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Hey guys, so $P(A) = 0.8$ and $P(B) = 0.6$. Since we know $P(A \cup B) \leq 1$ then if we add $P(A) + P(B)$ we get $1$.

So because it exceeds 1, that means $A$ and $B$ are NOT mutually exclusive? Is that right? And since they are not mutually exclusive, that means we can use the property: $$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$

So $$1.4 - P(A\cap B) \leq 1$$ $$-P(A\cap B) \leq -.4$$ $$P(A \cap B) \geq .4$$

Is my proof correct?

If not, please explain how I should approach this problem.

Stuy
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Since we know $P(A\cup B)\leq 1$ then if we add $P(A)+P(B)$ we get $1$.

You can't conclude $P(A)+P(B)=1$ from $P(A \cup B) \leq 1$.

So because it exceeds $1$, that means $A$ and $B$ are NOT mutually exclusive? Is that right?

This is correct, but an unnecessary step.

And since they are not mutually exclusive, that means we can use the property: $$P(A∪B)=P(A)+P(B)−P(A∩B)$$

This property is always true for any events $A$ and $B$, regardless of whether $A$ and $B$ are mutually exclusive. Your proof should start with this property, and the fact that $P(A \cup B) \leq 1$, and then the rest of your proof is correct.

kccu
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  • But in the textbook, it says "If A and B are mutually exclusive, then P(A u B) = P(A) + P(B)". – Stuy Feb 03 '19 at 20:17
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    That's right. You are given $P(A)=0.8$ and $P(B)=0.6$. You don't know if they're mutually exclusive or not, a priori. You do know that $P(A \cup B) \leq 1$. Therefore if they were mutually exclusive, you would have $P(A)+P(B) \leq 1$, i.e., $0.8+0.6 \leq 1$. This is not true, so $A$ and $B$ cannot be mutually exclusive. However, none of that is necessary to solve the problem, because whether or not they are mutually exclusive we can use the property $P(A \cup B)=P(A)+P(B)-P(A \cap B)$. – kccu Feb 03 '19 at 20:22