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Say the probability of event A happening is 0.3, event B is 0.2, event C is 0.3, the probability of (A and B) is 0.15, (A and C) is 0.2 and (B and C) is 0.22, and (A and B and C) is 0.05.

Is the probability of at least one happening:

P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) - P(A and B and C)?

Because I know P(A or B) = P(A) + P(B) - P(A and B), and I'm dealing with a P(A or B or C) here right, so I just adapted it.

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Then, after you have improved your accept rate, draw a Venn diagram. It might help you sort things out. –  Gerry Myerson Jan 21 '13 at 23:36
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Oh wow, you're right, thanks for the heads up, I've been behind on that. –  Doug Smith Jan 21 '13 at 23:40

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up vote 1 down vote accepted

This is just the inclusion-exclusion principle. You almost have it correct, save for the last sign. It should be as follows:

$P(A\cup B \cup C) = P(A) + P(B) + P(C) - P(A\cap B) - P(A\cap C) - P(B\cap C) + P(A\cap B\cap C) = .3 + . 2 + .3 - .15 - .2 - .22 + .05 = 0.28.$

See Inclusion-Exclusion Principle

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Why do you add the last one? –  Doug Smith Jan 21 '13 at 23:51
    
Doug, did you draw that Venn diagram I suggested before? –  Gerry Myerson Jan 22 '13 at 11:37
    
So you can split $A\cup B \cup C$ into different parts; a venn diagram is very useful here. But basically, if $x\in A\cap B\cap C,$ you add $x$ for each of $A,\, B,\,$ and $C,$ remove it from each of $A\cap B,\, A\cap C,\, B\cap C,$ and add it back for $A\cap B\cap C,$ for a net total of adding it once. This is exactly what you need for each element of $A\cup B\cup C$ - check that this actually works. Another example: if $x\in (A\cap B)\backslash C,$ then you add $x$ for $A$ and $B$ and remove it when you subtract $A\cap B,$ for a net total of adding it once. –  lyj Jan 22 '13 at 15:32

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