B. P. Lathi, Zhi Ding - Modern Digital and Analog Communication Systems-Oxford University Press (2009)

406 FUNDAMENTALS OF PROBABILITY THEORY
Furthennore, the data packet has been incorrectly decoded. Therefore
n
LP; = l
i=l
Applying the total probability theorem, we find that
n
P(B) = L P(BIE;)P(E;) = L q;p;
i=I
i=l
n
Isolating a Particular Cause: Bayes' Theorem
The total probability theorem facilitates the probabilistic analysis of a complex event by using a
divide-and-conquer approach. In practice, it may also be of interest to determine the likelihood
of a particular cause of an event among many disjoint possible causes. Bayes' theorem provides
the solution to this problem.
Bayes' Theorem: Let n disjoint events A 1, ... , A n form a partition of the sample space
S. Let B be an event with P(B) > 0. Then for j = 1 , ... , n,
P(A · IB) __ P _
(B _ IA J_ )P _
( A 1_·) _ _ P_( _ B _ IA J ) _ P _ (A 1 )_
1
- P(B) - L7=1 P(BIA;)P(A i )
The proof is already given by the theorem itself.
Bayes' theorem provides a simple method for computing the conditional probability of Aj
given that B has occurred. The probability P(Aj IB) is often known as the posterior probability
of event AJ . It describes, among n possible causes of B, the probability that B may be caused
by Aj. In other words, Bayes' theorem isolates and finds the relative likelihood of each possible
cause to an event of interest.
Example 8.1 1 A communication system always encounters one of three possible interference waveforms:
F1 , F2 , or F 3 . The probability of each interference is 0.8, 0.1 6, and 0.04, respectively. The
communication system fails with probabilities 0.01, 0.1, and 0.4 when it encounters F1 , F2 ,
and F 3 , respectively. Given that the system has failed, find the probability that the failure is a
result of F1 , F2, or F 3 , respectively.
Denote B as the event of system failure. We know from the description that
Furthermore, the effect of each interference on the system is given by
P(BIF1) = 0.01
Now following Bayes' theorem, we find that
P(Fi IB) = ; (BIF1)P(F1) (0.01)(0.8) 0 2
Li=l P (B IF;)P(F;) = (0.01 ) (0.8) + (0.1)(0.1 6) + (0.4)(0.04) =