Question

A speaks truth in 60% of the cases, while B in 90% of the cases. In what percent of
cases are they likely to contradict each other in stating the same fact? In the cases of
contradiction do you think, the statement of B will carry more weight as he speaks
truth in more number of cases than A?

Answer 1

Let the probability that A and B speak truth be P(A) and P(B) respectively.

Therefore, $P\left(A\right)=\frac{60}{100}=\frac{3}{5}$ and $P\left(B\right)=\frac{90}{100}=\frac{9}{10}$

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 = $P\left(A\right)\times (1-P(B\left)\right)=\frac{3}{5}\times (1-\frac{9}{10})=\frac{3}{50}$

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

Required Probability = $(1-P(A\left)\right)\times P\left(B\right)=(1-\frac{3}{5})\times \frac{9}{10}=\frac{9}{25}$

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

=$(\frac{3}{50}+\frac{9}{25})\times 100\%=\left(\frac{3+18}{50}\right)\times 100\%=42\%$

From case 1, it is clear that is not necessary that the statement of B will carry more weight as he speaks truth in more number of cases than A.