Question

Three prisonersA,B, andCon death row know that exactly two of them are going to be executed, but they do not know which two. PrisonerAknows that the jailer will not tell him whether or not he is going to be executed. He, therefore, asks the jailer to tell him the name of one Prisoner other thanAhimself who will be executed. The jailer responds thatBwill be executed. Upon receiving this response, PrisonerAreasons as follows: Before he spoke to the jailer, the probability was \(\frac{2}{3}\) that he would be one of the two prisoners executed. After speaking to the jailer, he knows that either he or PrisonerCwill be the other one to be executed. Hence, the probability that he will be executed is now only \(\frac{1}{2}\). Thus, merely by asking the jailer his question, the Prisoner reduced the probability that he would be executed \(\frac{1}{2}\)because he could go through exactly this same reasoning regardless of which answer the jailer gave. Discuss what is wrong with prisonerA’s reasoning.

Short answer

Hence the probability is \( - 0.75\) . And the information about Prisoners \(A\) to jailers is wrong.

Step-by-step solution

Given information

Hence the probability is\( - 0.75\). And the information aboutPrisoners\(A\)to jailers is wrong.

There are three prisoners \(A,B\) and \(C\). The Prisoner \(A\) knows that the jailer will not tell him whether or not he is going to be executed. The jailer responds that Prisoner \(B\) will be executed.

State the event  

The Prisoner \(A\) has the probability \(p\left( A \right) = \frac{1}{3}\)

The Prisoner\(B\)has the probability\(p\left( B \right) = \frac{1}{3}\)

The Prisoner\(C\)has the probability\(p\left( C \right) = \frac{1}{3}\)

The Prisoner \(A\) spoke to the jailer, one of two Prisoners executed. Then the probability is defined as \(p\left( {B \cap C} \right) = \frac{2}{3}\)

After speaking to the jailer, the Prisoner\(A\)knows that either he or the Prisoner\(C\)will be another one to be executed. Then the probability is given by\(p\left( {A \cap C} \right) = \frac{1}{2}\)

Thus, merely by asking the jailer his question, the Prisoner reduced the probability that he would be executed\(\frac{1}{2}\). Then the probability of asking the jailer questions and the Prisoner reduced is defined by\(p\left( {A \cap B} \right) = \frac{1}{6}\)

Therefore, the probability of one of two Prisoner executed or either Prisoner\(A\)will be executed define that\(p\left( {A \cap B \cap C} \right) = \frac{1}{{27}}\)

If no prisoner not is executed, then the probability is given by:

\(\begin{aligned}{}p\left( {A \cup B \cup C} \right) &= p\left( A \right) + p\left( B \right) + p\left( C \right) - \left( {p\left( {A \cap B} \right) + p\left( {B \cap C} \right) + p\left( {A \cap C} \right)} \right) + p\left( {A \cap B \cap C} \right)\\ &= \frac{1}{3} + \frac{1}{3} + \frac{1}{3} - \left( {\frac{1}{6} + \frac{2}{3} + \frac{1}{2}} \right) + \frac{1}{{27}}\\ &= \frac{{28}}{{27}} - \frac{4}{3}\\& = - 0.44\end{aligned}\)

Compute the probability

To compute the probability by the theorem of conditional probability. So then it is defined as

\(\begin{aligned}{}p\left( {A\left| {A \cup B \cup C} \right.} \right) &= \frac{{p\left( {A \cap \left( {A \cup B \cup C} \right)} \right)}}{{p\left( {A \cup B \cup C} \right)}}\\& = \frac{{p\left( A \right)}}{{p\left( {A \cup B \cup C} \right)}}\\ &= \frac{{0.33}}{{ - 0.44}}\\& = - 0.75\end{aligned}\)

Hence jailer gave that Prisoner \(A\) was wrong.

And the wrong with Prisoner \(A\) - He thinks that he only gains information of Prisoner \(B\) but not gain information of Prisoner \(C\). Because the jailer would like to name Prisoner \(C\), the probability will be negative, and the Prisoner \(A\) is wrong.