Question

$A$ speaks truth in $70\%$ cases and $B$ in $80\%$ cases. In what percentage of cases they are likely to contradict each other in giving the same statement.

• A. $\frac{38}{100}$
• B. $\frac{62}{100}$
• C. $\frac{72}{100}$
• D. $\frac{18}{100}$

Answer 1

The correct option is A $\frac{38}{100}$

Let the probability that $A$ and $B$ speak truth be $P\left(A\right)$ and $P\left(B\right)$ respectively.

Therefore, $P\left(A\right)=\frac{70}{100}=\frac{7}{10}$ and $P\left(B\right)=\frac{80}{100}=\frac{8}{10}$

$A$ and $B$ can contradict in stating a fact when one is speaking the truth and other is not speaking the truth.

Case 1: $A$ is speaking the truth and $B$ is not speaking the truth.

Required probability $=P\left(A\right)\times (1-P(B\left)\right)=\frac{7}{10}\times (1-\frac{8}{10})=\frac{14}{100}$

Case 2: $A$ is not speaking the truth and $B$ is separately the truth.

Required Probability $=(1-P(A\left)\right)\times P\left(B\right)=(1-\frac{7}{10})\times \frac{8}{10}=\frac{24}{100}$

Therefore, Percentage of cases in which they are likely to contradict in stating the same fact

$=(\frac{14}{100}+\frac{24}{100})=\left(\frac{14+24}{100}\right)=\frac{38}{100}$