1. Independent Prob Proof

If A and B are independent events each with positive probability, prove that they cannot be mutually exclusive.

$\displaystyle \displaystyle P(A\cap B)=P(A)P(B)\Rightarrow P(A|B)=P(A) \ \ \text{and} \ \ P(B|A)=P(B)$

This is all I have.

2. Originally Posted by dwsmith
If A and B are independent events each with positive probability, prove that they cannot be mutually exclusive.

$\displaystyle \displaystyle P(A\cap B)=P(A)P(B)$

This is all I have.
Okay so,

The events are mutually exclusive if $\displaystyle P(A \cap B)=0$ . Here you have $\displaystyle P(A) \;and\; P(B)$ are both greater than 0, so $\displaystyle P(A \cap B) \neq 0$.

3. Originally Posted by harish21
Okay so,

The events are mutually exclusive if $\displaystyle P(A \cap B)=0$ . Here you have $\displaystyle P(A) \;and\; P(B)$ are both greater than 0, so $\displaystyle P(A \cap B) \neq 0$.

$\displaystyle P\land\sim Q$

$\displaystyle P(A\cap B)=P(A)P(B)>0$

We have reached a contradiction since $\displaystyle P(A\cap B)=P(A)P(B)\neq 0$.

Therefore, A and B cannot be mutually exclusive.