Then
P(T1) =
2
3
4
3
1
1
2
, P(T )
,P(T )
and P(T )
10
5
10
5
=
=
=
(given)
P(E|T1) = Probability that the doctor arriving late comes by train = 1
4
Similarly, P(E|T2) = 1
3 , P(E|T3) = 1
12  and P(E|T4) = 0, since he is not late if  he
comes by other means of transport Therefore, by Bayes' Theorem, we have
P(T1|E) = Probability that the doctor arriving late comes by train
=
1
1
1
1
2
2
3
3
4
4
P(T )P(E|T )
P(T )P(E|T )+P(T )P(E|T )+P(T )P(E|T )+P(T )P(E|T )
=
3
1
10
4
3
1
1
1
1
1
2
0
10
4
5
3
10
12
5
 
 
 
 
 
 
 
 
 = 3
120
1
40
18
2
×
=
Hence, the required probability is 1
2 © NCERT
not to be republished
PROBABILITY      555
Example 21 A man is known to speak truth 3 out of 4 times He throws a die and
reports that it is a six