Question

$X$ speaks truth in $60\%$ and $Y$ in $50\%$ of the cases. The probability that they contradict each other in narrating the same incident is

• A. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{2}{3}$

Answer 1

The correct option is C

$\frac{1}{2}$

Explanation for the correct option:

Step $1$: Finding the probability of the events

Formulas:

Probability of an event $E$,

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

$P\left(E\text{'}\right)=1-P\left(E\right)$

Given,

The probability that $X$ speaks truth is:

$P\left(X\right)=\frac{60}{100}$

$\Rightarrow$ $P\left(X\right)=\frac{3}{5}$

The probability that $Y$ speaks truth is:

$P\left(Y\right)=\frac{50}{100}$

$\Rightarrow$ $P\left(Y\right)=\frac{1}{2}$

Step $2$: Finding the probability of contradiction

$X$ and $Y$ contradict each other, if $X$ speaks truth and $Y$ speaks false or $Y$ speaks truth and $X$ speaks false

Hence, the required probability that they contradict each other in narrating the same incident is,

$P\left(contradict\right)=P\left(XtruthandYfalse\right)+P\left(YtruthandXfalse\right)$

$\Rightarrow$ $P\left(contradict\right)=P\left(X\right)·P\left(Y\text{'}\right)+P\left(Y\right)·P\left(X\text{'}\right)$

$\Rightarrow$ $P\left(contradict\right)=\left(\frac{3}{5}\right)·\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}\right)·\left(1-\frac{3}{5}\right)$ $\left[\because P\left(E\text{'}\right)=1-P\left(E\right)\right]$

$\Rightarrow$ $P\left(contradict\right)=\left(\frac{3}{5}\right)·\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)·\left(\frac{2}{5}\right)$

$\Rightarrow$ $P\left(contradict\right)=\left(\frac{3}{10}\right)+\left(\frac{2}{10}\right)$

$\Rightarrow$ $P\left(contradict\right)=\frac{5}{10}$

$\Rightarrow$ $P\left(contradict\right)=\frac{1}{2}$

Hence, the correct option is $C)\frac{1}{2}$.