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From the question, we observe that three prisoners are informed by their jailer that one of them has been chosen at random to be executed and the other two are to be freed.

We have to find the jailor's reasoning.

Let's consider $A,B$and $C$are the three events.

The probability of each prisoners will be die is equal and they are given below:

$P\left(A\right)=\frac{1}{3}$

$P\left(B\right)=\frac{1}{3}$

$P\left(C\right)=\frac{1}{3}$

If $A$is to die he could be told either $B$ is to be forced or $C$is to be forced, each with a probability of $\frac{1}{2}$.

Consider $D$remains the event that represents the jailor told $B$to be freed.

That is, the conditional probability of jailor told $B$to be freed given $A$dies is as follows.

$P\left(D\right|Adies)=\frac{1}{2}$

If $B$is to die, $A$would not be told $B$is to be freed.

Therefore, the conditional probability of jailor told $B$to be freed given $B$dies as follows.

$P\left(D\right|Bdies)=0$

If $C$is to be die, $A$would must be told $B$is to be freed. Therefore, the conditional probability of jailor told $B$told to be freed given $C$dies as follows:

$P\left(D\right|Cdies)=1$

By using Bayes theorem, to calculate the probability of prisoner $A$dies given that jailor told prisoner $B$to be freed.

$P(A\text{dies}鈭�D)=\frac{P\left(A\text{dies}\right)P(D鈭�A\text{dies})}{\left\{\begin{array}{l}P\left(A\text{dies}\right)脳P(D鈭�A\text{dies})+P\left(B\text{dies}\right)脳\\ P(D鈭�B\text{dies})+P\left(C\text{dies}\right)脳P(D鈭�C\text{dies})\end{array}\right\}}$

$=\frac{\frac{1}{3}脳\frac{1}{2}}{\left(\frac{1}{3}脳\frac{1}{2}\right)+\left(0脳\frac{1}{3}\right)+\left(1脳\frac{1}{3}\right)}$

$=\frac{\frac{1}{6}}{\frac{1}{6}+0+\frac{1}{3}}$

Simplify,

$=\frac{\frac{1}{6}}{\frac{1}{2}}$

$=\frac{2}{6}$

Therefore,

$=\frac{1}{3}$

From the probability, we observe that the Jailor is wrong.