Let $E,F$ and $G$ be three events such that the events $E$ and $F$ are mutually exclusive, $P(E\cup F)=1$, $P(E\cap G)=1/4$ and $P(G)=7/12$. Then $P(F\cap G)=?$
My attempt: Since $P(E\cup F)=1$. It means $E\cup F$ is the entire sample space($S$). So, $E\cup F=S$. So, $P(E\cup F\cup G)=P(S\cup G)=P(S)+P(G)-P(S\cap G)=1+7/12-7/12=1$ (Since the intersection of Sample space and the set G is the set G itself.)
And then using the formula for $P(E\cup F\cup G)$ and putting all the values. I got $P(F\cap G)=1/3$.
My question: Is my approach correct? Or is there any other better way to do it?