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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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If $R_1'$ is the complement of $R_1$ then $P(R_1'|Q) = 1-P(R_1|Q)$, but from your question I can't tell if that's what you are asking. –  Chris Taylor May 6 '11 at 10:45
    
Yeps,that's what I am asking but how is it working? I know that $P(R_1') = 1-P(R_1)$ but i could not understand howz it holding for the conditional also? –  Max May 6 '11 at 10:50
    
Go back to the definitions. What is $P(R_1|Q)$? What is $P(R_1'|Q)$? What is their sum? –  Did May 6 '11 at 11:16

1 Answer 1

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:

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

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