p{011 h] psm
1 °-3°8
136
By the definition of conditional probability we have
nro 1 def P{01 and M
P{0i and fs} we find from
p(f L 1 def p{01 and
Pit* 161) p{01}
P{6x and <2} P{0i} P}<2 I 61}
ft is given that P{0i} 0.2. P{t2 i 0i} is the probability to
observe t2 if 0i is the state of nature. This probability is, according
to table 3
P{t2 I 0x} 0.3830
Hence: P{0i and h) 0.2 X 0.3830 0.0766
P{h} is the probability to observe t2. According to table 3, and
taking account of the probabilities of 0i and 02, it is:
0.2 X 0.3830 0.8 X 0.0428 0.1108
Hence: P{0i |f2}=-^|= 0.692.
We also compute P{02 h)
P{tz I ©2} 0.0428, and P{02} 0.8.
Hence: P{02 and t2} 0.8 x 0.0428 0.0342
P{k} 0.1108
It follows that
We execute a check:
P{0i I h) P{02 I h) 0.692 0.308 1
We have now a "no data" problem with the a posteriori proba
bilities
P'^Oi} 0.692 and P'{®2} 0.308
The risks are:
for «1: 0.692 x 20 0.308 X 25 7.70
for a2: 0.692 x 17 0.308 X o 11.76
The Bayes action is a\.
