A transmitter is sending a message by using a binary code, namely, a sequence or

Question

A transmitter is sending a message by using a binary code, namely, a sequence or 0s and 1s. Each transmitted bit (0 or 1) most pass through 3 relays to reach the receiver. At each relay, the probability is .2 that the bit sent will be different from the bit received (a reversal). Assume that the relays operate independently of one another.

Transmitter -> Relay 1-> Relay 2 -> Relay 3 -> Receiver

(a) If a 1 is sent from the transmitter, what is the probability that a 1 is sent by all 3 relays?

(b) If a 1 is sent from the transmitter, what is the probability that a 1 is received by the receiver?

(c) Suppose 70% of all bits sent from the transmitter are 1s. If a 1 is received by the receiver, what is the probability that a 1 was sent?

Explanation / Answer

a. *given each relay has an error rate of 0.2 at each relay, the probability that the correct code makes it through is 0.8 so 1 needs to make it through 3 times... the probability is = 0.8 * 0.8 * 0.8 = 0.512 so 51.2 % chance b. if a 1 is sent, there are mutliple possilities that 1 is received for example R is used to show the relay 1 ---> R1 --> 0 ---> R2 --> 0 --->R3 --> 1 probability = 0.2 * 0.8 * 0.2 1 0 1 1 probability = 0.2 *0.2 *0.8 1 1 0 1 probabliity = 0.8 * 0.2 * 0.2 1 1 1 1 probablility = 0.8*0.8 * 0.8 so total probability of receiving a 1 = sum of all these probabilities = 60.8 % part c. 1 is received , what is the probability that a 1 was sent? Bayes rule : P(A | B) = P(B | A ) P(A)/P(B) the event B is "1 is received" the even A is "1 is sent" so P(A |B ) is what you want to calculate but first we need to calculate P(B) where P(B) is the probability that a 1 is received. P(B) = P(B|1 is sent) * 0.7 + P(B|0 is sent)*0.3 = .608 * 0.7 + ((0.8^2*0.2)*3+0.2^3)*0.3 = 0.5432 P(A | B) = P(B | A ) P(A)/P(B) = 60.8 % * P(A) / P(B) , P(B|A) is calculated in part a. = 60.8% * 70% / p(B) = 0.608*0.7/0.5432 = 78 % chance!

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