# classic probability question...

• Jan 16th 2011, 07:20 AM
WeeG
classic probability question...
8 balls are being put in 4 cells, randomly.

1. What is the probability that all 8 balls will be put in 1 cell (doesn't matter which one) ?

2. What is the probability that each cell had at least 1 ball

3. What is the probability that in each cell there were exactly 2 balls ?

4. What is the probability that all 8 balls were put in 2 cells only (doesn't matter which ones) ?

Any help and guidance will be very appreciated !

cheers !
• Jan 16th 2011, 08:08 AM
Unknown008
The total number of ways the balls can arrange themselves in the 4 cells is $4^8$

Ball 1 has 4 possible ways, ball 2 has 4 possible ways, etc.

1.

Ball 1 can enter any cell. (4 ways)
Ball 2 - 8 must go in that particular cell. (1 way each)

2.

This is 1 - P(one cell has no ball at all)

Let's name the cells, cell 1, cell 2, cell 3 and cell 4.

Ball 1 - 8 can enter any cell besides cell 1 (3^8 ways)
Ball 1 - 8 can enter any cell besides cell 2 (3^8 ways)
Ball 1 - 8 can enter any cell besides cell 3 (3^8 ways)
Ball 1 - 8 can enter any cell besides cell 4 (3^8 ways)

Total 4(3^8) ways

Can you try those for now?

• Jan 16th 2011, 08:12 AM
ahaok
do you have the final solution to the qustion?
Quote:

Originally Posted by WeeG
8 balls are being put in 4 cells, randomly.

1. What is the probability that all 8 balls will be put in 1 cell (doesn't matter which one) ? (1/4)^8

2. What is the probability that each cell had at least 1 ball? 1-P( one cell has no ball)=1-(3/4)^8

3. What is the probability that in each cell there were exactly 2 balls ?

4. What is the probability that all 8 balls were put in 2 cells only (doesn't matter which ones) ? (1/2)^8

Any help and guidance will be very appreciated !

cheers !

• Jan 16th 2011, 08:19 AM
Plato
I think the wording of question clearly implies that the balls are identical but the cells are distinct. If we read it in that way, then there are $\dbinom{8+4-1}{8}$ ways to place the balls in the cells.

There $4$ ways to place all balls in one cell.

There are $\dbinom{4+4-1}{4}$ ways to place the balls in the cells so that each cell has at least one ball.

Can you finish?
• Jan 16th 2011, 12:43 PM
WeeG
Hi Guys, I solved 1 and 2, but stuck with 3 and 4.

Maybe my question wasn't clear, I'll tell you how I solved 2, it will help you get the point.

I said:

A - Cell 1 is empty B - Cell 2 is empty C-Cell 3 is empty D-Cell 4 is empty

Then I calculated the Union of events A,B,C and D (using the entire long formula of union with 4 events, with intersection and all that). Then the probability I was looking for is 1-the union. I did it, and got the correct answer.

On 3 I need to get 0.038 and on 4 I need to get 0.023. I have no idea how....
• Jan 16th 2011, 12:59 PM
Plato
Are you counting the balls as if they are all distinct?
Say the balls are numbered 1 to 8.

What were the first two answers?
• Jan 16th 2011, 01:07 PM
WeeG
I think they are numbered, does it matter ?

The answer for "1" is 1/4^7 , since the first ball can go anywhere and then the rest has to go that 1 cell that was chosen.

The answer for "2" is 0.62292, like I said, coming from the formula of union with 4 events reducing all intersections and adding some....(1-union)

I am really lost about "3" and "4"...no clue
• Jan 16th 2011, 01:15 PM
Plato
I should have asked, "How do you know those answers are correct?"

I disagree with this way of counting these. But I know how they are gotten.
The number of onto functions from a set of M to a set of N, $M\ge N$, is $\text{Surj}(M,N) = \sum\limits_{k = 0}^N {( - 1)^k \binom{N}{k}\left( {N - k} \right)^M }$.

The answer to #4 is $\dfrac{\binom{4}{2}\text{Surj}(8,2)}{4^8}$.

The answer to #3 is $\dfrac{8!}{2^4\cdot 4^8}$.
• Jan 16th 2011, 09:50 PM
WeeG