Question

A man speaks the truth $6$ out of $9$ times. What is the probability that he would not narrate an incident correctly?

Answer 1

Step 1: Find the number of times the event is favored:

Let $E$ be the event of the man speaking not the truth.

Given that, the man speaks the truth $6$ out of $9$ times

That implies he speaks the lie $9-6=3times$

The sample space$\left(S\right)$ of the experiment is $=9times$

So, $n\left(E\right)=3;n\left(S\right)=9$

Step 2: Find the probability of the man not speaking truth:

$P\left(E\right)=\frac{NumberoffavourableoutcomesofeventE}{Totalnumberofoutcomes}$

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

$$\begin{gathered} \therefore P\left(E\right)=\frac{3}{9} \\ \Rightarrow P\left(E\right)=\frac{1}{3} \end{gathered}$$

Hence, the probability of the man not speaking the truth is $\frac{1}{3}.$