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This a sub-part of an big question,If I have $P(R_1|Q)$ how can we compute P(R_1'|Q) ?

It is given $R_1$,$R_2$ and $R_3$ are mutually exclusive events I computed $P(R_1|Q)$ using baye's theorem.

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    Go back to the definitions. What is $P(R_1|Q)$? What is $P(R_1'|Q)$? What is their sum?2011-05-06

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By conditioning on $Q$ you are simply restricting your attention to the worlds where $Q$ has already happened. All of the normal laws of probability hold in this world, which is why you have P(R'|Q) = 1 - P(R|Q).

Alternatively you could express it mathematically:

P(R'|Q) = \frac{P(R'\wedge Q)}{P(Q)} = \frac{P(Q) - P(R\wedge Q)}{P(Q)} = 1 - \frac{P(R\wedge Q)}{P(Q)} = 1 - P(R|Q)