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

Solution

✅ Correct Option: 3

A(truth)=75%→34;A(not truth)=25%→14B(truth)=80%→45;B(not truth)=20%→15\begin{aligned} & A(\text {truth})=75 \% \rightarrow \frac{3}{4} ;\quad A(\text {not truth})=25 \% \rightarrow\frac{1}{4} \\ & B(\text {truth})=80 \% \rightarrow \frac{4}{5} ;\quad B(\text {not truth})=20 \%\rightarrow \frac{1}{5} \end{aligned}

They will contradict each other when one is telling the truth & the other one is not.

So, two cases are possible:

  1. A telling truth, B lying.

  2. A lying, B telling truth.


∴A(truth)×B(lie)+B(truth)×A(lie)⇒34×15+45×14=320+15=320+420=720=35%\begin{aligned} & \therefore A(\text {truth}) \times B(\text {lie})+B(\text {truth}) \times A(\text {lie}) \\ \\ & \Rightarrow \frac{3}{4} \times \frac{1}{5}+\frac{4}{5} \times \frac{1}{4} \\ \\ & =\frac{3}{20}+\frac{1}{5}=\frac{3}{20}+\frac{4}{20}=\frac{7}{20} = \boxed{35\%} \end{aligned}

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