# A speaks th

Let us assume U1, U2 and A be the events as follows:

U1 = Getting 5 on throwing a die

U2 = Getting other than 5 on throwing a die

A = Reporting 5 after throwing the die

From the problem,

P(A|U1) = P(Reporting 5 on actually getting 5 on throwing a die)

P(A|U1) = P(Telling the truth)

P(A|U2) = P(Reporting 5 but not getting 5 on throwing a die)

P(A|U2) = P(Not telling the truth)

Now we find

P(U1|A) = P(the die actually shows 5 given that man reports 5)

Using Baye’s theorem:

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 :

A bag A contains RD Sharma - Volume 2

The contents of tRD Sharma - Volume 2

A girl throws a dMathematics - Board Papers

Three urns contaiRD Sharma - Volume 2

Two groups are coMathematics - Board Papers

The contents of tRD Sharma - Volume 2

Suppose a girl thMathematics - Board Papers

There are three cMathematics - Board Papers

Given three identMathematics - Board Papers