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 3
1 An urn contains 5 red and 5 black balls