probabilities of combinations

[rikari]

Hi,

I'm thinking if you can help me think of a better way to solve this.
Here's the situation

I have probabilities of the number of bags I can sell in a day

2 - 30 %
3 - 28%
4 - 20%
5 - 16%
6 - 6%

Now, I want to know the probability of selling, for example, 8 bags after 3 days.
What I did so far is this,

I get all the permutations of 3 numbers out of {2, 3, 4, 5, 6}, where the sum is 8, so I have the following with their respective probabilities

2,2,4 = .30*.30*.20 = .018
2,3,3 = .30*.28*.28 = .02352
2,4,2 = .30*.20*.30 = .018
3,2,3 = .28*.30*.28 = .02352
3,3,2 = .28*.28*.30 = .02352
4,2,2 = .20*.30*.30 = .018

thus, the probability of selling 8 bags after 3 days is .018+.02352+.018+.02352+.02352+.018 = .12456

My problem is, I need to get all the possible number of bags sold in 3 days with their probabilities of occurring. Is there a way that I can compute for that easily? The true situation is working with more than 5 possible number of bags in a day.

[Itai]

I dont think there is an elegant way to solve this, like I understand you are looking for, primarily since this is a random discrete distribution. One way in which I can think you could solve this easier is to compute the probability of selling LESS than 8 bags in more than 5 days, and substitute that from 1. There are much less cases in this scenario.

[rikari]

Can someone help me think on how I can use generating functions on this one? THanks

[SmoothJohn]

A quick thought. This is an example of the multinomial theorem in action. You need to expand

(a+b+c+d+e)^3, where a,b,c,d,e are the probabilities listed. Each term of the expansion will correspond to the probability of each combination of 3 probabilities.