Question

A speaks truth in 75% 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?

Answer 1

let P(A) be the probability that A speaks truth.then P(A') be the probability that A does not speak truth.
let P(B) be the probability that B speak truth.
then P(B') be the probability that B does not speak truth.
P(A) = 75/ 100 = 0.75
P(A') = 1-0.75= 0.25
P(B) = 90/100 = 0.9
P(B') = 1- 0.9 = 0.1
the probability that A speak truth and B does not speak truth = P(A)*P(B')
= 0.75*0.1
=0.075
the probability that B speak truth and A does not = P(B)*P(A')
=0.9*0.25
=0.225
therefore the probability that contradiction arise
=.225+.075
=.3
thus the required percentage = 0.3*100 = 30%