# Law of total probability or Bayes?

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, $p_1p + p_0(1-p)$?

-
"w.p. p"? with probability $p$? – Henry Oct 4 '12 at 18:53
The word "given" should tell you that what you're looking for here is a conditional probability. – Michael Hardy Oct 4 '12 at 19:33

## 1 Answer

Yes, you have to use the law of total probability

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

-
Where you say "when", you seem to mean "given that", i.e. this is a conditional probability $\Pr(\text{event happens}\mid\text{it is predicted})$. I'd have preferred standard nomenclature and notation. – Michael Hardy Oct 4 '12 at 19:32
Yes that is what I mean. I updated my answer. – wnvl Oct 4 '12 at 19:35