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 ?

Solution(By Examveda Team)

Let $${{{\text{E}}_1}}$$ = event that A speaks the truth
And $${{{\text{E}}_2}}$$ = event that B speaks the truth
Then,
$$\eqalign{ & P\left( {{E_1}} \right) = \frac{{75}}{{100}} = \frac{3}{4}, \cr & P\left( {{E_2}} \right) = \frac{{80}}{{100}} = \frac{4}{5}, \cr & P\left( {{{\overline E }_1}} \right) = \left( {1 - \frac{3}{4}} \right) = \frac{1}{4}, \cr & P\left( {{{\overline E }_2}} \right) = \left( {1 - \frac{4}{5}} \right) = \frac{1}{5} \cr} $$
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)]
$$\eqalign{ & = P\left[ {\left( {{E_1} \cap {{\overline E }_2}} \right)or\left( {{{\overline E }_1} \cap {E_2}} \right)} \right] \cr & = P\left[ {\left( {{E_1} \cap {{\overline E }_2}} \right) + \left( {{{\overline E }_1} \cap {E_2}} \right)} \right] \cr & = P\left( {{E_1}} \right).P\left( {{{\overline E }_2}} \right) + P\left( {{{\overline E }_1}} \right).P\left( {{E_2}} \right) \cr & = \left( {\frac{3}{4} \times \frac{1}{5}} \right) + \left( {\frac{1}{4} \times \frac{4}{5}} \right) \cr & = \left( {\frac{3}{{20}} + \frac{1}{5}} \right) \cr & = \frac{7}{{20}} \cr & = \left( {\frac{7}{{20}} \times 100} \right)\% \cr & = 35\% \cr} $$