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

A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

Open in App
Solution

Let E be the event that the man reports that six occurs in throwing of the die and let S1 be the event that six occurs and S2 be the event that six does not occur.

Then,
P(S1)= Probability that six occurs =16

P(S2)= Probability that six doesn't occur =56

P(E|S1)= Probability that the man reports six when six has actually occurred
= ​​Probability that the man speaks the truth =34

P(E|S2)= Probability that the man reports six when six has not actually occurred
= ​​Probability that the man does not speaks the truth
=134=14

Therefore, by Bayes' Theorem,
Probability that the report of the man that six has occurred is actually a six is given by,

P(S1|E)=P(S1)P(E|S1)P(S1)P(E|S1)+P(S2)P(E|S2)
=16×3416×34+56×14
=38


flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Binomial Experiment
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon