# How would I go about finding the probability of one of these happening, but not the other two?

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.

What's the probability of event A happening, but neither B nor C?

What about (neither A nor B) or C?

Not looking for the answer necessarily, but how to do it.

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Your data are inconsistent: the numbers given for $P(A\text{ and }B\text{ and }C)$, $P(A\text{ and }B)$, and $P(B\text{ and }C$ imply that the probability of $B$ is at least $0.32$. – Brian M. Scott Jan 22 '13 at 2:09
Do you know what the principle of inclusion and exclusion (PIE) is? In particular, $|A \cup B| = |A| + |B| - |A \cap B|$. – Calvin Lin Jan 22 '13 at 2:09
@CalvinLin Not sure how that's relevant here, though. – Doug Smith Jan 22 '13 at 2:50

You can do this very easily with a Venn diagram, provided that you start with consistent data. The probability of all three is $0.05$; fill that in in the centre of the diagram. The probability of $A$ and $B$ is $0.15$, of which $0.05$ is accounted for by the probability of all three, so the probability of $A$ and $B$ but not $C$ is $0.15-0.05=0.10$; fill that in as shown below. Continuing in that fashion you can fill in all of the overlaps. The three regions that have probabilities assigned and that are part of $A$ already total $0.30$, so the probability of $A$ but neither $B$ nor $C$ must be $0$. And when you try the same thing with what’s left of $B$, you find that the probability of $B$ can’t be $0.20$: it must be at least $$0.10+0.05+0.17=0.32\;.$$ Similarly, the probability of $C$ must be at least $$0.15+0.05+0.17=0.37\;.$$