Problem #1 Consider a 5-bit word with 4 data bits and 1 parity bit. Assume that

Question

Problem #1
Consider a 5-bit word with 4 data bits and 1 parity bit. Assume that the Pr[bit error] = 10-2 and that bit errors are independent. Answer the following questions.
a) What is the probability of a received word having no errors?
b) What is the probability of a received word having a single bit in error? Would this be a detectable error?
c) What is the probability of a received word having a two bits in error? Would this be a detectable error?
d) What is the probability of a received word having a detectable error?

Explanation / Answer

Given word size is = 5 bit [ 4 data bits and 1 parity bit ]

         Probability of bit error is = 1/100 implies

         Probability of no bit error is = 1 - 1/100 = 99/100

         All bit errors are independent.

a)

     Here all the 5 bits in the word should recieve properly with out error,

     Hence probability of a received word having no errors

               = 99/100 * 99/100 * 99/100 * 99/100 * 99/100

               = (99/100)5

     Hence probability of a received word having no errors = (99/100)5

b)

      The single bit can be any of the 5 bits and

  Probability of that bit error is = 1/100

      Probability of no bit error for remaining 4 bits is = (99/100)4

       Here that one bit can be any of the 5 bits and hence

     Probability of a received word having a single bit in error

                    = 5 * 1/100 * (99/100)4

                           = 1/20 * (99/100)4

     Hence Probability of a received word having a single bit error = 1/20 * (99/100)4.

    This is a detectable error, if the error happened in 4 data bits.

   This is a Undetectable error, if the error happened in 1 parity bit.

c)

    The two bits can be any of the 5 bits and selected in 5c2 = 10 ways

  Probability of that 2 bit errors is = (1/100)2

      Probability of no bit error for remaining 3 bits is = (99/100)3

     Probability of a received word having a two bit errors

                    = 10 * (1/100)2 * (99/100)3

                           = 1/1000 * (99/100)3

     Hence Probability of a received word having a single bit error = 1/1000 * (99/100)3.

    This is a detectable error, if the 2 errors happened in 4 data bits.

   This is a Undetectable error, if there is an error in 1 parity bit.

d)

     If there is error in 4 data bits then it is detectable error

   hence the probability of a received word having a detectable error

= 4/5

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