Let us have A1,A2,..,An - random events. Let S0 = 1, Sk = (sum for 1<=i1<..<ik<=n) P(Ai1 and Ai2 and .. and Aik). Let Bm be a event, that exactly m events from (A1,A2,..,An) would happen. How to prove, that P(Bm)=(sum k from m to n) ((-1)^(k-m)*C(k,m)*Sk)
Where C(k,m) is binomial coefficient
C(k,m) = k!/((k-m)!*m!)
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Anonymous2009-12-19 13:22
Hmm this is a hard one to describe with text. To get a better idea of what's happening, consider the case with just 3 events, A1,A2,A3, and m = 2. The possible things that can happen are:
No event happens
A1
A1A2
A1A3
A2
A2A3
A3
A1A2A3
We're interested in the probability that A1A2, A1A3, or A2A3 happens. Draw a standard Venn diagram with circles representing A1,A2, and A3... we want to add up the areas of the Venn diagram where circles overlap, EXCEPT the area in the middle where all three overlap... you have to subtract that area off from each of the sections where 2 circles overlap. So, the probability is [P(A1 and A2) - P(A1 and A2 and A3)] + [P(A1 and A3) - P(A1 and A2 and A3)] + [P(A2 and A3) - P(A1 and A2 and A3)]
= P(A1 and A2) + P(A1 and A3) + P(A2 and A3) - 3*P(A1 and A2 and A3).
You'll see that this is exactly what you get by using the formula. You'll have to fill in the details for the general case but it's the same basic idea.