3) Binary digits 0 and 1 are transmitted over a communications channel. If a 1 i

Question

3) Binary digits 0 and 1 are transmitted over a communications channel. If a 1 is sent, it will be received as a 1 with probability .95 and as a 0 with probability .05; if a 0 is sent, it will be received as a 0 with probability .99 and as a 1 with probability.01. If the probabilities that a 0 or a 1 is sent are equal, what is a) [4] the probability that a 1 is received? b) [3] the probability that a 1 is received given that a 1 was sent? c) [3] the probability that a 0 was sent given that a 0 was received?

Explanation / Answer

Here we are given that:

P( 1 received | 1 sent ) = 0.95,
P( 0 received | 1 sent ) = 0.05
P( 0 received | 0 sent ) = 0.99
P( 1 received | 0 sent ) = 0.01

Also we are given that P( 0 sent ) = P( 1 sent ) = 0.5

a) Therefore using law of total probability, we get:

P( 1 received ) = P( 1 received | 1 sent )P( 1 sent ) + P( 1 received | 0 sent )P( 0 sent )

P( 1 received ) = 0.5*(0.95 + 0.01) = 0.48

Therefore 0.48 is the required probability here.

b) Given that 1 is sent, probability that 1 is received is already given to us as:

P( 1 received | 1 sent ) = 0.95

Therefore 0.95 is the required probability here.

c) Now we already know from part a) that P( 1 received ) = 0.48 This means that P( 0 received ) =1 - 0.48 = 0.52

Therefore given that 0 was received, probability that 0 was sent is computed using Bayes theorem as:

P( 0 sent | 0 received ) = P( 0 received | 0 sent )P( 0 sent ) / P( 0 received )

P( 0 sent | 0 received ) = 0.99*0.5 / 0.52 = 0.9519

Therefore 0.9519 is the required probability here

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