Answer7

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WHAT'S YOUR ENGINEERING QUOTIENT?

Problem 7—What is the probability of finding more than two byte errors in a block of 200, given a BER of 10^-3?

Answer:

The general expression for finding exactly k errors in a block of n bytes, given a probability p of a single error, is:

C(n, k) × p^k × (1–p)^n–k

Where C(n, k) is the binomial coefficient, defined as:

C(n, k) = n!
k! × (n–k)!

If p = 0.0079721, n = 200, the values for the first few values of k are:

k Probability of k errors
C(n, k) × p^k × (1–p)^n–k Probability of k
or fewer errors
(cumulative sum) Probability of more
than k errors
(1 – sum)

0 0.20173 0.20173 0.79827

1 0.32423 0.52597 0.47403

2 0.25925 0.78522 0.21478

The probability of more than two errors is 0.21478, meaning that more than one in five blocks will be uncorrectable.

Contributor: Dave Tweed

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