# 4 balls and 4 cells probabillity

1. Jul 8, 2015

### diracdelta

1. The problem statement, all variables and given/known data
Four balls are distributed randomly and independently in to 4 cells.
What is probability that:
a) In every cell there is exactly one ball?
b) All four balls be in one cell?
c)Two balls in one cell ( any) , other two balls in other cell (any)?
(assume that the balls are different one from each other)

3. The attempt at a solution

For a)
P=4! / (4^4)
4! because there is 4! ways that we could distribute them, and 4^4 is all possible ways to distribute balls.
b)
$$\frac{\binom{4}{1} \cdot 4!}{4^{4}}$$
Out of four cells i choose one where i will put them in, and since balls are not the same 4! to permutate them all

c)
$$\frac{\binom{4}{1} \cdot \frac{4!}{2!}\cdot \binom{3}{1}\frac{2!}{2!}}{4^{4}}$$
Out of four cells choos one where i will put two balls and out of three cells choose one where i will put two balls.

Is this ok?

2. Jul 8, 2015

### Orodruin

Staff Emeritus
No, this is wrong. You have 4 possibilities to chose the cell. The order in which the balls enter is irrelevant (there is only one order - all balls must go into the same cell).

No, you need to chose two cells in which to put two balls. Then you need to consider in how many different ways you can put 4 different balls into two different cells such that each cell has two balls (i.e., you need to pick which two balls goes into the first cell and which two goes into the second).

3. Jul 8, 2015

### PeroK

Using the counting method doesn't seem like the best way to tackle this problem. Can you think of a more direct approach?

4. Jul 8, 2015

### verty

By this logic, there are $4! * 4^4$ ways of distributing the balls. I can put any ball anywhere but I can also permute the order that I pick them.

You should assume that the balls are numbered 1-4 and you pick them in number order. Why, because it's simpler and that's usually what these questions want.

5. Jul 8, 2015

### diracdelta

Alright. So I only choose one out of 4 cells, which yields p= 4/256

As far as this, 4 choose 2 (cells) and (4 choose 2) for balls?

Last edited: Jul 8, 2015
6. Jul 8, 2015

Yes.

7. Jul 9, 2015

### verty

You need to be super careful to handle the permutations correctly. This answer is correct though (as the number of ways).

8. Jul 9, 2015

### diracdelta

Ok. Thanks for help!
I noticed.
By the way, could someone help me with PeroK's suggestion?
I have no clue what other acces should I aproach this problem with. Only thing that pops into my mind is number of number of ways "A" occurs / number of all possible ways.

9. Jul 9, 2015

### verty

The more direct way is to learn probability and the rules of probability and then to apply them to the problem. If you know the rules, you just apply them and it's more direct. For example, you might have heard about inclusion-exclusion, that is one of the rules. I think there are about 5 rules, you learn them, you apply them. Any good probability book will help you to do that (they usually cover them in the early chapters).

Last edited: Jul 9, 2015
10. Jul 9, 2015

### haruspex

I'm intrigued. I cannot think of a more direct way (only equally direct ways) to do (c) than that in the OP. Would you be so kind as to elaborate?

11. Jul 10, 2015

### verty

This is as much as I'll say on this topic.

The balls being distributed are independent trials. We can think of it as a 4-sided die being thrown 4 times. Call a throw acceptable if the desired outcome is still possible. Let $P(i)$ be the probability that throw $i$ was acceptable. We want to find $P(\cap_i)$.

Now if you know rules like the multiplication rule and complimentary events, you will be able to find a formula for the answer.

12. Jul 10, 2015

### haruspex

That's more direct!?

13. Jul 11, 2015

### verty

Perhaps not in this problem but I'll give another problem here and perhaps it'll be more obvious that it is more direct (as a method to find the answer).

This is from the book by Sheldon Ross, it's a very good book actually. Most methods fail here but my approach works.

14. Jul 11, 2015

### haruspex

The key difference with this 'council' question is that the process is inherently serialised, so requires the serial analysis you propose.

15. Jul 12, 2015

### verty

Hmm, that is a good point. And I'm starting to see that the counting method can also work here, by which I mean, thinking of it as a set and counting the elements. And that is much easier for 99% of problems than what was my approach. Like who is ever going to want to know how decisive a vote is? It's a silly question.

I think I was too critical of the other methods. There is definitely merit in them. My approach of course works but it is complicated and why shouldn't easy questions have easy solutions?