Q22 of 45 Page 1

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?

Let the probability that A speak truth be P(A)

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


Therefore,



and


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(A) × [1 – P(B)]


= 0.6 × [1 – 0.9]


= 0.6 × 0.1


= 0.06


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


Required probability = P(B) × [1 – P(A)]


= 0.9 × [1 – 0.6]


= 0.9 × 0.4


= 0.36


∴Percentage of cases in which they are likely to contradict in stating the same fact = (0.06 + 0.36) × 100%


= (0.42) × 100%


= 42%


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


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