[shikaari]

5 keys are randomly placed among 3 places designated as p1,p2 and p3 Find the probability of the following events
1- place p1 is empty
2- place p1 and only p1 is empty
3- exactly 1 place is empty
4- At least 1 place is empty
5- No place is empty
6- 2 places are empty
7- 3 places are empty
8- either p1 or p2 is empty

Repeat the above experiment with n keys and three places Check you general expression against your answer to above experiment
The persons p1 p2 p3 are among those 10 persons who have placed their cheques on a counter.Each person take a cheque random from the cheques on the counter. What is the probability that at least one either p1 p2 or p3 will get his own cheque?

[Unknown008]

1. Place p1 is empty:

This combines the cases where you have the distributions:
0 5 0
0 4 1
0 3 2
0 2 3
0 1 4
0 0 5

Can you get each probability?

2. Place p1 and only p1 is empty

This is simpler, and includes:
0 4 1
0 3 2
0 2 3
0 1 4

3. Exactly 1 place is empty:

0 4 1
0 3 2
0 2 3
0 1 4
4 0 1
3 0 2
2 0 3
1 0 4
4 1 0
3 2 0
2 3 0
1 4 0

4. At least 1 place is empty:

0 0 5
0 5 0
5 0 0
0 4 1
0 3 2
0 2 3
0 1 4
4 0 1
3 0 2
2 0 3
1 0 4
4 1 0
3 2 0
2 3 0
1 4 0

5. No place is empty:

1 1 3
1 2 2
1 3 1
2 1 2
2 2 1
3 1 1

6. Two empty places:

0 0 5
0 5 0
5 0 0

7. Can't you seriously not figure this out?

8. Either p1 or p2 is empty:

0 0 5
0 5 0
5 0 0
0 4 1
0 3 2
0 2 3
0 1 4
4 0 1
3 0 2
2 0 3
1 0 4

[galactus]

Assuming the keys are identical and the boxes are different.

If we are distributing r identical objects into n different boxes, then there are

ways

In this case, ways

I would think keys are distinct, though, but perhaps not. It is not specified.

i.e. If no space is to be empty, then place one key in each of the 3 boxes. Then, there are only a matter of distributing the 2 remaining keys in the 3 boxes in ways. Which is what U has in his post. Thus, the probability no box is empty would be 6/21=2/7.

If the keys are distinct, that is another matter.