Question
Question: For narrating some incidents A speaks truth in \(75\% \) cases and B in \(80\% \) of the cases. In n...
For narrating some incidents A speaks truth in 75% cases and B in 80% of the cases. In narrating some cases both A and B are likely to contradict each other. What is the percentage of the cases in which they both likely contradict each other, narrating the same incident?
(A) 25%
(B) 40%
(C) 35%
(D) 10%
Solution
In this we apply the probability of happening an event and not happening of that particular event. So, the probability of A speaking truth is P (A) = 75%or 43and not speaking truth or lie P(A) = 25% or41. Similarly the probability of B speaking truth is P (B) = 80% or 54and not speaking truth is P(B)= 20% or51. In case of contradiction with each other is when A speaks truth and B speaks lie or A speaks lie and B speaks truth. For finding the probability we add both the situation and find the percentage.
Complete step-by-step answer:
Let A be Event that speaks the truth and B be Event that speaks the truth
So, the probability of A speaking truth P(A)= 75%
=10075=43
The probability of B speaking truth P(B)= 80%
=10080=54
We know P(A)+P(A)=1 Hence,
Probability of A not speaking truth P(A)= 1−43=41
Probability of B not speaking truth P(B) = 1−54=51
Now, A and B contradict each other=[A speaks truth and B speaks lie] or [A speaks lie and B speaks truth]
We will add both situation
= P(A)×P(B)+P(A)×P(B)
=(43×51)+(41×54)
=203+204
=207
For percentage we multiply 207 by 100%
=(207×100)%
=35%
So, the case when A and B contradicts is 35%.
The correct answer is C=35%.
So, the correct answer is “Option C”.
Note: When there is a case of contradiction we must consider both the cases i.e. probability of happening of the event and probability not happening of the event. The probability of the event lies between 0 and 1, where 0 and 1 are included.