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BAYES THEOREM

PB A P A( i ) ( i )
PA B( ) = (10.42)
i
PB A P A( ) ( ) + P B A P A( ) ( ) +…+ PB A P A( ) ( )
1 1 2 2 n n
PROOF From the definition of the conditional probability [Eq. (10.37)], we
can write:

PB( ∩ A ) = P A B PB( ) ( ) (10.43)
i i

Again, using Eqs. (10.37) and (10.43), we have:

PB A P A( ) ( )
PA B( ) = i i (10.44)
i
PB()
Now, substituting Eq. (10.41) in the denominator of Eq. (10.44), we obtain Eq.
(10.42).

Example 10.10
A digital communication channel transmits the signal as a collection of ones
(1s) and zeros (0s). Assume (statistically) that 40% of the 1s and 33% of the 0s
are changed upon transmission. Suppose that, in a message, the ratio
between the transmitted 1 and the transmitted 0 was 5/3. What is the proba-
bility that the received signal is the same as the transmitted signal if:
a. The received signal was a 1?
b. The received signal was a 0?

Solution: Let O be the event that 1 was received, and Z be the event that 0 was
received. If H is the hypothesis that 1 was received and H is the hypothesis
1
0
that 0 was received, then from the statement of the problem, we know that:
PH( ) 5
1 = and PH( ) + PH( ) = 1
PH( ) 3 1 0
0
giving:

PH( ) = 5 and PH( ) = 3
1 0
8 8
Furthermore, from the text of the problem, we know that:

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