A barometer manufacturer has discovered that one of its cheaper products is faulty. On rainy days, it incorrectly predicted ‘no rain’ 10% of the time. On fine days it incorrectly predicted ‘rain’ 30% of the time. In Manchester, it rains 40% of the time. If, on a particular day, one of these barometers predicts ‘rain’, what is the probability that it will actually rain?
My idea is that the probability of rain is 40% and if the barometer predicts 'no rain' for 10%, it's actually raining 90% of the time. So the probability should be $0.9\times 0.4=0.36$. However, it's wrong. I want to know why
Best Answer
You have to use Bayes' Theorem. Your result is wrong because you have to condition your probability on the total predict rain probability, thus the correct result is
$$\mathbb{P}[\text{Rain}|\text{Predict Rain}]=\frac{0.36}{0.36+0.18}=\frac{2}{3}$$