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 in narrating the same incident?

Answer 1

$$\begin{gathered} P\left(\mathrm{Both}\mathrm{narrating}\mathrm{different}\mathrm{incident}\right)=P\left(\mathit{\text{B}}\text{lies and}\mathit{\text{A}}\text{speaks the truth}\right)+P\left(\mathit{\text{A}}\text{lies and}\mathit{\text{B}}\text{speaks the truth}\right) \\ =P\left(A\cap \overline{B}\right)+P\left(\overline{A}\cap B\right) \\ =P\left(A\right)P\left(\overline{B}\right)+P\overline{\left(A\right)}P\left(B\right) \\ =0.75\left(1-0.8\right)+\left(1-0.75\right)0.8 \\ =0.75\times 0.2+0.25\times 0.8 \\ =0.15+0.2 \\ =0.35 \\ =35\% \end{gathered}$$