1. ## I am lost!:(

You have 10 balls and 4 boxes. The balls are place din boxes at random.
a) What is the probability that 2 of the boxes contain four or more balls?
b) What is the probability that no box contians more than 3 balls?
c) What is the probability that 2 of the boxes contain exactly one ball and that each of the other 2 contain more than 2 balls?

2. Given the way you have posted the problem there is no possible answer.
There are four possible ways to read this: the balls are different and the boxes are different; the balls are identical and the boxes are different; the balls are different and the boxes are identical; both the balls are identical & boxes are identical.
Which of the four possibilities is the one that fits you question?

3. Thanks - I will ask my professor this morning.

4. Originally Posted by wvlilgurl
You have 10 balls and 4 boxes. The balls are place din boxes at random.
a) What is the probability that 2 of the boxes contain four or more balls?
b) What is the probability that no box contians more than 3 balls?
c) What is the probability that 2 of the boxes contain exactly one ball and that each of the other 2 contain more than 2 balls?
a) We will assume the boxes are equally likely. If we disregard (temporarily) the order of the boxes, the 10 balls can be placed in the boxes in 4 ways with at least 2 boxes containing at least 4 balls: 5+5+0+0, 5+4+1+0, 4+4+2+0, and 4+4+1+1.

Let's consider the 5+5+0+0 case first. If, for example, boxes 1 and 2 contain 5 balls each, then the probability of this event comes from a multinomial distribution: $\binom{10}{5,5,0,0} (1/4)^{10}$. But boxes 1 and 2 aren't the only choice: all together, there are $\binom{4}{2,2}$ choices of the boxes that result in a 5+5+0+0 mix. So the 5+5+0+0 pattern occurs with probability $\binom{4}{2,2}\binom{10}{5,5,0,0} (1/4)^{10}$.

The other cases are similar:

The pattern 5+4+1+0 occurs with probability $\binom{4}{1,1,1,1} \binom{10}{5,4,1,0} (1/4)^{10}$.

The pattern 4+4+2+0 occurs with probability $\binom{4}{2,1,1} \binom{10}{4,4,2,0} (1/4)^{10}$.

The pattern 4+4+1+1 occurs with probability $\binom{4}{2,2} \binom{10}{4,4,1,1} (1/4)^{10}$.