p}_i=1$

Now we define the indicator random variable

$I_i=\left\{\begin{array}{l}1鈥呪赌呪赌呪赌�;1^{\text{th}}\text{box has one ball}\\ 0鈥呪赌呪赌呪赌�;\text{otherwise}\end{array}\right.$

Now the indicator random variable follows a binomial distribution

$\mathrm{P}\left({\mathrm{I}}_i=1\right)=\left(\begin{array}{c}10\\ 1\end{array}\right){\left(p_i\right)}^1{\left(1鈭�p_i\right)}^9$

Now, the number of boxes contains one ball is given by $\underset{i=1}{\overset{5}{鈭�}}鈥�I_i$

Expected number of boxes that have exactly 1 ball $=E\left[\underset{i=1}{\overset{5}{鈭�}}鈥�I_i\right]$

$=\underset{i=1}{\overset{5}{鈭�}}鈥�E\left[I_i\right]$

$=\underset{i=1}{\overset{5}{鈭�}}鈥�\left(\begin{array}{c}10\\ 1\end{array}\right){\left(p_i\right)}^1{\left(1鈭�p_i\right)}^9$

Expected number of boxes that have exactly 1 ball is$\underset{i=1}{\overset{5}{鈭�}}鈥�\left(\begin{array}{c}10\\ 1\end{array}\right){\left(p_i\right)}^1{\left(1鈭�p_i\right)}^9$