Find $P(A \cap B)$ with $P(A)=0.4$, $P(B) = 0.3$ and $P(A \cup B) = 0.6$ 
Find $P(A \cap B)$ with $P(A)=0.4$, $P(B) = 0.3$  and $P(A \cup B) =
 0.6$.

My professor doesn't specify whether these events are mutually exclusive. If I solve this as if they are, I get
$$P(A\cap B) = P(A)*P(B) = 0.12$$
If I solve this as if they are not, I get
$$P(A\cup B) = P(A)+P(B)-P(A\cap B) \\
\Leftrightarrow 0.6 = 0.4+0.3 -P(A\cap B)\Leftrightarrow \\
0.6 -0.4-0.3 = -P(A \cap B) \Leftrightarrow \\
0.1 = P(A\cap B)$$
The values are close, kind of. My professor solved this as if they are not mutually exclusive. Is there a way of telling, or did he simply forget to mention that?

J.W. Tanner is exactly right. In the case that A and B were mutually exclusive, both can not happen at the same time so   $P(A∩B)$ would just be zero. 
