Q. 164.7( 3 Votes )

A, B, and C are independent witness of an event which is known to have occurred. A speaks the truth three times out of four, B four times out of five and C five times out of six. What is the probability that the occurrence will be reported truthfully by majority of three witnesses?

Answer :

Given:


A speaks truth three times out of four


B speaks truth four times out of five


C speaks truth five times out of six


P(TA) = P(A speaks truth)



P(TB) = P(B speaks truth)



P(TC) = P(C speaks truth)



P(NA) = P(A speaks lies)




P(NB) = P(B speaks lies)




P(NC) = P(C speaks lies)



.


It is told that the occurrence is will be reported by majority of witness. This is only possible when at least two of the witnesses speaks truth.


P(M) = P(TATBNC) + P(TANBTC) + P(NATBTc) + P(TATBTC)


Since speaking by a person is independent the probabilities will multiply each other.


P(M) = (P(TA)P(TB)P(NC)) + (P(TA)P(NB)P(TC)) + (P(NA)P(TB)P(Tc)) + (P(TA)P(TB)P(TC))




.


The required probability is .


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :