# Thread: Probability and expected value! Urgent

1. ## Probability and expected value! Urgent

There are 11 boxes and 7 balls. The balls are dropped independently and uniformly at random in to the boxes. Find the expected number of boxes which contain no more than 1 ball.

The key part of this problem is finding the probabilty of boxes that containes no more than 1 ball. But I dont know how to get there...

Thanks very much for the help.

2. Originally Posted by Betty Jan
There are 11 boxes and 7 balls. The balls are dropped independently and uniformly at random in to the boxes. Find the expected number of boxes which contain no more than 1 ball.

The key part of this problem is finding the probabilty of boxes that containes no more than 1 ball. But I dont know how to get there...

Thanks very much for the help.
Hi Betty Jan,

Let $X_i$ be the number of balls in box number $i$, where $1 \leq i \leq 11$.

Then $X_i$ has a Binomial distribution with $n = 7 \text{ and } p = 1/11$, so
$P(X_i \leq 1) = P(X_i=0) + P(X_i=1) = (10/11)^7 + \binom{7}{1} (1/11) (10/11)^6 \qquad(*)$.

Now define
$Y_i = 1 \text{ if } X_i \leq 1$,
$\quad = 0$ otherwise, for $1 \leq i \leq 11$,
so $E[Y_i] = P(X_i \leq 1)$.

Then the expected number of boxes with no more than one ball in each box is
$E(Y_1 + Y_2 + \dots + Y_{11}) = E(Y_1) + E(Y_2) + \dots + E(Y_{11}) \qquad(**)$
$= 11 \; P(X_i \leq 1)$

where $P(X_i \leq 1)$ is given by (*) above.

The theorem that E(X+Y) = E(X) + E(Y) plays a crucial role at equation (**).
It's important to know that this theorem is true even if X and Y are not independent, which is good for us here because the $Y_i$'s are not independent.