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If $P(A) = 0.2, P(B) = 0.45$, what is $P(A \cup B)$ assuming $A$ and $B$ are mutually exclusive.

So $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

So it will be $0.2 + 0.45 - 0.0$? As the intersection probability is $0$?

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    Possible duplicate of [Probability and independent vs mutually exclusive events](http://math.stackexchange.com/questions/54495/probability-and-independent-vs-mutually-exclusive-events) – 2017-01-20

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Yes, it will be $P(A\cup B)=0.65$. If two events are mutually exclusive that means that if one event happens the other must not happen. Your answer is correct.