Q. 33.9( 10 Votes )
A speaks truth in 75% and B in 80% of the cases. In what percentage of cases are they likely to contradict each other in narrating the same incident?[CBSE 2013]
⇒ A speaks truth in 75% of cases
⇒ B speaks truth in 80% of cases
⇒ P(TA) = P(A speaking truth) = 0.75
⇒ P(NA) = P(A not speaking truth) = 1-0.75
⇒ P(NA) = 0.25
⇒ P(TB) = P(B speaking truth) = 0.80
⇒ P(NB) = P(B not speaking truth) = 1-0.80
⇒ P(NB) = 0.20
We need to find the probability for case in which A and B contradict each other for narrating an incident.
This happens only when A not telling truth while B is telling truth and vice-versa.
⇒ P(CAB) = P(A and B contradict each other)
⇒ P(CAB) = P(A tells truth and B doesn’t) + P(B tells truth and A doesn’t)
Since the speaking of A and B are independent events their probabilities will multiply each other.
⇒ P(CAB ) = (P(TA)P(NB)) + (P(NA)P(TB))
⇒ P(CAB) = (0.75×0.20) + (0.25×0.80)
⇒ P(CAB) = 0.15 + 0.20
⇒ P(CAB) = 0.35
∴ The required probability is 0.35.
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