A communications system consists of two transmitters and receivers. The transmit

Question

A communications system consists of two transmitters and receivers. The transmitters transmit bytes consisting of 8 independent bits, which can each be a 0 or a 1. In Transmitter A, the probability that a bit is a 1 is 0.5. While in Transmitter B, the probability that a bit is a 1 is 0.8. Transmitter A sends three times as many bytes as Transmitter B. The receiver receives bytes from both transmitters.

a. Calculate the probability that a received byte has exactly five 1’s.

b. If a byte has exactly five 1’s, what is the probability that it was sent from Transmitter B?

Explanation / Answer

P(1 in transmitter A) = 0.5

P(1 in transmitter B) = 0.8

P(A) = 3*P(B)

P(A)+P(B) = 1

P(A) = 0.75,P(B) =0.25

a)probability that a received byte has exactly five 1’s

= P(A)*P(5 1's in transmitter A)+ P(B)*P(5 1's in transmitter B)

=0.75*(8C5)(0.5)5(0.5)3 +0.25*(8C5)(0.8)5(0.2)3

= 0.20076

b)

probability that it was sent from Transmitter B given that byte has exactly five 1’s = P(byte was sent from Transmitter B and byte has exactly five 1’s)/P(byte has exactly five 1’s)

= 0.25*(8C5)(0.8)5(0.2)3/0.25 = 0.1468

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