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A digital signal “1” or “0” is transmitted through a noisy channel, the received data may be different from the signal sent out. Suppose the transmitter sends out “0” with probability 0.7, and “1” with probability 0.3. When “0” is transmitted, the receiver receives “0” with probability 0.9, and “1” with probability 0.1. When “1” is transmitted, the receiver receives “0” with probability 0.2, and “1” with probability 0.8.

  • Find the probability that “1” was transmitted, when error occurs.

My attempt is the following:

We need to calculate P(S=1|error).

P(error) = P(S=0) P(R=1|S=0) + P(S=1) P(R=0|S=1)= (0.7)(0.1)+(0.3)(0.2)=0.13

P(S=1|error)= P(error|S=1) P(S=1) / P(error)

I have all the information except for P(error|S=1) which I am not sure how to calculate.

Any clue?

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  • $\begingroup$ Isn't $P(\text{error}|S=1)=.2?$ That's how I interpret the hypothesis that when $1$ is transmitted $0$ is received with probability $.2$ Do I misunderstand? $\endgroup$
    – saulspatz
    Feb 9, 2019 at 23:51
  • $\begingroup$ I think you are right! I missed it $\endgroup$
    – HaneenSu
    Feb 10, 2019 at 0:07

1 Answer 1

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$P(error|S=1)$ is just the chance of an error when the digit 1 is sent - it is given in the question as $0.2$.

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  • $\begingroup$ Ops! I missed that. So P(S=1|error) = 0.2*0.3/0.13 = 0.46. Is this correct? $\endgroup$
    – HaneenSu
    Feb 10, 2019 at 0:03
  • $\begingroup$ That looks fine to me. $\endgroup$ Feb 10, 2019 at 0:06

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