$\frac{2}{3}$ of the people on Weird Island tell the truth all the time and the rest lie all the time. You are sitting in a room with no windows and two people come in from outside.
Person 1 says: "It is raining outside"
Person 2 says: "Person 1 is telling the truth"
What is the probability that it is raining outside?
My progress:
On
According to Bayes theorem.
$P(E1 = truth) = \frac{1}{2}$
$P(E2 = lie) = 1- \frac{1}{2}$
$P(E2 = lie) = \frac{1}{2}$
$P\left(\frac{A}{E1}\right) = $people speaks truth $= \frac{2}{3}$
$P\left(\frac{A}{E2}\right) = $people speaks lie $= 1- \frac{2}{3}$
$P\left(\frac{A}{E2}\right) = $people speaks lie $= \frac{1}{3}$
$Probability = \frac{\frac{A}{E1} * E1}{\frac{A}{E1} * E1 + \frac{A}{E2} * E2}$
$$= \frac{\frac{2}{3} * \frac{1}{2}}{\frac{2}{3} * \frac{1}{2} + \frac{1}{3} * \frac{1}{2}}$$
$= \frac{1}{3} * \frac{2}{1}$
$= \frac{2}{3}$
This question has no solution. Here's why...
Define the following four cases:
Case 1: Both are liars. Probability = 1/9 Case 2: Both are truthers. Probability = 4/9 Case 3: P1 is liar, P2 is truther. Probability = 2/9 Case 4: P1 is truther, P2 is liar. Probability = 2/9
If it's truly raining outside, then after a little contemplation, you'll realize that only Case 2 could occur. Otherwise, there would be some form of contradiction.
If it isn't really raining, then only Case 1 could occur. But this information actually doesn't matter.
We can re-express the probability of rain as follows: $$P(Rain) = P(Rain \cap Case 1) + P(Rain \cap Case 2) + P(Rain \cap Case 3) + P(Rain \cap Case 4) = P(Case 1)P(Rain | Case 1) + P(Case 2)P(Rain | Case 2) + P(Case 3)P(Rain | Case 3) + P(Case 4)P(Rain | Case 4) = 1/9*P(Rain | Case 1) + 4/9*P(Rain | Case 2)$$ Since we are given no conditional information about the probability of rain given Case 1 or Case 2, the overall probability of rain can't be determined.