# Recall that in chapter 2 we dened a random variable

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Unformatted text preview: how to bound the probability of undetected error, assuming that each bit is in error with probability 0.001, independently of all other bits. Let Y denote the number of bits in error. As already discussed, there can be undetected bit errors only if Y ≥ 4. So the probability there exist undetected bit errors is less than or equal to P {Y ≥ 4}. We have Y = X1 + · · · + X64 , where Xi is one if the ith bit in the array is in error, and Xi = 0 otherwise. By assumption, the random variables X1 , . . . , X64 are independent Bernoulli random variables with parameter p = 0.001, so Y is a binomial random variable with parameters n = 64 and p = 0.001. The event {Y ≥ 4} can be expressed as the union of 644 events of the form {Xi = 1 for all i ∈ A}, indexed by the subsets A of {1, . . . , n} with |A| = 4. For such an A, P {Xi = 1 for all i ∈ A} = i∈A P {Xi = 1} = p4 . So the union bound with m = 644 yields P {Y ≥ 4} ≤ 644 p4 . Thus, 64 4 p = (953064)p4 = 0.953064 × 10−6 . 4 P (undetected errors) ≤ P {Y ≥ 4} ≤ A considerably tighter bound can be obtained...
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## This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.

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