Subject: General Aptitude · Chapter: Probability · Exam: · Marks: · Difficulty:
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?
(a)5\%
(b)15\%
(c)35\%
(d)45\%
(e)None of these
Answer
Answer (as printed): C
Explanation
Let $E_{1}=$ Event that A speaks the truth and $E_{2}=$ Event that B speaks the truth. Then, $$\begin{aligned} & P\left(E_{1}\right)=\frac{75}{100}=\frac{3}{4}, P\left(E_{2}\right)=\frac{80}{100}=\frac{4}{5}, P\left(\bar{E}_{1}\right)=\left(1-\frac{3}{4}\right) \\ & \quad=\frac{1}{4}, P\left(\bar{E}_{2}\right)=\left(1-\frac{4}{5}\right)=\frac{1}{5} . \end{aligned}$$ $P(A$ 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)] $$\begin{aligned} & =P\left[\left(E_{1} \cap \bar{E}_{2}\right) \text { or }\left(\bar{E}_{1} \cap E_{2}\right)\right]=P\left(E_{1} \cap \bar{E}_{2}\right)+P\left(\bar{E}_{1} \cap E_{2}\right) \\ & =P\left(E_{1}\right) \cdot P\left(\bar{E}_{2}\right)+P\left(\bar{E}_{1}\right) \cdot P\left(E_{2}\right) \\ & =\left(\frac{3}{4} \times \frac{1}{5}\right)+\left(\frac{1}{4} \times \frac{4}{5}\right)=\left(\frac{3}{20}+\frac{1}{5}\right)=\frac{7}{20} \\ & =\left(\frac{7}{20} \times 100\right) \%=35 \% . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.