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Three events $E$, $F$, and $G$ cannot occur simultaneously. It is known that $$P(E F) = P(F G) = P(E G) = \frac{1}{3}.$$ What is the value of $P(E)$?

Thanks in advance!

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    $\begingroup$ Are those given probabilities $1/3$? Please edit your question to make that clear. $\endgroup$ – Rory Daulton Feb 15 '15 at 11:24
  • $\begingroup$ Inclusion-Exclusion-Principle? $\endgroup$ – fgp Feb 15 '15 at 11:39
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If $E$, $F$, and $G$ can't all occur simultaneously, then these events $EF$, $EG$, and $FG$ are essentially mutually exclusive (the probability of the intersection of any two of them is zero). You'll need to argue that, but it's pretty quick. So then since each of $EF$, $EG$, and $FG$ also have probabilities of $\frac{1}{3}$, they are essentially exhaustive (meaning that there's probability 1 of being in exactly one of them). So being in $E$ has the same probability as being in $EF\cup EG$. And since being in $EF\cap EG=EFG$ is probability zero, you can finally conclude that $P(E)=P(EF)+P(EG)$.

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Hint: Draw a $3$-circle Venn diagram with circles representing $E$, $F$, and $G$. Fill in the triple intersection with $0$ (probability). Then see what other regions you can fill in with the given set difference probabilities.

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