Given the following question:

putting 4 balls $\{ball_i\}_{i=1}^4$ into 4 boxes $\{box_k\}_{k=1}^4$.

Each ball $ball_i$ have a fair probability to fall into each $box_k$, independently to other balls.

Calculate the expected value, of number of boxed will get empty.

I tried by finding the probability for number of empty boxes, random variable $X$ denotes the number of empty boxes.

I used a permutation $(n_1, n_2, n_3, n_4)$, where $n_i$ tells number of balls in box $box_i$.

Assume $n_1, .. , n_4 \ne 0.$

$P(X = 1) = \frac{(0, n_2, n_3, n_4)}{(n_1, n_2, n_3, n_4)} = \frac{3!}{4!} = \frac{6}{24} = \frac{1}{4}.$

$P(X = 2) = \frac{(0, 0, n_3, n_4)}{(n_1,..,n4)} = \frac{2!}{4!} = \frac{2}{24} = \frac{1}{12}.$

$P(X = 3) = \frac{(0, 0, 0, n_4)}{(n_1,..,n4)} = \frac{1!}{4!} = \frac{1}{24}.$

If it was right, then the expected value was: $\sum_{i=1}^3 P(X = i)\cdot i = \frac{1}{4} + \frac{1}{6} + \frac{3}{24} \approx 0.54.$

According to possible answers, its wrong. What is the right way to solve it? Thanks in advance.

  • $\begingroup$ $\sum P(X=i)$ mus be $1$. $\endgroup$ – Math-fun Mar 20 '15 at 10:53
  • $\begingroup$ @Math-fun Where is my mistake? $\endgroup$ – Billie Mar 20 '15 at 10:54

Define Bernoulli random variables $A_i$, $i=1,2,3,4$ where $A_i=1$ if box $i$ is empty; and $A_i=0$ if box $i$ has at least one ball. Then you want $E[\sum_{i=1}^4 A_i]=\sum_{i=1}^4 E[A_i]$

We have $E[A_i]=P(A_i=1)=\left(\frac34 \right)^4=\frac{81}{256}$.

So $\sum_{i=1}^4 E[A_i]=4\cdot \frac{81}{256}=\frac{81}{64}$


When $X=1$ then one box should be empty, say box $1$. Let's see the number of ways you could distributed 4 balls into 3 boxes:

$(3,0,0), (0,3,0), (0,0,3),(2,1,0) ... $

which will be the number of positive solutions to $$x_1+x_2+x_3=4$$ or $\binom {4-1}{3-1}=15$, but then you have to choose which box should be left empty. This is possible in $\binom41=4$ ways. Thus there are $3 \times 4 =12$ ways to distribute the balls with exactly one box empty. But the number of ways we could distribute 4 balls into 4 boxes is the number of non-negative solutions to $$x_1+x_2+x_3+x_4=4$$ or $\binom {4+4-1}{4-1}=35$, hence $$P(X=1)=\frac{12}{35}$$

Using the same logic you get $$P(X=2)=\frac{\binom42 \binom {4-1}{2-1}}{\binom {4+4-1}{4-1}}=\frac{18}{35}$$ $$P(X=3)=\frac{4}{35}$$ ... and finally there could be no box empty, i.e. $X=0$ with probabilty $$P(X=0)=\frac{1}{35}$$ The expected value is then $$E(X)=0\times \frac{1}{35} + 1 \times \frac{12}{35} + 2 \times \frac{18}{35} + 3 \times \frac{4}{35} = \frac{12}{7}.$$

  • $\begingroup$ First of all thank you, but are you sure about the final answer? It's not one of the choices $\endgroup$ – Billie Mar 20 '15 at 11:47
  • $\begingroup$ Many thanks for writing back. ;aybe I misunderstood the question. I will check it again. $\endgroup$ – Math-fun Mar 20 '15 at 12:10

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