Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

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:

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.