Question

A man is known to speak the truth 3 out of 5 times. He throws a die and reports that it is ‘6’. The probability that it is actually ‘6’. The probability that is actually ‘6’ is $\dfrac{3}{k}.$ Find the value of k.

Answer 1

Hint: To solve the given question, we will use the concept of Bayes theorem. First, we will find out the probability of buying which will be obtained by subtracting the probability of speaking truth from 1. Then, we will calculate the probability that the outcome on the dice will be 6 and the probability that the outcome is not 6. Then we will use the formula for Bayes theorem which is,
$P\left( A|B \right)=\dfrac{P\left( B|A \right).P\left( A \right)}{P\left( B \right)}$

Complete step-by-step solution:
We are given in the question that a man speaks the truth 3 out of 5 times. Thus, the probability of speaking truth is,
$P\left( T \right)=\dfrac{3}{5}$
Now, the probability that the man lies is given by,
$P\left( L \right)=1-P\left( T \right)$
$\Rightarrow P\left( L \right)=1-\dfrac{3}{5}$
$\Rightarrow P\left( L \right)=\dfrac{2}{5}$
Now, it is given that when the man throws a dice, there comes an outcome. The man reports that the outcome was 6. Now, we have to find the probability that when the dice is thrown, 6 comes on it. For this, we will use the formula of finding the probability which is shown below,
$P\left( E \right)=\dfrac{\text{Favorable Outcomes}}{\text{Total Number of Outcomes }}$
Now the probability that 6 will come on the dice is,
$P\left( 6 \right)=\dfrac{1}{6}$
Similarly, the probability that any other number than 6 will appear on the dice when it is thrown is given by,
$P\left( \text{not 6} \right)=\dfrac{5}{6}$
Now, we will use the concept of Bayes theorem which says that if an event A can occur with one of the n mutually exclusive and exhaustive events ${{B}_{1}},{{B}_{2}},{{B}_{3}}.....{{B}_{n}}$ and the probability $P\left( \dfrac{A}{{{B}_{1}}} \right),P\left( \dfrac{A}{{{B}_{2}}} \right).....P\left( \dfrac{A}{{{B}_{n}}} \right)$ are known. Then
$P\left( \dfrac{{{B}_{i}}}{A} \right)=\dfrac{P\left( {{B}_{i}} \right).P\left( \dfrac{A}{{{B}_{i}}} \right)}{\sum\limits_{i=1}^{n}{P\left( {{B}_{i}} \right)P\left( \dfrac{A}{{{B}_{i}}} \right)}}$
In our case, we have,
$P\left( \text{actually 6} \right)=\dfrac{P\left( T \right).P\left( 6 \right)}{P\left( T \right).P\left( 6 \right)+P\left( L \right).P\left( \text{not 6} \right)}$
$\Rightarrow P\left( \text{actually 6} \right)=\dfrac{\dfrac{3}{5}\times \dfrac{1}{6}}{\left( \dfrac{3}{5}\times \dfrac{1}{6} \right)+\left( \dfrac{2}{5}\times \dfrac{5}{6} \right)}$
$\Rightarrow P\left( \text{actually 6} \right)=\dfrac{\dfrac{3}{30}}{\dfrac{3}{30}+\dfrac{10}{30}}$
$\Rightarrow P\left( \text{actually 6} \right)=\dfrac{\dfrac{3}{30}}{\dfrac{13}{30}}$
$\Rightarrow P\left( \text{actually 6} \right)=\dfrac{3}{13}$
Now, it is given that, $P\left( \text{actually 6} \right)=\dfrac{3}{k}.$ Thus, we have,
$\Rightarrow \dfrac{3}{13}=\dfrac{3}{k}$
$\Rightarrow k=13$.

Note: The alternate method of solving the question is shown below.
$P\left( \text{actually 6} \right)=1-P\left( \text{not actually 6} \right)$
$\Rightarrow P\left( \text{actually 6} \right)=1-\dfrac{\dfrac{2}{5}\times \dfrac{5}{6}}{\left( \dfrac{2}{5}\times \dfrac{5}{6} \right)+\left( \dfrac{3}{5}\times \dfrac{1}{6} \right)}$
$\Rightarrow P\left( \text{actually 6} \right)=1-\dfrac{\dfrac{10}{30}}{\dfrac{10}{30}+\dfrac{3}{30}}$
$\Rightarrow P\left( \text{actually 6} \right)=1-\dfrac{\dfrac{10}{30}}{\dfrac{13}{30}}$
$\Rightarrow P\left( \text{actually 6} \right)=1-\dfrac{10}{13}=\dfrac{3}{13}$
Thus, we will get the value of k as 13.