1
$\begingroup$

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!

  • 0
    You’re looking for the [inclusion-exclusion principle](http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle); why don’t you have a look at that article and then see what questions you still have.2012-11-18
  • 0
    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?2012-11-18

2 Answers 2