Question

$A$ speaks truth in $60\%$ cases and $B$ in $90\%$ cases. In what percentage of cases are they likely to contradict each other in stating the same fact?

• A. $0.42$
• B. $0.58$
• C. $0.46$
• D. None of these
  

Answer 1

The correct option is A $0.42$

Let $E=$ The event of $A$ speaking truth and
$F=$ event of $B$ speaking truth

$\therefore P\left(E\right)=\frac{60}{100}=\frac{6}{10}$ and $P\left(F\right)=\frac{90}{100}=\frac{9}{10}$

Probability of $A$ and $B$ contradicting each other $P(E\overline{F}$ or $\overline{F}E$)

$P\left(E\overline{F}\right)+P\left(\overline{E}F\right)=P\left(E\right)P\left(\overline{F}\right)+P\left(\overline{E}\right)P\left(F\right)$

$=P\left(E\right)(1-P\left(F\right))+(1-P\left(E\right))P\left(F\right)$

$=\frac{6}{10}(1-\frac{9}{10})+(1-\frac{6}{10})\frac{9}{10}$

$=\frac{6}{10}\times \frac{1}{10}+\frac{4}{10}\times \frac{9}{10}=\frac{42}{100}$

$\therefore A$ and $B$ are likely to contradicts each other in $42\%$ cases.