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Maximum allowable probability of error for the (8, 4) Hamming Code

  1. May 6, 2014 #1
    1. The problem statement, all variables and given/known data

    What is the maximum allowable probability of error is a typical digit in order that the (8, 4) Hamming Code can be used with probability .999 that the receiver will not be misled (i.e., 3 or more errors occur) in a single word?

    2. Relevant equations

    http://en.wikipedia.org/wiki/Hamming_code#Hamming_codes_with_additional_parity_.28SECDED.29

    3. The attempt at a solution

    Using the (8, 4) extended Hamming code, what is the maximum allowable probability of error in a typical digit (p) so that the receiver will know with certainty .999 that she has not been misled. In other words, what value of p do we have to take so that the probability of 3 or more errors in transmission is smaller than .001.
     
  2. jcsd
  3. May 7, 2014 #2

    haruspex

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    That's not an attempt at a solution.
    I assume wherever you wrote digit you meant binary digit, or bit.
    It's not clear to me that the probability of an undetected error is the same as the probability of 3 or more single bit errors in the same word. Certainly the code will detect all errors of one or two bits, and not all errors of three or more bits, but it might detect some errors of more than two bits.
    Putting that aside for now, suppose the prob of a single bit error is p. What is the probability of three or more errors in the same word? (How long is a word in this context?)
     
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