Question

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

Answer 1

Let $E$ be the event that $A$ speaks truth and $F$ be the event that $B$ speaks truth. Then $E$ and $F$ are independent events such that
$P\left(E\right)=\frac{60}{100}=\frac{3}{5}$ and $P\left(F\right)=\frac{90}{100}=\frac{9}{10}$
$A$ and $B$ will contradict each other in narrating the same fact in the following mutually exclusive ways:
(i) $A$ speaks truth and $B$ tells a lie i.e. $E\cap \overline{F}$
(ii) $A$ tells a lie and $B$ speaks truth lie i.e. $\overline{E}\cap F$
$\therefore$ $P$($A$ and $B$ contradict each other)
$=P\left(IorII\right)=P(I\cup II)=P[(E\cap \overline{F})\cup (\overline{E}\cap F)]$
$=P(E\cap \overline{F})+P(\overline{E}\cap F)$ [$\because$ $E\cap \overline{F}$ and $\overline{E}\cap F$ are mutually exclusive]
$=P\left(E\right)P\left(\overline{F}\right)+P\left(\overline{E}\right)P\left(F\right)$ [$\because$ $E$ and $F$ are in dep.]
$=\frac{3}{5}\times (1-\frac{9}{10})+(1-\frac{3}{5})\times \frac{9}{10}=\frac{3}{5}\times \frac{1}{10}+\frac{2}{5}\times \frac{9}{10}=\frac{21}{50}$

$=\frac{21}{50}\times 100=42$%