Independent Prob Proof

**dwsmith** · February 3rd 2011, 04:49 PM

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

Not to sure about this one but this is what I have:

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

This is all I have.

**harish21** · February 3rd 2011, 05:03 PM

Originally Posted by dwsmith

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

Not to sure about this one but this is what I have:

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

This is all I have.

Okay so,

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

**dwsmith** · February 3rd 2011, 05:09 PM

Originally Posted by harish21

Okay so,

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

Proof by contradiction:

$P\land\sim Q$

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

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

Therefore, A and B cannot be mutually exclusive.

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