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