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)
$=P\left(A\bar{B}or\bar{A}B\right)=P\left(A\bar{B}\right)+P\left(\bar{A}B\right)$
$=P\left(A\right)P\left(\bar{B}\right)+P\left(\bar{A}\right)P\left(B\right)=\frac{3}{4}.\frac{1}{5}+\frac{1}{4}.\frac{4}{5}$
$\Rightarrow P\left(A\bar{B}or\bar{A}B\right)=\frac{7}{20}$ $\therefore eachother=\frac{7}{20}\times 100=35$%