Question

A man is known to speak the truth $3$ out of $4$ times. He throws a die arid report that it is six. The probability that it actually a six is

• A. $\frac{3}{8}$
• B. $\frac{1}{5}$
• C. $\frac{3}{4}$
• D. None of these

Answer 1

The correct option is A

$\frac{3}{8}$

Explanation for the correct option:

Finding the probability:

Given that, the probability that man speaks truth

$P\left(A\right)=\frac{3}{4}$

Hence, the probability that man speaks a lie

$\begin{array}{rcl}P(A\text{'})& =& 1-\frac{3}{4}\\ & =& \frac{1}{4}\end{array}$

The probability of getting $6$

$P\left(B\right)=\frac{1}{6}$

The probability of not getting $6$

$\begin{array}{rcl}P(B\text{'})& =& 1-\frac{1}{6}\\ & =& \frac{5}{6}\end{array}$

By using Bayes theorem, the probability that is actually a six is:

$\begin{array}{rcl}P\left(A\right|B)& =& \frac{\mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}}{\mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{(}\mathbf{B}\mathbf{\text{'}}\mathbf{)}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\text{'}}\mathbf{)}}\\ & =& \frac{{\displaystyle \frac{1}{6}}.{\displaystyle \frac{3}{4}}}{\left(\frac{1}{6}.\frac{3}{4}\right)+\left(\frac{5}{6}.\frac{1}{4}\right)}\\ & =& \frac{3}{8}\end{array}$

So, the required probability $=\frac{3}{8}$

Hence, the correct answer is option (A).