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*A: Prisoner A is selected for release. *B: The guard informs Prisoner A that Prisoner B is going to be released. Are A and B independent?

Mathematics Asked by T536373 on December 3, 2020

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 the event that 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 us then consider the following events:

A: Prisoner A is selected for release.

B: The guard informs Prisoner A that Prisoner B is going to be released.

Are A and B independent? Justify your answer.

Can someone help me? I know that there are these possible events in the 1st part:

E1: A and B are released

E2: A and C are released

E3: B and C are released

With a probability of 2/3 in total because there are only 3 possibilities and in 2 it is released.

And in B I don’t understand what it means to be independent but if it means that if one passes to forces the other passes, the answer is that they are independent because the events exist:

E1: a and b are released

E2: b and c are released

But I’m not sure if the answer is that or if something else is missing.

One Answer

Chance of A being released before guard spoke is 2/3.

If A is going to be released, guard will automatically tell him the name of the other prisoner B or C who is going to be released.

So, the guard will not lie.

E = event that A will be released.

F = event that B will be named.

$$p(E|F) = frac{p(EF)}{p(F)}.$$

$$p(EF) = (1/3).$$

$$p(F) = (1/3) + [(1/3)times p].$$

Immediately above, the first term is the chance that C is not released. The 2nd term is what happens if A is not released. Note that if B is not released, B never gets named.

Unless $p$ happens to exactly $= (1/2)$, events E and F are not independent.

Addendum
Intuitive explanation for why A's chances of release improve as $p$ decreases.

For smaller $p$, when the guard names B, it becomes more likely than normal that it was because C is not being released, rather than because A is not being released.

Answered by user2661923 on December 3, 2020

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