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Q. $ P $ speaks truth in $ 70\% $ cases and $ Q $ speaks in $ 80\% $ of the cases. In what percentage of cases are they likely to contradict each other in stating the same fact?

J & K CETJ & K CET 2017Probability - Part 2

Solution:

$P$ ($P$ speaks truth) $=70\%=0.7$
$P$ ($Q$ speaks truth) $=80\%=0.8$
$\therefore P\left(\bar{P}\right)=1-P\left(P\right)=0.3$
and $P\left(\bar{Q}\right)=1-P\left(Q\right)=0.2$
Required probability $=P\left(P\right)P\left(\bar{Q}\right)+P\left(\bar{P}\right)P\left(Q\right)$
$=0.7\times0.2+0.3\times0.8$
$=0.14+0.24=0.38$
$=38\%$
Thus, in $38\%$ cases they are likely to contradict each other in stating the same fact