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

Most of the problems in my textbook have numeric solutions in the back of the book except the proofs. Is this proof correct?

Prove that if $A \cup B$ and $A \cap B$ are independent events, then either $P(A \cap B)=0$ or $P(A \cup B) = 0)$.

  1. $P((A \cup B) \cap (A \cap B)) = P(A \cup B)P(A \cap B)$ (definition of independence)
  2. $P(A \cap B)=P(A \cup B)P(A \cap B)$ (properties of sets and intersections)
  3. Either $P(AB)=0$ or $P(A\cup B)=1$. In the first case we already proved the proposition. In the second case $P(A\cup B)=1\implies P(\bar{A} \cap \bar{B})=0$. $\square$
share|cite|improve this question
What does $P(AB)$ mean? Do you mean $P(A \cap B)$? – Fly by Night Jan 19 '13 at 22:29
If $A$ and $B$ are each equal to the whole sample space, then $A\cup B$ and $A\cap B$ are independent and each has probability $1$. – Michael Hardy Jan 19 '13 at 22:29
Yes. Some older probability texts write $AB$ instead of $A\cap B$. – ncmathsadist Jan 19 '13 at 22:32
@ncmathsadist The OP used both $AB$ and $A\cap B$. – Fly by Night Jan 19 '13 at 22:42
I noticed that. That is indeed a gaffe Mr. Dong has perpetrated. He should use the $A\cap B$ notation, this being the early 21st century. – ncmathsadist Jan 19 '13 at 23:24
up vote 1 down vote accepted

The second line should read

$$(A \cap B) = (A\cap B) \cap (A\cup B) \Rightarrow P(A \cap B) = P[(A\cap B) \cap (A\cup B)]$$

This is the properties of sets mentioned, and we do not apply the probability aspect as yet. Your second line currently isn't explained, but you'd see that it is true by taking probabilities on the sets above.

The third line should then read

$$ P(A\cap B) = P(A\cup B) P(A\cap B) \Rightarrow P(A\cap B) = 0 \mbox{ or } P(A \cup B) = 1$$

share|cite|improve this answer

Let $X$ and $Y$ be any events such that $X\subseteq Y$. If $X$ and $Y$ are independent, then $P(X\cap Y)=\Pr(X)\Pr(Y)$. But clearly $\Pr(X\cap Y)=\Pr(X)$, since $X\cap Y=X$. It follows that $$\Pr(X)=\Pr(X)\Pr(Y).$$ If $\Pr(X)\ne 0$, by cancellation we obtain $\Pr(Y)=1$.

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.