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.

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

I'm taking a graduate course in probability and statistics using Larsen and Marx, 4th edition and I'm struggling with a seemingly basic question.

If A and B are any two events, not mutually exclusive: $$P((A \cup B) ^\complement) = 0.6, P(A \cap B) = 0.2$$

What is the probability that A or B occurs, but not both? Or in other words: $$P((A \cap B)\ ^\complement \cap (A \cup B)) = ?$$

So far, I've been able to infer the following: $$ P(A \cup B) = 1 - P((A \cup B) ^\complement ) = 1 - 0.6 = 0.4 $$ and $$P((A \cap B) ^\complement) = 1 - P(A \cap B) = 1 - 0.2 = 0.8$$

Can someone kindly give a hint as to how to approach from here? Am I heading in the right direction? How should I think about these types of problems in general? I feel like the text gives you the basic set of axioms to define things but then neglects the showing of solutions for slightly more complicated examples of compound probability equations such as the above.

share|cite|improve this question
Have you tried drawing a Venn diagram? – Qiaochu Yuan Jun 11 '12 at 15:00
Try finding another expression of $P((A\cap B)^c\cap(A\cup B))$, maybe with Venn diagrams. – Gregor Bruns Jun 11 '12 at 15:01
Drawing the Venn diagram should really be the first thing you do in a situation like this. It makes this problem basically a one line computation. (See Cameron Buie's answer.) – rschwieb Jun 11 '12 at 15:12
I guess it wasn't intuitive to me that probabilities would be additive in that sense. I did draw the Venn Diagram and was tempted but I thought better of it and decided to ask. – PatternMatching Jun 11 '12 at 15:19
up vote 3 down vote accepted

Hint: Find the probability of $A$ only plus the probability of $B$ only.

Let $x= P(A~ \text{only})$, $y=P(B ~\text{only})$. Then $x +y + 0.2 =0.4$. Thus $x+y= 0.2$

share|cite|improve this answer

THe probability that at least one of them occurs is $0.4$ and the probability that both occur is $0.2$. Thus, the probability that exactly one occurs is $0.4-0.2=0.2$.

share|cite|improve this answer

$P((A \cup B)\setminus(A \cap B)) = P(A\setminus B) + P(A\setminus B) = P(A) + P(B) − 2P(A \cap B)$ you will get the answer!

share|cite|improve this answer
what is $P((A \cup B)(A \cap B))$? – Surb May 3 '15 at 13:40
@timmbob: There was a backslash in the source text; yagmur intended and tried to write $(A\cup B)\setminus (A\cap B)$ as one would expect based on what followed. – Jonas Meyer May 3 '15 at 17:21
I very much doubt that you meant $A\setminus B$ in both cases. – Cameron Buie Sep 18 '15 at 5:02

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.