Independent Prob Proof

• Feb 3rd 2011, 04:49 PM
dwsmith
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.
• Feb 3rd 2011, 05:03 PM
harish21
Quote:

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 $\displaystyle P(A \cap B) \neq 0$.
• Feb 3rd 2011, 05:09 PM
dwsmith
Quote:

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 $\displaystyle P(A\cap B)=P(A)P(B)\neq 0$.