Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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 $$0.15+0.05+0.17=0.37\;.$$

enter image description here

share|cite|improve this answer

Your Answer


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.