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 E₁ be the event that A speaks truth and E₂ be the event that B speaks truth. Then E and f are independent events such that

P(E₁) = 60/100 = 3/5, P(E₁) = 90/100 = 9/10

⇒ P(bar E₁) = 2/5, P(bar E₂) = 1/10

P (A and B contradict each other) = P(E₁)P(bar E₂) + P(bar E₁)P(E₂)

= 3/5 x 1/10 + 2/5 x 9/10 = (3 + 18)/50 = 21/50

Yes, the statement of B will carry more weight as the probability of B to speak truth is more than that of A.