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Question : 22
Total: 29
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?
Solution:
Let the probability that A and B speak truth be P(A) and P(B) respectively.
Therefore, P (A) =
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) =
× ( 1 −
)
Case 2: A is not speaking the truth and B is speaking the truth.
Required probability = 1 - P (A) × P (B) =( 1 −
)
∴ Percentage of cases in which they are likely to contradict in stating the same fact =(
+
) (
)
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.
Therefore, P (A) =
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) =
Case 2: A is not speaking the truth and B is speaking the truth.
Required probability = 1 - P (A) × P (B) =
∴ Percentage of cases in which they are likely to contradict in stating the same fact =
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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