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as 7 message bits are given

so   $2^{P}$>=m+P+1.  {P=number of parity bits ; m=number of message bits ; 1 is the case when there is no error}

$2^{P}$>=7+P+1. 

minimum value of P is 4 which is satisfying it.

sent message format will be

P1 P2 1 P3 0 0 1 P4 0 0 0

odd parity is given (here odd parity is for odd number of zeros)

for P1:

               [P1, 1, 0, 1, 0, 0]. => P1 is ''1'' 

for P2:

               [P2, 1, 0, 1, 0, 0]. => P2 is "1"   

for P3:

               [P3, 0, 0, 0, 0]. => P3 is "0"   

for P4:

               [P4, 0, 0, 0]. => P4 is "1"

so final code that will be sent is

1 1 1 0 0 0 1 1 0 0 0

 

(C) is the answer.

  

by Active (1.7k points)
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@newdreamz a1-z0 where to read about it?

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error correcting codes in computer networks.(forouzan)

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