91Ó°ÊÓ

Initially, since the execution decision is random, the probabilities for each prisoner being executed are equal; that is, for each prisoner, the probability of being executed is \(\frac{1}{3}\). Formally, we can write: - \(P(A) = P(B) = P(C) = \frac{1}{3}\) Now, let's consider what would happen if the jailer tells prisoner A which of his fellow prisoners (B or C) will be set free.

Suppose the jailer tells A that B will be set free (without the loss of generality, as the same reasoning would apply if the jailer tells A that C will be set free). The situation for prisoner A then becomes a new conditional probability problem. We have to calculate the probability of A being executed, given that B is set free. We can write this as: - \(P(A|B')\) (Here, \(B'\) denotes the event that B is set free.) We will use the Bayes' theorem to calculate this probability: \[ P(A|B') = \frac{P(B'|A)P(A)}{P(B')} \] To find \(P(B'|A)\), we know that if A is executed, then either B or C will be set free. So we have: - \(P(B'|A) = \frac{1}{2}\) Next, we need to find the probability P(B'), that is B will be set free regardless of who will be executed. To find P(B'), we consider all possible outcomes where B is set free: 1. A is executed, and B, C is set free: \(P(A)P(B'|A) = \frac{1}{3}\cdot\frac{1}{2}\) 2. C is executed, and A, B is set free: \(P(C)P(B'|C) = \frac{1}{3}\cdot1\) So the unconditional probability of B being set free is: - \(P(B') = P(A)P(B'|A) + P(C)P(B'|C) = \frac{1}{3}\cdot\frac{1}{2} + \frac{1}{3}\cdot1 = \frac{1}{2}\) Plugging these values into the Bayes' theorem, we get: - \(P(A|B') = \frac{\frac{1}{2}\cdot\frac{1}{3}}{\frac{1}{2}} = \frac{1}{3}\) The probability of A being executed, given that B is set free, is still \(\frac{1}{3}\). Therefore, the jailer's reasoning is incorrect, and the probability for A being executed doesn't change to \(\frac{1}{2}\) if the jailer reveals the information about one of the fellow prisoners being set free.