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6 votes

//This is what I feel:

Information provided:       * dataword is of 2 bits

                                                * parity mechanism is used.

                                                * p(bit is flipped)          = .1

                                                 =>p(bit not flipped)    = .9 

Also

                                                Flipping of bits is an independent activity.

Ans 36.

                Code will arrive without any errors if neither of the bits is flipped.

                 i.e. P(bit 1 didn't flip and bit 2 didn't flip) = P(bit 1 didn't flip) P(bit 2 didn't flip)

                                                                                   = (.9)(.9)

                                                                                    = .81


Ans 37.        Code arrives with an error that is correctly detected.

                     In parity bit mechanism correct error detection takes place only if odd number of bits have been flipped.

                     So, we are asked:

                    P(bit 1 is flipped but not bit 2) + P(bit 2 is flipped but not bit 1)

                    = (.9)(.1) + (.9)(.1)

                    = .09 + .09

                    = .18


Ans 38.   Code arrives with an error that is undetected.

               In parity mechanism, error can't be detected if even no. if bits have been flipped.

               Therefore, basically we are asked:

               P(both bits are flipped) = P(bit 1 flipped)P(bit 2 flipped)

                                                     = (.1)(.1)

                                                      = .01 


0 votes
0 votes
Dataword Codeword
0 00
1 11

 

Odd number of bits flipped then error can be detected. 00 → 01

Even number of bits flipped then error will not be detected. eg. 00 → 11

  1. both bits unflipped = .9*.9
  2. one bit flipped = .1*.9 + .9*.1 = .18
  3. both bits flipped = .1*.1 = .01

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