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]**

**[CBSE 2013]**

Answer :

Given:

⇒ A speaks truth in 75% of cases

⇒ B speaks truth in 80% of cases

Now,

⇒ P(T_{A}) = P(A speaking truth) = 0.75

⇒ P(N_{A}) = P(A not speaking truth) = 1-0.75

⇒ P(N_{A}) = 0.25

⇒ P(T_{B}) = P(B speaking truth) = 0.80

⇒ P(N_{B}) = P(B not speaking truth) = 1-0.80

⇒ P(N_{B}) = 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(C_{AB}) = P(A and B contradict each other)

⇒ P(C_{AB}) = 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(C_{AB} ) = (P(T_{A})P(N_{B})) + (P(N_{A})P(T_{B}))

⇒ P(C_{AB}) = (0.75×0.20) + (0.25×0.80)

⇒ P(C_{AB}) = 0.15 + 0.20

⇒ P(C_{AB}) = 0.35

∴ The required probability is 0.35.

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