Two out of three prisoners are chosen at random to be released. Prisoner $A$ asks the guard to investigate the selected names and to tell him one of them that is not the same. Suppose that the guard agrees to do what the prisoner asks of him and that, in case $A$ is not going to be released, he will tell him the name of prisoner B with probability $p$ and that of prisoner $C$ with probability $1 - p$.
Let's consider then the following events: $A$: Prisoner $A$ is selected for release B: Guard informs Prisoner $A$ that Prisoner $B$ is going to be released
Are $A$ and $B$ independent?
I have been stuck in this problem, I know it like a sort of variation of the classic 3 Prisioner porblem, but in this case how can I find $p$ that will prove that $A$ and $B$ are indepent.
Is it like the classic problem. Do I have only to use Bayes to prove that $P(A) = \frac{2}{3} = P(A|B)$ ?
Hint
For independence you require $\ P(A\cap B)=P(A)P(B)\ $. You have already correctly calculated that $\ P(A)=\frac{2}{3}\ $.
The event $\ A\cap B\ $ occurs if and only if prisoner $\ C\ $ is the one that won't be released. What is the probability of that event?
The event $\ B\ $ occurs if and only if either $\ C\ $ is the one that won't be released, or $\ A\ $ is the one that won't be released and gets told that $\ B\ $ will be released. What is the probability of that event? Your answer should depend on $\ p\ $.
When you plug your answers to these questions into the equation $\ P(A\cap B)=P(A)P(B)\ $ you should obtain a linear equation in $\ p\ $ which you can solve to obtain its value.