Prove P( A intersect B) is greater than 0.4

Question

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.

Answer 1

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.