# How to calculate some probabilities?

1. Nov 17, 2004

### Mastermind

I need help calculating some probabilities.

Here is my case:

There are 5 possibilities for an "experiment" : Right, Centre and Left, Up and down.

Chance of R = Chance of L = 26%
Chance of C = 38%
Chance of U = 8%
Chance of D = 2%

What is the possibility of each distribution in a group of 10 "experiments". The order of appearance doesn't matter.

i.e what's the probability of R=2, C = 4,L=4,U=1,D=1 (R+C+L+U+D=10), etc?????

If it is too difficult/timeconsuming please explain how am I to calculate probability of each distribution when only R,C,L are possible. Group must be 10!!!

2. Nov 17, 2004

### da_willem

2+4+4+1+1=12...

Lets call the probabalility for R pR, the probabalility for C pC etc...

The probabalility for a certain outcome, lets say RRCCLLLUUD is just pR*pR*pC*pC*pL*pL*pL*pU*pU*pD = 2,196E-8

If the order doesn't matter, so you just want to know what the chance is of getting 2 outcomes R, C and U, 3 times L and one D and you dont care in what order. You have to multiply the above calculated chance by the number of ways you can get this result. Because, and this is important, the chance on every such outcome is the same. This is, the chance to have RRCCLLLUUD as an outcome is exactly the same as the chance to get CRUULDLRCL...

But how many ways are there? RRCCLLLUUD, RCRCLLLUUD, LRRCCULLUUD, .... well a lot... If all 10 letters were different, it would be easy, it would be just 10!=10*9*8*7*6*5*4*3*2*1. But they are not the same and you have to account for that. If they are all different but two (for example the numer of ways to distribute the letters AABCDEFGHI where the order does not matter) there are 10!/2 ways to do it, because the sequences with the two A's interchanged are the same sequence. The number of ways to distribute AAABCDEFGH where order does not matter is 10!/6 or 10!/3! because there are 3! ways to interchange the A's

In general: The number of ways to distribute a sequence of 10 letters, where there are n letters who are the same and m other letters who are the same and o other letters who are the same (e.g. AAABBCCCDE here m=3, n=2 and o=2) is 10!/(m!*n!*o!*...).

So the number of ways to distribute RRCCLLLUUD where the order does not matter is 10!/(2!*2!*3!*2!*1!)= 48. So the chance for a sequence with 2 times R, 2 times C 3 times L, 2 times U and one time D is 48*2,196E-8=1,0541E-6

Good luck with calculating the rest!