Bayes' theorem doubt

Question

Bayes' theorem states If $A_1,A_2,...,A_n$ are mutually exclusive and exhaustive events in the sample space $S$ and $E$ is any event in $S$, then $P(A_k/E)=\frac{P(A_k)P(E/A_k)}{P(A_1)P(E/A_1)+P(A_2)P(E/A_2)+...+P(A_n)P(E/A_n)}$.

A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

In my solution, $A$ is taken as event that man is saying truth and $B$ as the event that the man is lying. And the another event $E$ as the die landed 6. And it worked and I understood the solution.

But my question is, for Bayes' theorem to apply we need two events to be mutually exclusive and exhaustive. Clearly the intersection of $A$ and $B$ is empty and it can be said exhaustive also. But my question is how this set $E$ will be contained in $S$, as we want $E$ also to be an event in $S$.

Someone please answer this, so that I will be relaxed.. Thanks in advance!!

Answer 1

I assume the events are: $E$ = man is saying truth, $B$ = man is lying, $A_i$ = man is saying die landed $i$.

You are correct, all events must be defined in the same sample space, but you need a more detailed sample space: not just $\{A_i\}$, but $\{A_i \cap E, A_i \cap B\}$.

Sample space is a theoretical concept. You need not think about sample space to solve your problem using Bayes rule, events are enough.

Answer 2

• In a Bayesian situation. you will always have something like true/false,$\;\;$ present/ not present

• It is in addition to this that the sample space must consist of mutually exclusive and exhaustive events.

• Here these mutually exclusive and exhaustive events are the various possible results of a die toss.

• You can read more about it in Bayes'Theorem generalised, where various examples will also be shown

Answer 3

$P(A)=\frac{3}{4},P(B)=\frac{1}{4}$

$P(E|A)=\frac{1}{6}$

$P(E|B)=(1-\frac{1}{6})(\frac{1}{5})=\frac{1}{6}$.

We multiply by (1/5) because he choses to say a particular lie out of 5 possible lies. Here it is not sufficient to say that he lies but he chooses to speak a particular lie.

$$P(A|E)=\frac{P(A)P(E|A)}{P(A)P(E|A)+P(B)P(E|B)}=\frac{3/4\times 1/6}{3/4\times 1/6+1/4\times 1/6}=\frac{3}{4}$$