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Data is transmitted over a very bad data channel with a bit error rate p = 0,04. A data sequence consist of 8 bits.

1.) Only the first three bits of the sequence are transmitted correctly. 2.) Only a single bit is transmitted correctly, either the first or the last.

How do I solve this problem. The chance to receive exactly 3 bits should be ${8 \choose 3}(0.96)^3/(0.04)^5$ but I do not know where to go from there.

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  • $\begingroup$ It should be 0.96 and there should be no division in your formula. $\endgroup$
    – karmanaut
    Commented Sep 16, 2015 at 14:02

1 Answer 1

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${8 \choose 3}{(0.96)^{3} (0.04)^{5}}$ represents the total probability of recieving any 3 bits correctly and not just first 3 bits correctly.

That probability is $(0.96)^3(0.04)^5$. You can now solve the second part using this idea and the fact that either first or last bit is correct but not both.

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