Subject: General Aptitude · Chapter: Probability · Exam: 2006 · Marks: · Difficulty:
A speaks truth in 60\% cases and B speaks truth in 70\% cases. The probability that they will say the same thing while describing a single event, is
(a)0.54
(b)0.56
(c)0.68
(d)0.94
(e)None of these
Answer
Answer (as printed): A
Explanation
Let $E_{1}=$ Event that A speaks the truth and $E_{2}=$ Event that B speaks the truth. Then, $P\left(E_{1}\right)=\frac{60}{100}=\frac{3}{5}, P\left(E_{2}\right)=\frac{70}{100}=\frac{7}{10}, P\left(\bar{E}_{1}\right)$ $$=\left(1-\frac{3}{5}\right)=\frac{2}{5}, P\left(\bar{E}_{2}\right)=\left(1-\frac{7}{10}\right)=\frac{3}{10} .$$ $P(\mathrm{A}$ and B say the same thing) $=P[(A$ speaks the truth and $B$ speaks the truth $)$ or ( $A$ tells a lie and $B$ tells a lie)] $$\begin{aligned} & =P\left[\left(E_{1} \cap E_{2}\right) \text { or }\left(\bar{E}_{1} \cap \bar{E}_{2}\right)\right]=P\left(E_{1} \cap E_{2}\right)+P\left(\bar{E}_{1} \cap \bar{E}_{2}\right) \\ & =P\left(E_{1}\right) \cdot P\left(E_{2}\right)+P\left(\bar{E}_{1}\right) \cdot P\left(\bar{E}_{2}\right) \\ & =\left(\frac{3}{5} \times \frac{7}{10}\right)+\left(\frac{2}{5} \times \frac{3}{10}\right)=\frac{27}{50}=0.54 . \end{aligned}$$