Thanks to Mr. Raghava G D’s comments on the post discussing Question 38 on Communication from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper, realized that I had made an error in the solution. Have the updated the post with the right answer and additional explanations.

## Q38. A binary symmetric channel (BSC) has a transition probability of 1/8. If the binary transmit symbol X is such that P(X=0)=9/10, then the probability of error for an optimum receiver will be

## (A) 7/80

## (B) 63/80

## (C) 9/10

## (D) 1/10

I had earlier claimed that the right answer is 7/8 and given that this answer is not listed in the options provided, there might have been a typo in the question paper. However, after digesting the comments from Mr. Raghava GD, realized that 7/8 is the probability that the received symbol is not equal to the transmitted symbol, and not the probability of error for an optimum detector.

After applying **Maximum A Posteriori Detector (MAP)** rule to the problem, the right answer seems to be **(D) 1/10**.

**Please refer to the Update section in the original post to see the relevant discussion. **

http://www.dsplog.com/2012/11/03/gate-2012-ece-q38-communication/#update

Thanks !

## References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

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