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An event happens in a given hour with probability $p$. In advance, somebody predicts the event (successfully) with probability $p_1$ when it does occur and (incorrectly) predicts it with probability $p_0$ when it does not occur. Given this person predicts the event in the next hour, what's the probability that it will actually happen?

Is this a place where one should use the law of total probability? So, $p_1p + p_0(1-p)$?

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    The word "given" should tell you that what you're looking for here is a _conditional_ probability.2012-10-04

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Yes, you have to use the law of total probability

$P(\text{event happens | event is predicted})=\frac{P(\text{event happens and is predicted})}{P(\text{event is predicted})}$ $P(\text{event happens | event is predicted})=\frac{p_1 \cdot p}{p \cdot p_1+(1-p) \cdot p_0}$

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    Yes that is what I mean. I updated my answer.2012-10-04