Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 1 down vote favorite

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.

share|improve this question
2  
Then, after you have improved your accept rate, draw a Venn diagram. It might help you sort things out. – Gerry Myerson Jan 21 at 23:36
1  
Oh wow, you're right, thanks for the heads up, I've been behind on that. – Doug Smith Jan 21 at 23:40

1 Answer

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

share|improve this answer
Why do you add the last one? – Doug Smith Jan 21 at 23:51
Doug, did you draw that Venn diagram I suggested before? – Gerry Myerson Jan 22 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 at 15:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.