Conditional probability find P(C)

Question

A and B are mutually exclusive events, and $P (A)=0.2$ and $P (B) = 0.8$ Find &P (AIC) when $P (CIA) = 0.4 $ and $P (ClB) = 0.5. $

From $ P (CIA) , P(C \cap A) =0.08$ From $P (CIB) , P (C \cap B) =0.4$ How can I find $P(C)$?

Answer 1

Since $P(A)+P(B)=1$ and $A \cap B =\emptyset$ it follows that $$P(C)=P(C\cap A)+P(C \cap B)$$ $$=P(C|A)P(A)+P(C|B)P(B)=(0.4)(0.2)+(0.5)(0.8)=0.48$$