In this problem, we will modify the communication system described in Exercise 2.46 so that the detector at the receiver is allowed to make one of three possible decisions: “0” the detector decides the received signal was a 0, “1” the detector decides the received signal was a 1, “E” the detector is not sure and declares the received signal an erasure (i. e., the receiver chooses not to choose). The operation of the detector is described by the following set of conditional probabilities:
Pr (0 received | 0 transmitted) = 0.90, Pr (0 received | 1 transmitted) = 0.04,
Pr (1 received | 0 transmitted) = 0.01, Pr (0 received | 1 transmitted) = 0.80,
Pr (E received | 0 transmitted) = 0.09, Pr (0 received | 1 transmitted) = 0.16.
Again, assume that 0s and 1s are equally likely to be transmitted.
(a) What is the probability that a symbol is erased at the receiver?
(b) Given that a received symbol is declared an erasure, what is the probability that a 0 was actually transmitted?
(c) What is the probability of error of this receiver? That is, what is the probability that a 0 was transmitted and it is detected as a 1 or a 1 was transmitted and it is detected as a 0?