Question

A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six is:
${\text{A}}{\text{. }}\dfrac{3}{8} \\\ {\text{B}}{\text{. }}\dfrac{1}{8} \\\ {\text{C}}{\text{. }}\dfrac{1}{4} \\\ {\text{D}}{\text{. }}\dfrac{1}{2} \\\$

Answer 1

Hint: In order to find the probability that it is actually a six, we find the probabilities of the man speaking lies and the probability of not getting a 6 on a die, by using the total probability of an event condition. Using all the obtained data, we apply Bayes theorem to determine the answer.

Complete step-by-step answer:
Given Data,
The probability of a man speaking truth is $\dfrac{3}{4}$

We know the probability of an event is given by –
${\text{Probability = }}\dfrac{{{\text{favorable outcomes}}}}{{{\text{Total outcomes}}}}$
Also, we know that Bayes theorem states that
${\text{P}}\left( {{\text{A| B}}} \right) = \dfrac{{{\text{P}}\left( {{\text{B| A}}} \right){\text{P}}\left( {\text{A}} \right)}}{{{\text{P}}\left( {\text{B}} \right)}}$ , where P (A| B) is the probability of A given B is true, P (B| A) is the probability of B given A is true, P (A), P (B) are the independent probabilities of A and B respectively.

Rolling of a dice is an independent event which has 6 possibilities of outcomes ranging from 1 – 6. So the probability of getting a number 6 when a dice is rolled is $\dfrac{1}{6}$ .
We know the sum of probabilities of all the events in a sample space is always equal to 1.
Therefore the probability of not getting a six when a dice is rolled is given by, $1 - \dfrac{1}{6} = \dfrac{5}{6}$

Given the probability of the man speaking the truth is $\dfrac{3}{4}$
So the probability of the man not speaking the truth is given by, $1 - \dfrac{3}{4} = \dfrac{1}{4}$

Let the probability of the man speaking truth be A and the probability of man speaking lies be B. Also let the probability of 6 appearing on the dice be E.
Now from the above we have –
${\text{P}}\left( {\text{A}} \right) = \dfrac{3}{4} \\\ {\text{P}}\left( {\text{B}} \right) = \dfrac{1}{4} \\\ {\text{P}}\left( {{\text{E| A}}} \right) = \dfrac{1}{6} \\\ {\text{P}}\left( {{\text{E| B}}} \right) = \dfrac{5}{6} \\\$
Now we are supposed to find the probability of the man speaking truth given the dice shows the value 6, which is P (A| E) which is given by, using Bayes theorem:
$\Rightarrow {\text{P}}\left( {{\text{A| E}}} \right) = \dfrac{{{\text{P}}\left( {{\text{E| A}}} \right){\text{P}}\left( {\text{A}} \right)}}{{{\text{P}}\left( {{\text{E| A}}} \right){\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {{\text{E| B}}} \right){\text{P}}\left( {\text{B}} \right)}} \\\ \Rightarrow {\text{P}}\left( {{\text{A| E}}} \right) = \dfrac{{\dfrac{1}{6} \times \dfrac{3}{4}}}{{\dfrac{1}{6} \times \dfrac{3}{4} + \dfrac{5}{6} \times \dfrac{1}{4}}} \\\ \Rightarrow {\text{P}}\left( {{\text{A| E}}} \right) = \dfrac{3}{{3 + 5}} = \dfrac{3}{8} \\\$

The probability that it is actually a six is $\dfrac{3}{8}$.
Option A is the correct answer.

Note: In order to solve this type of question the key is to know the concept of probability and its formula, also the sum of events in a sample space is always one.
Knowing the concept of Bayes theorem and applying its formula appropriately is the most important step of this type of problem. The probability of an event given another event is known as conditional probability. It includes a basic principle which is given as P (A| B) P (B) = P (B| A) P (A), where A and B are two independent events.