I just want to make sure I'm doing this correctly.
Here is the problem:
Let $A$ and $B$ be sets such that $P(A \cap B)=\frac{1}{4}, P(\tilde{A})=\frac{1}{3},$ and $P(B)=\frac{1}{2}$. What is $P(A \cup B)$?
My answer is $\frac{11}{12}$ and here is my reasoning:
$P(\tilde{A})=\frac{1}{3}$, therefore, $P(A)=\frac{2}{3}$.
$P(A \cup B)=P(A)+P(B)-P(A \cap B)=\frac{2}{3} + \frac{1}{2} - \frac{1}{4}=\frac{11}{12}$
Question:
Does this make sense or I am just making stuff up?