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Let E,F and G be three events such that E and F are mutually exclusive and $P(E\cup F)=1$, $P(G)=\frac{7}{12}$, $P(E \cap G)=\frac{1}{4}$. Then probability that $P(F \cap G)$ is:

My approach

To obtain the part of G which does not contain E, we can subtract $\frac{1}{4}$ from $\frac{7}{12}$ which will yield $\frac{1}{3}$. The book is showing this $\frac{1}{3}$ as answer but I am not able to understand why since there can be some part of G alone left in this value and we have to calculate the probability of simultaneously occurrence of F and G. Please throw light on how this will be an answer.

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Because E and F are mutually exclusive, and $P(E \cup F) = 1$, exactly one of them must happen. Thus, if $E$ doesn't happen, $F$ happens. G happens $\frac 7{12}$th of the time, and G and E happen $\frac14$ of the time. F and G happens whenever G happens but E doesn't happen, which is $\frac 7{12} - \frac 14 = \boxed{\frac 13}$ of the time.

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  • $\begingroup$ I understand now. Thank you so much. $\endgroup$
    – userNoOne
    Apr 7, 2018 at 3:24

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