Question

$A$ speaks truth in $75$ percent cases, and $B$ in $80$ percent of the cases. In what percentage of cases are they likely to contradict each other in stating the same fact?

• A. $60$
• B. $75$
• C. $35$
• D. $5$

Answer 1

Answer: (C)

$35$

Let $E_1$ denotes the event that $A$ speaks the truth and $E_2$ denotes the event that $B$ speaks the truth.

Given $P\left(E_1\right)=\frac{75}{100}=\frac{3}{4}$

$P\left({\overline{E}}_1\right)=1-\frac{3}{4}=\frac{1}{4}$

$P\left(E_2\right)=\frac{80}{100}=\frac{4}{5}$

$\Rightarrow P\left(\overline{E_2}\right)=1-\frac{4}{5}=\frac{1}{5}$

Let $E$ be the event that $A$ and $B$ contradict each other.

They will contradict each other if one speaks the truth and other does not.

Hence $E=E_1\overline{E_2}+\overline{E_1}{E}_2$

Now $P\left(E_1\overline{E_2}\right)$= the probability that $A$ speaks the truth and $B$ tells a lie =$P\left(E_1\right)P\left(\overline{E_2}\right)\frac{3}{4}\times \frac{1}{5}=\frac{3}{20}$

Similarly $P\left(\overline{E_1}{E}_2\right)=\frac{1}{4}\times \frac{4}{5}=\frac{1}{5}$

Since, $E_1$and $\overline{E_2}$are independent events and so are $\overline{E_1}$and$E_2.$

Since $E_1\overline{E_2}and\overline{E_1{E}_2}$ are mutually exclusive events, we have

$P\left(E\right)=P(E_1\overline{E_2}+E_1\overline{E_2})=P\left(E_1\overline{E_2}\right)+P\left(E_1\overline{E_2}\right)=\frac{3}{20}+\frac{1}{5}=\frac{7}{20}=\frac{35}{100}$ i.e., 35%

Hence in $35\%$ cases, $A$ and $B$ will contradict each other.