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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!!

Ethan Bolker
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math student
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3 Answers3

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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.

kludg
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  • 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

true blue anil
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  • The link you’ve provided stresses out that conditioning and Bayes analysis is done for sequential events, which I find highly misleading. The whole idea of Bayes is disintegration which is symmetric in what’s condition and what to be found – SBF Feb 04 '23 at 19:17
  • @Ilya: I referred OP to a particular part of the link where the student's doubt lay – true blue anil Feb 04 '23 at 19:24
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$P(A)=\frac{3}{4},P(B)=\frac{1}{4}$

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

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

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.

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

Maverick
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  • You have computed the probability that he is telling the truth given that the die came up six. Telling the truth is independent of the outcome of the die which is why you got $\frac{3}{4}$. We want the probability that the die came up six given that he said it did. – John Douma Feb 04 '23 at 20:11
  • Please take a look at my computation of $P(E|B)$ – Maverick Feb 04 '23 at 20:48