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Suppose that R and S are events in a certain probability space such that $P[R∣S]=0.52$, $P[S∣R]=0.65$ and $P[R\cap S]=0.26$.

I have found that $P[R]=0.4$ and $P[S]=0.5$.

I thought this was asking for $P[R \cup S]$ so I calculated $P[R]+P[S]-P[R\cap S]=0.4+0.5-0.26 = .64$, but this answer isn't write. I think I may be misinterpreting the question. Any help would be appreciated.

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What you want is not the union of the two sets; that would mean finding the probability of one or the other of the events or both. Rather you want what's called the symmetric difference of the events: That's the event $(R \cup S) - (R \cap S)$. (You can visualize this as the Venn diagram showing the two events, but with the intersection removed.) Therefore, you can simply calculate $P[R \cup S] - P[R \cap S ] = 0.64 - 0.26 = 0.38$.

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  • $\begingroup$ +1 for simplest and most direct way to solve the problem $\endgroup$ – Newb Feb 13 '14 at 22:41
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Here's a hint:

If you know the probabilities of $R$ and $S$ occuring, then you know the probabilities of $\lnot R$ and $\lnot S$: $P(\lnot R) = 1 - P(R)$, and $P(\lnot S) = 1 - P(S)$.

Then the probability of one, but not both of them occurring is $(P(R)\cap P(\lnot S)) \cup (P(\lnot R) \cap P(S))$.

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  • $\begingroup$ The simpler way to do this is actually just by symmetric difference. See dmk's answer. $\endgroup$ – Newb Feb 13 '14 at 22:40

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