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

Answer: Option C

Let $${\text{E}}_1$$ = event that A speaks the truth
And $${\text{E}}_2$$ = event that B speaks the truth
Then,
$$\begin{array}{rl}& P\left(E_1\right)=\frac{75}{100}=\frac{3}{4},\\ & P\left(E_2\right)=\frac{80}{100}=\frac{4}{5},\\ & P\left({\overline{E}}_1\right)=\left(1-\frac{3}{4}\right)=\frac{1}{4},\\ & P\left({\overline{E}}_2\right)=\left(1-\frac{4}{5}\right)=\frac{1}{5}\end{array}$$
P (A and B contradict each other)
= P [(A speaks the truth and B tells a lie) or (A tells a lie and B speaks the truth)]
$$\begin{array}{rl}& =P\left[\left(E_1\cap {\overline{E}}_2\right)or\left({\overline{E}}_1\cap E_2\right)\right]\\ & =P\left[\left(E_1\cap {\overline{E}}_2\right)+\left({\overline{E}}_1\cap E_2\right)\right]\\ & =P\left(E_1\right).P\left({\overline{E}}_2\right)+P\left({\overline{E}}_1\right).P\left(E_2\right)\\ & =\left(\frac{3}{4}\times \frac{1}{5}\right)+\left(\frac{1}{4}\times \frac{4}{5}\right)\\ & =\left(\frac{3}{20}+\frac{1}{5}\right)\\ & =\frac{7}{20}\\ & =\left(\frac{7}{20}\times 100\right)\%\\ & =35\%\end{array}$$