Question

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

(a) Find the expected number of boxes that do not have any balls.

(b) Find the expected number of boxes that have exactly $1$ball.

Short answer

(a) The expected number of boxes that do not have any balls is$\sum _{i=1}^5{\left(1-p_i\right)}^{10}$

(b) The expected number of boxes that have exactly 1 ball is$\sum _{i=1}^{5}10p_i{\left(1-p_i\right)}^9$

Step-by-step solution

Given information(part a)

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 $p_i,\sum _{i=1}^5{p}_i=1$.

Step 2:Explanation (Part a)

Given:

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

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

$\sum _{i=1}^5{p}_i=1$

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

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

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

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

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

$=1·1·{\left(1-p_i\right)}^{10}$

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

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 $)=\sum xP(x)$

$=\sum _{i=1}^{5}1\times {\left(1-p_i\right)}^{10}$

$=\sum _{i=1}^5{\left(1-p_i\right)}^{10}$

Final answer (Part a)

The expected number of boxes that do not have any balls is$\sum _{i=1}^5 {\left(1-p_i\right)}^{10}$

Given information(part b)

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

Step 5:  Explanation (Part b)

Now we define the indicator random variable

$I_i=\left\{\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em};1^{\text{th}}\text{box has one ball}\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em};\text{otherwise}\end{array}\right.$

Now the indicator random variable follows a binomial distribution

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

Now, the number of boxes contains one ball is given by $\sum _{i=1}^5 I_i$

Expected number of boxes that have exactly 1 ball $=E\left[\sum _{i=1}^5 I_i\right]$

$=\sum _{i=1}^5 E\left[I_i\right]$

$=\sum _{i=1}^5 \left(\begin{array}{c}10\\ 1\end{array}\right){\left(p_i\right)}^1{\left(1-p_i\right)}^9$

Step 6:Final answer

Expected number of boxes that have exactly 1 ball is$\sum _{i=1}^5 \left(\begin{array}{c}10\\ 1\end{array}\right){\left(p_i\right)}^1{\left(1-p_i\right)}^9$