# Math Help - Simple probability question

1. ## Simple probability question

If 12 balls are thrown at random into 20 boxes, what is the probability that no box will receive more than one ball?

Please give only a hint, and not the full solution.

2. ## Re: Simple probability question

Originally Posted by alexmahone
If 12 balls are thrown at random into 20 boxes, what is the probability that no box will receive more than one ball?Please give only a hint, and not the full solution.
If we know that the boxes are distinct then there are $\binom{12+20-1}{12}$ ways to put 12 identical balls into twenty different boxes.
That is a start for you.

Now if there is a different reading of this question, let us know.

3. ## Re: Simple probability question

Originally Posted by Plato
If we know that the boxes are distinct then there are $\binom{12+20-1}{12}$ ways to put 12 identical balls into twenty different boxes.
That is a start for you.
Thanks, but how did you get $\binom{12+20-1}{12}$?

4. ## Re: Simple probability question

Originally Posted by alexmahone
Thanks, but how did you get $\binom{12+20-1}{12}$?
That is a standard occupancy formula.
The number of ways to put N identical objects into K distinct cells is
$\binom{N+K-1}{N}$.

BTW. Why are you asked to do problems that you seem totally unprepared to attempt?

5. ## Re: Simple probability question

Originally Posted by Plato
BTW. Why are you asked to do problems that you seem totally unprepared to attempt?
I'm really sorry but I suck at combinatorics. I'm currently teaching myself Probability and Statistics from DeGroot and got stuck at this problem.

6. ## Re: Simple probability question

Originally Posted by Plato
If we know that the boxes are distinct then there are $\binom{12+20-1}{12}$ ways to put 12 identical balls into twenty different boxes.
The 1st ball can be put into any of the 20 boxes, the 2nd ball can be put into any of the 20 boxes and so on. So, isn't the total number of ways to put 12 balls into 20 boxes $20^{12}$?

7. ## Re: Simple probability question

Originally Posted by alexmahone
The 1st ball can be put into any of the 20 boxes, the 2nd ball can be put into any of the 20 boxes and so on. So, isn't the total number of ways to put 12 balls into 20 boxes $20^{12}$?
Are you assuming that the balls are all different?
That does not seem to be implied in the statement of the question.

8. ## Re: Simple probability question

Originally Posted by Plato
Are you assuming that the balls are all different?
That does not seem to be implied in the statement of the question.
How does that make a difference, when all we're asked is to find the probability that no box will receive more than one ball?

9. ## Re: Simple probability question

My solution:

No. of favourable arrangements:
The 1st ball can be put into any of the 20 boxes, the 2nd ball can can be put into any of the other 19 boxes and so on. So, the number of ways to put 12 balls into 20 boxes so that no box receives more than one ball is $20*19*\cdots*9=\frac{20!}{8!}$.

Total no. of arrangements:
The 1st ball can be put into any of the 20 boxes, the 2nd ball can be put into any of the 20 boxes and so on. So, the total number of ways to put 12 balls into 20 boxes is $20^{12}$.

So, $P=\frac{20!}{8!20^{12}}$

10. ## Re: Simple probability question

Originally Posted by alexmahone
My solution:

No. of favourable arrangements:
The 1st ball can be put into any of the 20 boxes, the 2nd ball can can be put into any of the other 19 boxes and so on. So, the number of ways to put 12 balls into 20 boxes so that no box receives more than one ball is $20*19*\cdots*9=\frac{12!}{8!}$.

Total no. of arrangements:
The 1st ball can be put into any of the 20 boxes, the 2nd ball can be put into any of the 20 boxes and so on. So, the total number of ways to put 12 balls into 20 boxes is $20^{12}$.

So, $P=\frac{12!}{8!20^{12}}$
I think you have a great deal to learn about modeling counting problems.
You have answered this question: If we pick a function from a set of ten to a set of twenty, what is the probability that function is an injection?

Does that model what this question is asking?

11. ## Re: Simple probability question

Originally Posted by Plato
You have answered this question: If we pick a function from a set of twelve to a set of twenty, what is the probability that function is an injection?

Does that model what this question is asking?
Yes.

BTW, my answer matches with the one given at the back of the textbook. Why do you think it is wrong?

12. ## Re: Simple probability question

Originally Posted by alexmahone
Yes.
BTW, my answer matches with the one given at the back of the textbook. Why do you think it is wrong?
I don't think that the answer is wrong. I think the model is wrong.
Why not assume that the boxes are identical and the balls are different?
OR, Why not assume that the boxes are identical and the balls are identical?

These are legitimate questions asked in counting theory.

How does one model these sorts of questions.