3. Consider a wireless channel with errors as shown in Figure 1. The probability
ID: 3074176 • Letter: 3
Question
3. Consider a wireless channel with errors as shown in Figure 1. The probability of receiving a "0" when "0" was sent is 0.8 and the probability of receiving a "1 when "1" was sent is also 0.8. bit sent bit received 0.8 0 Q.2 0.2 0.8 Figure 1: A wireless channel with errors (1) [10pt] Assume that "1" was sent with probability 0.6 and "0" was sent with probability 0.4. (2) [10pt] Assume that "1" was sent with probability 0.6 and "" was sent with probability 0.4 ) 10pt Assume "O00" (i.e., three consecutive zeros) are sent over the wireless channel. what is Compute the probability that you receive a "1". Compute the probability that "0" was sent conditioned on that you received a "1". the probability to receive two or more "0s?Explanation / Answer
Terminology:
P(1r): bit "1" recieved
P(0r): bit "0" recieved
P(1s): bit "1" sent
P(0s): bit "0" sent
Given: P(0r | 0s) = 0.8, P(1r | 0s) = 0.2, P(1r | 1s) = 0.8, P(0r | 1s) = 0.2
(1)
P(1s) = 0.6, P(0s) = 0.4
P(1r) = ?
P(1r) = P(1s)P(1r | 1s) + P(0s)P(1r | 0s) = (0.6 x 0.8) + (0.4 x 0.2) = 0.56
(2)
P(1s) = 0.6, P(0s) = 0.4
P(0s | 1r) = P(0s 1r) / P(1r)
P(1r) = 0.56 from previous part.
P(0s 1r) = P(0s) P(1r | 0s) = 0.4 x 0.2 = 0.08
P(0s | 1r) = 0.08 / 0.56 = 0.1429
(3)
P(Two or more 0 recieved) = P(Two 0 recieved) + P(Three 0 recieved)
P(Two 0 recieved) = P(Two 0 and One 1 recieved) = 0.8 x 0.8 x 0.2 = 0.128
P(Three 0 recieved) = 0.8 x 0.8 x 0.8 = 0.512
P(Two or more 0 recieved) = 0.128 + 0.512 = 0.640
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