# What is the Probability of a Best of five data bit transmision?

A telephone company uses a best-of-five encoder. In this system every bit is transmitted 5 times and at the receiver a majority vote decides the value of each bit. If the probability of an uncoded bit error is 10^-3 (0.003), what is the probability of a decoded bit error?

My estimate is, please correct me if im wrong, that it should be: Pe:probability of an uncoded bit to be transmitted with an error Pc=(1-Pe):probability of correct transmision of an uncoded bit Pc*Pc*Pe*Pe*Pe=> (1-0.003)*(1-0.003)*0.003*0.003*0.003 Thanks in advance

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Is it 10^-3 or 0.003? These are different numbers. – Superbest Dec 31 '13 at 13:28

Let p be the error probability for each, then the information is transmitted wrong if 3 or more bit is transmitted wrong.

so, it should be $10\times p^3(1-p)^2 + 5\times p^4(1-p) +p^5$

So: there are two things wrong with your calcuation:

(i) you did not count the number of different ways you can get 3 wrong and 1 correct. It could be

WWWCC WWCWC ... (w stand for wrong, c for correct)

(ii) you were not aware that making more than 3 errors also get your message wrong.

Notice that 10 and 5 comes from Binomial coefficients. They are respectively the number of ways to choose 3 objects from 5 and the number of ways getting 4 objects from 5. see http://en.wikipedia.org/wiki/Binomial_coefficient The cheapest way of getting these is via Pascal triangle (for simple problems like yours anyway)

What you are really asking is:

Let $X$ follow a binomial distribution with parameters $(5,p)$. What is $P(X\geq 3)$? http://en.wikipedia.org/wiki/Binomial_distribution might also be useful to you.

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Thank you very much! You are absolutely right. – Pokopik Jan 2 '14 at 8:24

Since we have 5 bits, corruption will only occur if at least 3 bits are incorrect for the same block.

I will call an incorrectly transmitted bit a "success" since that's what we are watching out for.

## Using the binomial distribution

The binomial distribution gives the probability of exactly $m$ successes in $n$ trials where the probability of each individual trial succeeding is $p$. With this method, you want the following:

• Binomial at $(m=3, n=5, p=10^{-3})$ for probability of corruption due to exactly 3 fails
• Binomial at $(m=4, n=5, p=10^{-3})$ for probability of corruption due to exactly 4 fails
• Binomial at $(m=5, n=5, p=10^{-3})$ for probability of corruption due to exactly 5 fails

And you can sum these up to get the answer. Since summing values of the PDF like this is a very common operation, we have the CDF, which gives probability of $n$ or less successes. You want $n$ or more successes, so just do $1-Probability(n-1 \, or \, less \, successes)$. Here's a handy binomial calculator: http://stattrek.com/online-calculator/binomial.aspx

## Without using the binomial distribution

You can also derive the expression yourself. Once again, you need:

• Probability of getting $3$ successes
• Probability of getting $4$ successes
• Probability of getting $5$ successes

The probability of getting $m$ successes in a row is $p\cdot p\cdot p ... = p^m$ and the probability of getting $n-m$ non-successes in a row is likewise $(1-p)^{n-m}$. But for corruption, they don't have to be in a row, they can be rearranged. So, if you are calculating for $n=3$, there are several ways of picking the 3 bits out of 5 which will be wrong. This is done with the combination function, also called "n choose k", written as ${n \choose k} = \frac{5!}{(5-3)!3!}$ which gives you the number of ways that $k$ things can be chosen out of $n$ things.

So for $3$ successes, there are ${5 \choose 3}$ arrangements of incorrect bits possible, and each one has probability $p^3\cdot (1-p)^{5-3}$. You multiply the two to get the probability of getting $3$ successes in any arrangement.

Then you repeat for $4$ and $5$ and sum them up.

If you pay a bit of attention here, you will see that you have just derived the expressions for the PDF and CDF of the binomial distribution.

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Thank you very much. You are both correct. The different ways of getting the same result eluded me. – Pokopik Jan 2 '14 at 8:22