Question

A speaks truth in 60% of the cases, while B in 90% of the cases. In what percent of cases are thy 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 then A?

Answer 1

Given:
$P\left(A_T\right)=\frac{60}{100},P\left(A_C\right)=\frac{40}{100}$

$P\left(B_T\right)=\frac{90}{100},P\left(B_C\right)=\frac{10}{100}$

$P\left(\text{contradiction}\right)=\frac{60}{100}\times \frac{10}{100}+\frac{40}{100}\times \frac{90}{100}$

= $\frac{600}{10000}+\frac{3600}{10000}=\frac{600+3600}{10000}$
=$\frac{4200}{10000}=42$

$\%P\left(\text{contradiction}\right)=42\%$

Yes, in cases of contradiction the statement of B carry more weight as he speaks truth in more than A.