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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?

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Yes, there is a slightly simpler way to solve it. Since $E$ and $F$ are mutually exclusive, and sum up to $1$, then $F=\bar{E}$. Via Law of Total Probability, we have $$P(G)=P(E\cap G)+P(\bar{E}\cap G)$$ and $$P(F\cap G)=P(\bar{E}\cap G)=7/12-1/4=1/3$$

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Much simpler is to use the law of total probability. Since $E$ and $F$ is a partition of the sample space you can write directly $P(G)=P(G\cap E) + P(G\cap F)$.

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