Answer to Question #166908 in Statistics and Probability for Bikram Chaudhary

Question

A speaks the truth in 80% of the cases and B in 70% of the cases. In what percentage of cases are they likely to contradict and do not contradict each other in stating the same fact?

Answer 1

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

Thus P(A) = 0.8 and P(B) = 0.7

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

A is speaking the truth and B is not speaking the truth:

P(A) * (1–P(B)) = 0.8 * (1–0.7) = 0.8 * 0.3 = 0.24

A is not speaking the truth and B is separately the truth:

(1–P(A))*P(B) = (1 – 0.8) * 0.7 = 0.2 * 0.7 = 0.14

Therefore, percentage of cases in which they are likely to contradict in stating the same fact: 0.24 + 0.14 = 0.38 which is 38%.

A and B do not contradict each other in stating the same fact when both of them are telling the truth or lying.

A and B tell the truth: P(A) * P(B) = 0.8 * 0.7 = 0.56

A and B lie: (1 – P(A)) * (1 – P(B)) = 0.2 * 0.3 = 0.06

Therefore, the percentage of cases in which they are likely to not contradict in stating the same fact: 0.56 + 0.06 = 0.62 which is 62%.

Answer: in 38% of cases A and B are likely to contradict each other in stating the same fact and in 62% of cases they do not.