Question

A speaks truth in $60\%$ and B in $80\%$ of the cases. In what percentage of cases are they likely to contradict each other narrating the same incident?

• A. $44\%$
• B. $36\%$
• C. $64\%$
• D. $48\%$

Answer 1

The correct option is A $44\%$

Let E = the event that A speaks the truth
and F = the event that B speaks the truth
Then $\overline{E}$ = the event that A tells a lie
And $\overline{F}$ = the event that B tells a lie
Clearly, E and F are independent events, So, E and $\overline{F}$ as well as $\overline{E}$ and F are independent.
Now, $P\left(E\right)=\frac{60}{100}=\frac{3}{5},P\left(F\right)=\frac{80}{100}=\frac{4}{5}$
$\therefore P\left(\overline{E}\right)=\frac{2}{5},P\left(\overline{F}\right)=\frac{1}{5}$
$\therefore P$ (A and B contradict each other)
= [ (A speaks the truth and B tells a lie)
or (A tells a lie and B speaks the truth)
$=P\left[\right(E\cap \overline{F})\cup (\overline{E}\cup F\left)\right]=P(E\cap \overline{F})+P(\overline{E}\cap F)=P\left(E\right)P\left(\overline{F}\right)+P\left(\overline{E}\right)P\left(F\right)=\frac{3}{5}\times \frac{1}{5}+\frac{2}{5}\times \frac{4}{5}=\frac{11}{25}\Rightarrow 44\%$
So A and B contradict each other in $44\%$ cases