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Basic Probability Concepts                                       25



             Theorem 2.2: Bayes’ theorem. Let  A   and  B  be  two  arbitrary  events  with





           P A) 6ˆ  0 and P B) 6ˆ  0. Then:
                                           P…AjB†P…B†
                                  P…BjA†ˆ             :                 …2:28†
                                              P…A†
             Combining this theorem with the total probability theorem we have a useful
           consequence:
                                                 n
                                             ,
                        P…B i jA†ˆ P…AjB i †P…B i †  X  P…AjB j †P…B j † :  …2:29†

                                                jˆ1
           for any i where events B j  represent a set of mutually exclusive and exhaustive
           events.
             The simple result given by Equation (2.28) is called Bayes’ theorem after the
           English philosopher Thomas Bayes and is useful in the sense that it permits us
           to evaluate a posteriori probability P BjA)  in terms of a priori information P(B)
           and P AjB), , as the following examples illustrate.

             Example 2.12. Problem: a simple binary communication channel carries
           messages by using only two signals, say 0 and 1. We assume that, for a given
           binary channel, 40% of the time a 1 is transmitted; the probability that a
           transmitted 0 is correctly received is 0.90, and the probability that a transmitted
           1 is correctly received is 0.95. Determine (a) the probability of a 1 being
           received, and (b) given a 1 is received, the probability that 1 was transmitted.
             Answer: let

                              A ˆ  event that 1 is transmitted;
                              A ˆ  event that 0 is transmitted;
                               B ˆ  event that 1 is received;
                               B ˆ  event that 0 is received:

           The information given in the problem statement gives us

                              P…A†ˆ 0:4;     P…A†ˆ 0:6;
                              P…BjA†ˆ 0:95;  P…BjA†ˆ 0:05;
                              P…BjA†ˆ 0:90;  P…BjA†ˆ 0:10:

           and these are represented diagrammatically in Figure 2.7.
             For part (a) we wish to find P(B). Since A and A  are mutually exclusive and
           exhaustive, it follows from the theorem of total probability [Equation (2.27)]








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