Equiprobable sample space: N Identical objects in K different containers

[Biker]

So we had an exam question which was the following:
Assume you have N identical balls and K different contains if there are N pre-Selected boxes and that N < k what is the probability that none of these pre-selected boxes are empty?
I answered it and it was the same as the professor's answer:
Total number of ways: (N+k-1)c(k) where c is combination
Then the prob is 1/total
But after the exam, we noticed that this is not equi-probable space. The professor didnt agree with us.

We needed to prove that for example if we have 2 different containers and we are distrbuting similar balls then for examples (3,1) isn't the same probability as (2,2)
We gave this simple example 4 different objects in 2 different containers and our event was that Box A and B have 2 and 2 thus we remove the idea of difference but we have an equiprobable space. Before we start distributing we choose one of the balls randomly and then distribute it. This ensures Equiprobablity.
Total ways = 2^4 * 4!
Our Event = 4c2 *4!
Thus this gives 3/8 (Not equiprobable)
Also this gives that the combination given by 2^4 are equiprobable so another quick method is 4c2/ 2^4 = 3/8

This is also similar to: Imagine we have 4 ones and 2 containers. Everytime we ask ourselves: add this one to one of the boxes. Gives the same answer as above.

So the answer to the exam's question would be: n!/k^n

If the above is true, Could you please inform me of a clear way to convince the professor? Otherwise where is the mistake in this argument?

 

[PeroK]

So we had an exam question which was the following:
Assume you have N identical balls and K different contains if there are N pre-Selected boxes and that N < k what is the probability that none of these pre-selected boxes are empty?

That, logically and grammatically, makes little sense to me. Can you state your problem more clearly?

 

[PeroK]

Is the problem this? Example N = 3, K =5:

There are 3 identical balls, each of which is put in one of the 5 containers (with equal probability).

You want to calculate the probability that one ball goes into each of the first three containers, say. It must be these three. Or, any set of three pre-selected containers. It clearly doesn't matter which ones they are.

You don't want to calculate the probability that you end up with any three containers with one ball each and the other two containers empty.

Is that it?

 

[Biker]

Is the problem this? Example N = 3, K =5:

There are 3 identical balls, each of which is put in one of the 5 containers (with equal probability).

You want to calculate the probability that one ball goes into each of the first three containers, say. It must be these three. Or, any set of three pre-selected containers. It clearly doesn't matter which ones they are.

You don't want to calculate the probability that you end up with any three containers with one ball each and the other two containers empty.

Is that it?

Yes you don't need to choose the 3 containers, For example assume they are the first 3

 

[PeroK]

Yes you don't need to choose the 3 containers, For example assume they are the first 3

In this case, the first ball must go into one of the three pre-selected; the second into one of the two remaining; and, the third into the one remaining. And the probability of that is:

(3/5)(2/5)(1/5) = 6/125

And, in general, it's $\frac{N!}{K^N}$

For the alternative version. The first ball can go anywhere; the second ball can go anywhere except where the first ball went; and the third ball can go anywhere else. The probability of that is:

(1)(4/5)(3/5) = 12/25

And, in general it's $\frac{K!}{(K-N)!K^N}$

 

[Biker]

So it does not matter whether the balls are identical or not ?

 

[PeroK]

So it does not matter whether the balls are identical or not ?

No. You could have an apple, an orange and a banana. Both versions of the problem are still the same.

 

[Biker]

No. You could have an apple, an orange and a banana. Both versions of the problem are still the same.

Okay so, do you have specific way or a clear question I could show that to the professor and convince him ?

 

[PeroK]

Okay so, do you have specific way or a clear question I could show that to the professor and convince him ?

Do a test with 2 containers and 1 ball (the simplest); or perhaps, 3 containers and 2 balls.

If you simulate the experiment this forces you to agree on what you mean by the problem statement.

 

[WWGD]

So we had an exam question which was the following:
Assume you have N identical balls and K different contains if there are N pre-Selected boxes and that N < k what is the probability that none of these pre-selected boxes are empty?
I answered it and it was the same as the professor's answer:
Total number of ways: (N+k-1)c(k) where c is combination
Then the prob is 1/total
But after the exam, we noticed that this is not equi-probable space. The professor didnt agree with us.

We needed to prove that for example if we have 2 different containers and we are distrbuting similar balls then for examples (3,1) isn't the same probability as (2,2)
We gave this simple example 4 different objects in 2 different containers and our event was that Box A and B have 2 and 2 thus we remove the idea of difference but we have an equiprobable space. Before we start distributing we choose one of the balls randomly and then distribute it. This ensures Equiprobablity.
Total ways = 2^4 * 4!
Our Event = 4c2 *4!
Thus this gives 3/8 (Not equiprobable)
Also this gives that the combination given by 2^4 are equiprobable so another quick method is 4c2/ 2^4 = 3/8

This is also similar to: Imagine we have 4 ones and 2 containers. Everytime we ask ourselves: add this one to one of the boxes. Gives the same answer as above.

So the answer to the exam's question would be: n!/k^n

If the above is true, Could you please inform me of a clear way to convince the professor? Otherwise where is the mistake in this argument?

Is the number N of balls the same as the number N of preselected boxes?Either way, you can count the way of leaving those empty and throwing your balls in k-N boxes and divide by the count when you consider all boxes. I think the equiprobability assumption is reasonable. What basis do you have to assume otherwise?