# probability set theory - probability of x out of n events

I am having an issue related to probability set theory with intersection/union terms.

When calculating the union of terms or in other words, the probability that at least one terms "fails", it can be written as the following for three terms:

P(A+B+C) = P(A)+P(B)+P(C)-P(AB)-P(AC)-P(BC)+P(ABC).

My question is how to assess a problem similar to this when we are looking at the probability at least x terms out of n terms fail. For example, at least 2 out of 4. Or at least 7 out of 10. I thought I came up with the correct answer when looking at a system of only 4 terms. For example:

P(at least 2 out of 4) = P(AB)+P(AC)+P(AD)+P(BC)+P(BD)+P(CD) -2*(P(ABC)+P(ABD)+P(ACD)+P(BCD)) +3*(P(ABCD))

P(at least 3 out of 4) = P(ABC)+P(ABD)+P(ACD)+P(BCD) -3*(P(ABCD))

This works here. In fact, it works for P(at least 2 out of n) and P(at least (n-1) out of n) for all cases of n. However it does not work for the situations in between.

I am looking for an analog formula that can evaluate any case for the probability of at least x out of n failure. Any help with be appreciate. Thanks!

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You’re looking for the inclusion-exclusion principle; why don’t you have a look at that article and then see what questions you still have. – Brian M. Scott Nov 18 '12 at 2:56
Hi - thanks for responding. I understand this principle as it applies to the union of events occuring. I cannot figure out how it applies to the probability that suppose 2 out of n events fail since it is not as simple as adding and subtracting the sum of each intersection terms. Does this make sense? For example, P(at least 2 out of 4) is not equal to sum(P(2-int terms)) - sum(P(3-int terms)) + sum(P(3-int terms)). Does this make more sense? – Holly Nov 18 '12 at 3:06

Update - I figured it out. Yes, it does have to do with the exclusion/inclusion principle but it becomes very tricky. Basically I went through each intersection level for n = 5 events and calculated how many of each term I am subtracting and adding. Doing this, I was able to get coefficients needed for subtracting and adding each subsequent intersection term. There isn't a pretty formula for it, but I put it into a code to be able to populate the coefficients for each x out of n case of interest.

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My apologies for not getting back to you: I got distracted and completely forgot. – Brian M. Scott Nov 19 '12 at 7:40

http://en.wikipedia.org/wiki/Binomial_distribution

Specfically, the cumulative distribution function (cdf). That formula tells the probability that less than or equal to x successes have occurred, so you'd want 1 - cdf for your "at least" question.

Example: 2 or fewer successes out of 3 trials where a success has probability 1/3

F(2,3,1/3) = ({3!/[0!(3-0)!]}[(1/3)^(0)][(2/3)^(3-0)]) + ({3!/[1!(3-1)!]}[(1/3)^(1)][(2/3)^(3-1)]) + ({3!/[2!(3-2)!]}[(1/3)^(2)][(2/3)^(3-2)])

= [(6/6)(8/27)] + [(6/2)(4/27)] + [(6/2)*(2/27)] ~= .963

In this case, you could have just found the probability of exactly 3 successes in 3 trails. Pr(X=3 | n=3, p=(1/3)) = (1/3)^3 ~= 0.037

I apologize for the lack of mathematical notation, but the formula on wikipedia is correct. For large n (trials), the formula is cumbersome and there are other methods described on wikipedia. There are also cumulative distribution tables for the binomial distribution.

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