Question: Please help with the Matlab code for this HW assignment. I know I can generate a binary random number using the x=rand(1) 2.5.1 The binary

Please help with the Matlab code for this HW assignment. I know I can generate a binary random number using the x=rand(1)

2.5.1 The binary communication channel A number of problems naturally involve theuse of conditional probability and/or Bayes' rule. The binary communication channel, which

2.5.1 The binary communication channel A number of problems naturally involve the use of conditional probability and/or Bayes' rule. The binary communication channel, which is an abstraction for a com- munication system involving binary data, uses these concepts extensively. The idea is illustrated in the following example. Example 2.12: The transmission of bits over a binary communication channel is repre- sented in the drawing below, where notation like OS, OR ... is used to denote events "0 Pr[OR0S] = 0.95 Os Pr[0] = 0.5 Pr[15] = 0.5 Transmitter = Pr[1R1s] = 0.90 = Channel = Pr[0R1s] = 0.10 Pr[1R 0s] = 0.05 OR sent," "O received," etc. When a 0 is transmitted, it is correctly received with proba- bility 0.95 or incorrectly received with probability 0.05. That is Pr[0r|0s] = 0.95 and Pr[1R|0S] 0.05. When a 1 is transmitted, it it is correctly received with probability 0.90 and incorrectly received with probability 0.10. The probabilities of sending a 0 or a 1 are denoted by Pr[0s] and Pr[1s] and are known as the prior probabilities. It is desired to compute the probability of error for the system. 1R Receiver This is an application of the principle of total probability. If two events A and A2 form a partition, then (2.15) can be written as Pr[B] = Pr[B|A] Pr[A1] + Pr[B|A2] Pr[A2] Since the two events 0s and 1s are mutually exclusive and collectively exhaustive, they can be identified with the events A and A2 and the event B can be taken to be the event that an error occurs. It then follows that Pr[error] Pr[error|0s] Pr[0s] + Pr[error|1s] Pr[1s] Pr[1R|0s] Pr[OS] + Pr[OR 1s] Pr[1s] (0.05) (0.5) + (0.10) (0.5) = 0.075

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