# Probability-6 balls in 12 boxes

1. Oct 29, 2011

### Bassalisk

1. The problem statement, all variables and given/known data
I recently posted a bingo probability problem, and you guys helped me get an correct answer. Now I have a different problem:

6 balls (can be) are distributed into 12 boxes. What is the probability that the 10 boxes are empty?
Simple enough, right? Well its giving me headache because, I just digested the problems with dice, bingo, and conditional probability.

2. Relevant equations
3. The attempt at a solution

Straight after I saw this problem, my stomach told me that there are 126 possible combinations. But that was about it.

I don't know how to find the number of ways you can fill only 2 boxes. I know from combinations that it would look something like this:
$\binom{6}{2}$

But I feel like im missing something, because these balls in boxes I was avoiding the whole time I was learning probability. I find them very confusing.

Any help?

Last edited: Oct 29, 2011
2. Oct 29, 2011

### murmillo

One way I think is helpful is first to fix 10 boxes that you want to be empty (let's say the first ten). How many ways can you get the 6 balls to go into the last 2 boxes? That gives you the number of total ways to get 6 balls into 12 boxes with the first 10 empty. That number is the same as the number of total ways to get 6 balls into 12 boxes with, say, the last 10 empty. So you have to multiply the initial number you got by the number of ways you can pick 10 empty boxes from 12 total boxes. And then divide by how many total possible ways there are of putting 6 balls into 12 boxes. Does that make sense? By the way, it makes a difference if you mean "at least 10 boxes" or "exactly 10 boxes." I can't tell from your question which one they mean.

3. Oct 29, 2011

### Bassalisk

I was thinking the same way. But didn't know how to put that in numbers. I will try that just now.

But another thing:

About that: number of ways that 6 balls can be put in 2 boxes.

I would use

$\binom{6}{2}$ But does number of ways, covers the case where I have

Box I Box II
oooooo xxxxxx
xxxxxx oooooo

I want to exlude that right? Because If I did have that, I would have in 2 cases 11 empty boxes right?

4. Oct 29, 2011

### murmillo

Yes, that's right--if the question says that exactly 10 boxes are to be left empty then you need at least one ball in one of the two remaining boxes. Usually when they say that 10 boxes are empty, they mean at least 10 boxes are empty -- so that having 11 empty boxes is OK.

5. Oct 29, 2011

### Ray Vickson

To get the probability that exactly 10 boxes are empty, look at the probability that the six balls fall into two boxes. For each different choice of two boxes from 12, the probability is the same that those two boxes have 6 balls; it can be found by looking at the problem as a binomial distribution, where we have p = 1/12 = 1/6 as the "success" probability (ball falls into one of the two given boxes) and a total of n = 6 trials; we want the probability that all 6 trials are "success".

To get the probability that 11 boxes are empty, just get instead the probability that all 6 balls fall into a single box, and use a method similar to the above.

RGV