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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? - Mathematics

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?

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Solution

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

Therefore, `P(A)=60/100=3/5 and P(B)=90/100=9/10.`

They can contradict in stating the 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)xx(1-P(B))=3/5xx(1-9/10)=3/50`

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

Required probability = `(1-P(A))xxP(B)=(1-3/5)xx9/10=9/25`

Therefore, the percent of cases in which they are likely to contradict in stating the same fact is equal to `(3/50+9/25)xx100=42%`

It is evident from the explanation 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, as in case 1.

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