Find the probability that it is actually a six Solution Let E be the event that the man reports that six occurs in the 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 = 1
6
P(S2) = Probability that six does not occur = 5
6
P(E|S1) = Probability that the man reports that six occurs when six has
actually occurred on the die
= Probability that the man speaks the truth = 3
4
P(E|S2) = Probability that the man reports that six occurs  when six has
not actually occurred on the die
= Probability that the man does not speak the truth 
3
1
1
4
4
 
 
 
Thus, by Bayes' theorem, we get
P(S1|E) = Probability that the report of the man that six has occurred is
actually a six
=
1
1
1
1
2
2
P(S )P(E |S )
P(S )P(E|S )+P(S )P(E|S )
=
1
3
1
24
3
6
4
1
3
5
1
8
8
8
6
4
6
4
 
 
 
 
 
 
 
Hence, the required probability is 3 8
EXERCISE 13