# Thread: Probability of a byte being recieved correctly?

1. ## Probability of a byte being recieved correctly?

If this belongs to basic stat. section, sorry... but I felt this was somewhat advanced...

In a system a byte (8 bits) is transmitted with a bit error probability of 0.1. If the system can correct at most one error made in each byte.
***What is the most probable number of errors(after correction) in a byte?

Okay, so this is a binomial distribution so n=8, k(or x)= 1 and p=0.1
To start I found what the probability of a byte being received correctly(after correction) and got P(A)=0.19. I used P(A)=1-P(k<= 1) which is B(x;8,0.1)
I used to binomial forumla just to make sure I didnt use the tables correctly.

I do not understand ***what is the most probable number of errors(after correction) in a byte? am i finding a value of x or k??

2. Originally Posted by mightydog78
If this belongs to basic stat. section, sorry... but I felt this was somewhat advanced...

In a system a byte (8 bits) is transmitted with a bit error probability of 0.1. If the system can correct at most one error made in each byte.
***What is the most probable number of errors(after correction) in a byte?

Okay, so this is a binomial distribution so n=8, k(or x)= 1 and p=0.1
To start I found what the probability of a byte being received correctly(after correction) and got P(A)=0.19. I used P(A)=1-P(k<= 1) which is B(x;8,0.1)
I used to binomial forumla just to make sure I didnt use the tables correctly.

I do not understand ***what is the most probable number of errors(after correction) in a byte? am i finding a value of x or k??
If we take it that the system can reliably correct a single-bit error,
then if only one of the 8 bits is in error, that will be corrected and the byte will be error-free after correction.

Hence, the maximum number of bit errors possible is 7.

Two bits in error will have one bit corrected and is a single error after error correction.

Hence the probability of the byte being correct after error correction is the sum of the
binomial terms for no error and a single-bit error, which is about 0.81.

The probability of the byte having exactly a single-bit error (after correction) is

$\binom{8}{2}(0.1)^2(0.9)^6$

which turns out to have the highest probability.

The probability of exactly 2 bits in error (after correction) is

$\binom{8}{3}(0.1)^3(0.9)^5$

since one of the 3 bits in error will be corrected........etc

3. that makes so much more sense. yea for the part where I got 0.19 I accident though it was P(k>=1) when its suppose to be P(k<=1)

so the most probable number of errors(after correction) in a byte would be a single error since the probability is the highest and around 0.148
prob of 2 bits in error is much less than that

thank you very much for the explanation, wish my teacher could speak better english so i can understand whats going on..

4. Originally Posted by Archie Meade
If we take it that the system can reliably correct a single-bit error,
then if only one of the 8 bits is in error, that will be corrected and the byte will be error-free after correction.

Hence, the maximum number of bit errors possible is 7.

Two bits in error will have one bit corrected and is a single error after error correction.
I think not (at least without specifying the code and algorithms), in fact what happens with more than one bit error is not clear.

It is not even clear if the code can detect if there is more than one error.

It is entirely possible that the number of errors goes up if there is more than one.

CB

5. It may be safe to assume that the EDC mechanism will not correct an erroneous bit
in a byte that contains more than one bit in error.

If one of multiple bit errors are corrected, the byte itself will still be in error.
Practically speaking, there is little purpose correcting one of multiple errors.
Also, there is the problem of accurately detecting multiple bit error numbers.

Hence, if the probability of the byte being error-free after correction is
the sum of the binomial probabilities of zero and one errors,
then the probability of 2 bits in error is

$\binom{8}{2}(0.1)^2(0.9)^6$

the probability of 3 bits in error is

$\binom{8}{3}(0.1)^3(0.9)^5$ and so on..

6. Originally Posted by Archie Meade
It may be safe to assume that the EDC mechanism will not correct an erroneous bit
in a byte that contains more than one bit in error.

If one of multiple bit errors are corrected, the byte itself will still be in error.
Practically speaking, there is little purpose correcting one of multiple errors.
Also, there is the problem of accurately detecting multiple bit error numbers.

Hence, if the probability of the byte being error-free after correction is
the sum of the binomial probabilities of zero and one errors,
then the probability of 2 bits in error is

$\binom{8}{2}(0.1)^2(0.9)^6$

the probability of 3 bits in error is

$\binom{8}{3}(0.1)^3(0.9)^5$ and so on..
The model I prefer is this: If there is more than one error then a random bit is flipped.

Then let n be the number of errors before corresction and n* the number after, then we have:

n=0, n*=0
n=1, n*=0
n=2, n*=1 with prob 1/4, n*=3 with prob 3/4
n=3, n*=2 with prob 3/8, n*=4 with prob 5/8
n=4, n*=3 with prob 1/2, n*=5 with prob 1/2
n=5, n*=4 with prob 5/8, n*=6 with prob 3/8
n=6, n*=5 with prob 3/4, n*=7 with prob 1/4
n=7, n*=6 with prob 7/8, n*=8 with prob 1/8
n=8, n*=7

CB