Question

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

• A. $\frac{1}{2}$
• B. $\frac{3}{4}$
• C. $\frac{3}{5}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{2}$

Let $X$ be the event that $X$ speaks the truth.

Let $Y$ be the event that $Y$ speaks the truth.
Then $P\left(X\right)=\frac{60}{100}=\frac{3}{5}$ and $P\left(Y\right)=\frac{50}{100}=\frac{1}{2}$

Also $P(X-lie)=1-\frac{3}{5}=\frac{2}{5}$ and $P(Y-lie)=1-\frac{1}{2}=\frac{1}{2}$

Now
$X$ and $Y$ contradict each other $=\left[XliesandYtrue\right]or\left[XtrueandYlies\right]$

$=P(X-lie)\times P\left(Y\right)+P\left(X\right)\times P(Y-lie)$

$=\frac{2}{5}\times \frac{1}{2}+\frac{3}{5}\times \frac{1}{2}$

$=\frac{2}{10}+\frac{3}{10}$

$=\frac{5}{10}$

$=\frac{1}{2}$

Therefore the probability that they contradict each other while narrating the some incident, is $\frac{1}{2}$