The events $A$, $B$, and $C$ occur with respective probabilities $0.26$ , $0.25$, and $0.41$. The events $B$ and $C$ are mutually exclusive; likewise the events $B$ and $A$ are mutually exclusive. The probability of the event $C\cap A$ is 0.02. Compute the probability of the event $C\cup (A \cap B')$.
I tried to solve it: $$P(C\cup(A\cap B')){= P((C\cup A) \cap (C\cup B')) \\=(P(C)+P(A)-P(C\cap A)) \cdot (P(C)+P(B')-P(C\cap B'))\\=(0.41 - 0.26 - 0.02)\cdot(0.41 + 0.74 - 0.3034)\\=0.65\cdot 0.8466\\=0.55029\\=}$$
I am not sure whether it is right or wrong. If it is wrong, then please tell me the correct answer. It would be much appreciated!
Hint: Since $A$ and $B$ are mutually exclusive, $A$ is completely contained in $B'$. In other words, $A \cap B' = A$.
So you can use the values of $P(A), P(C), P(C \cap A)$ to get the answer.
As to your answer, there's a problem with the 2nd step because effectively, you're writing $P(X \cap Y) = P(X) \cap P(Y)$, which makes no sense since $P(X)$ and $P(Y)$ are numbers, not sets.
Edit: $P(C \cup (A\cap B')) = P(C\cup A) = P(C) + P(A) -P(C \cap A) = 0.26 + 0.41 - 0.02 = 0.65$.