1.

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?

Answer»

Given:

⇒ A speaks truth in 75% of cases

⇒ B speaks truth in 80% of cases

Now,

⇒ 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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