MHB Max Value of $P(A \cap B): \min(P(A),P(B))

[Guest2]

Suppose $A$ and $B$ are events with $P(A)+P(B)>1$. Show that the largest possible value of $P(A \cap B)$ is $ \min(P(A), P(B))$.

I suspect I'm supposed to use $P(A \cap B) = P(A)+P(B) -P(A \cup B)$ but I've no idea how.

 

[johng1]

Rearrange your equation to:
$$P(A)-P(A\cap B)=P(A\cup B)-P(B)$$
I hope you now see that
$$P(A\cap B)\leq P(A)$$
Now finish.

 

[Guest2]

Rearrange your equation to:
$$P(A)-P(A\cap B)=P(A\cup B)-P(B)$$
I hope you now see that
$$P(A\cap B)\leq P(A)$$
Now finish.

Thank you. I get it now. :D