Probability Problem

[itzpetey]

Let's pretend this is a game. You are given 15 events, all with a different known probability of happening. You may select any number of those events. I need to figure out a formula for determining what the probability of exactly X out of Y events happening is. I'll assume that if the user is selecting 8 events, he will select the 8 with the highest probability.

So, for example, what is the probability that if you pick 9 events, 6 will occur? If you pick 9, 7 will occur? If you pick 11, 5 will occur? It's easy to determine what the probability of picking 9, and all 9 occurring is, or 0 occuring is.

Can't quite figure it out. Sorry if it's a simple question.

 

[Ishuda]

Initially, we assume you 'are given' the number of events considered for which you want the probability of so many occurring, i.e. "I'll assume that if the user is selecting 8 events, he will select the 8 with the highest probability." So what you want is the formula for, for example, given 4 events and their probabilities of occurrence, what is the probability that 3 will occur. The solution depends on whether the events are independent or not but, as the problem is stated, generally one would assume independence.

If the events all had the same probability, the formulas would be easier, i.e. the binomial distribution for 4 items with three successes, etc. If the events don't have the same probability, you will just have to go through the sets and add them up. Which sets you ask? Ah, maybe
http://www.mathsisfun.com/combinatorics/combinations-permutations.html
would help.
Going back to that 4 events with 3 occurring, how many way can we choose three items out of four. The answer is 4 choose three [=₄C₃] which is 4. What are the 4? Well we can leave out any single one so
p₁ p₂ p₃ q₄
p₁ p₂ p₄ q₃
p₁ p₃ p₄ q₂
p₂ p₃ p₄ q₁
where q = 1-p. Multiple them out and add them up. Note that we consider the events in order p₁ p₂ p₃ q₄ the same the events in order of p₂ p₁ p₃ q₄, for example. That is, order doesn't matter. Oh, and do you see why I said it would be much easier is all the probabilities were the same?

Edit: Some sort of weird something. Had to redo one of the equations, it was showing different in post than in edit preview. Oh, and also note that the example answer above is for 'exactly three' and not 'at least three'