A transmitter is sending a message by binary code (a sequence of 1\'s and 0\'s).

Question

A transmitter is sending a message by binary code (a sequence of 1's and 0's). Each transmitted bit (a 0 or 1) must pass through three relays to reach a receiver. At each relay, the probability is 0.20 that the bit that was sent will be different from the bit received (i.e., bit switching or reversal). Assume that the relays operate independently of one another. Transmitter Relay 1 Relay 2 Relay 3 Receiver If a 1 is sent from the transmitter, what is the probability that a 1 is sent by all three relays? If a 1 is sent by the transmitter, what is the probability that a 1 is received by the receiver? Suppose that 70% of all bits sent by the transmitter are 1's. If a 1 is received by the receiver, what is the probability that a 1 was sent from the transmitter?

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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