Question

A speaks truth in 75% and B in 80% of the cases. In what percentage of cases are they likely to contradict each other narrating the same incident ?

• A. 35%
• B. 45%
• C. 15%
• D. 5%

Answer 1

Answer: (A)

$P\left(A\right) = \frac{75}{100} = \frac{3}{4}, P\left(B\right) = \frac{80}{100}= \frac{4}{5} $
$ \Rightarrow P\left(\bar{A} \right) = 1 -P\left(A\right) $ and $P\left(\bar{B}\right) = 1 - P\left(B\right) $
$\Rightarrow P\left(\bar{A} \right)= 1 - \frac{3}{4} = \frac{1}{4}$ and $ P\left(\bar{B} \right) = 1 - \frac{4}{5} = \frac{1}{5} $
Now, prob. (contradict to each other)
$= P\left(A \bar{B} or \bar{A }B\right)=P\left(A \bar{B} \right) + P\left(\bar{A} B\right)$
$ = P\left(A\right)P\left(\bar{B} \right) + P\left(\bar{A} \right) P\left(B\right) = \frac{3}{4} . \frac{1}{5} + \frac{1}{4} . \frac{4}{5}$
$ \Rightarrow P\left(A\bar{B} or \bar{A} B\right) = \frac{7}{20} $ $\therefore % of cases are A and B likely to contradict each other = \frac{7}{20}\times 100 =35$%