1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A man is known to speak the truth 3 out of 5 times. He throws a die and reports that it is ′6′. The probability that it is actually '6' is 3k. Find the value of k.

Open in App
Solution

## Let E be the event that the man reports that one occurs in throwing of the die and let S1 be the event that one occurs and S2 be the event that one does not occur. Then, P(S1)=16P(S2)=56P(E|S1)= Probability that the man reports that 6 occurs when six has actually occured on die = 35= probability that man speaks truth.P(E|S2)= Probability that the man reports that 6 occurs when 6 has not actually occured on the die = 1−35=25= probability that man does not speak truth.By Baye's theorem,P(S1|E)=P(S1)P(E|S1)P(S1)P(E|S1)+P(S2)P(E|S2)=16×3516×35+56×25=313Hence, required probability is 313.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Binomial Experiment
MATHEMATICS
Watch in App
Join BYJU'S Learning Program