0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $\sum P(X=i)$ 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
2
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $$E(X)=0\times \frac{1}{35} + 1 \times \frac{12}{35} + 2 \times \frac{18}{35} + 3 \times \frac{4}{35} = \frac{12}{7}.$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.