Binomial Distrinution - is my answer reasonable?

[KataKoniK]

Q: The simplest error detection scheme used in data communication is parity-checking. Messages sent consists of characters, each character consisting of a number of bits, 0 or 1. In parity checking, a 0 or a 1 is appended to the end to make the number of 1's even. The receiver checks the number of 1's in every character received and if the result is odd, it signals an error. Suppose that each bit is receveid correctly with probability of 0.999 independently of other bits. What is the probability that a 7bit character is received in error and not dected by the parity check?

The equation I used is this one

http://mathworld.wolfram.com/BinomialDistribution.html

My answer:

p(no error in 7-bits) = (7 choose 0) * (.001)^0 (1 - .001)^7 = .99302
p(single error in 7-bits) = (7 choose 1) * (.001)^1 (1 - .001)^6 = .00696
p(undetected error in 7-bit transmission) = 1 - p(no error in 7-bits) - p(single error in 7-bits)
= 1 - .99302 - .00696
= .00002

Correct? Resonable?

Thanks in advance.

 

[matt grime]

no, that isn't what it's getting at. there can be two errors, or 4 and that woul'dnt be boticed. the q asks what is the chance of an even number of errors

 

[KataKoniK]

Hmm,

Can you give me a hint on what I need to do?

p(single error in 7-bits) = (7 choose 1) * (.001)^1 (1 - .001)^6 = .00696

Do 7 choose 2? and so forth? up to 7 choose 7? Then do 1 - the sum of all those values?

 

[matt grime]

yes that is what you ned to do

 

[KataKoniK]

Thank you matt. Also, one last question, we don't need this p(no error in 7-bits) = (7 choose 0) * (.001)^0 (1 - .001)^7 = .99302 right?

 

[matt grime]

wait on. why are youi working out the probablity of 1,3,5 or errors? they would be detected sicne they cange the parity. you need to know the probabilty of 2,4 or 6 errors, that is all. does the 7bit char include the error correciting bit? ie are 8 bits received?

 

[KataKoniK]

Thanks for the heads up. So, from what you are saying, I just need to determine

p(single error in 7-bits) = (7 choose 2) = answer
p(single error in 7-bits) = (7 choose 4) = answer
p(single error in 7-bits) = (7 choose 6) = answer

Then the ans after getting all those values would be

p(undetected error in 7-bit transmission) = 1 - (those 3 values added together)?
I am a little bit confused here.

does the 7bit char include the error correciting bit? ie are 8 bits received?

I don't think the 7bit char includes the error correcting bit, so 8 bits are not going to be received, if that's what you mean.

 

[matt grime]

no no no!

you receive 8 bits. no error is detected if an even number of errors occured. what is the probability of an even number of errors? basic binomial distribution question.


if p is prob of error on any given bit, P(two errors in 8) is (8choose2)p^2q^6 where q=1-p.

P(4errors) is (8choose4)p^4q^4

P(6errors) is (8choose6)p^6q^2

P(8errors) is (8choose8)p^8=p^8

 

[KataKoniK]

Hmm, I am bit more confused now, sorry. So we hvae to threat the question as 8 bits? Even though it says 7? Or was that just an example?

 

[NateTG]

Hmm, I am bit more confused now, sorry. So we hvae to threat the question as 8 bits? Even though it says 7? Or was that just an example?

Remember that the pairity bit is also sent across the line, and is subject to the same error possibilities.

 

[KataKoniK]

So, if I am correct, I must do the following?

p(2 errors in 7-bits) = (8 choose 2) binomial eqn cont. = answer
p(4 errors in 7-bits) = (8 choose 4) binomial eqn cont. = answer
p(6 errors in 7-bits) = (8 choose 6) binomial eqn cont. = answer
p(8 errors in 7-bits) = (8 choose 8) binomial eqn cont. = answer

Then 1 - (those four values added together)?

 

[NateTG]

So, if I am correct, I must do the following?

p(2 errors in 7-bits) = (8 choose 2) binomial eqn cont. = answer
p(4 errors in 7-bits) = (8 choose 4) binomial eqn cont. = answer
p(6 errors in 7-bits) = (8 choose 6) binomial eqn cont. = answer
p(8 errors in 7-bits) = (8 choose 8) binomial eqn cont. = answer

Then 1 - (those four values added together)?

Maybe it would be clearer if you distinguished between transmitted bits (8) and data bits (7) explicitly. Things might make a bit more sense to you then.

That said, adding up those numbers should give you the correct probability of an undetected error.

 

[KataKoniK]

Thank you. I understand the reason behind checking for 8-bits, since as you mentioned, "pairity bit is also sent across the line, and is subject to the same error possibilities". We start counting from 0 in this case, so 0-7, which equals 8 bits?

That said, adding up those numbers should give you the correct probability of an undetected error.

Hmm, understood.

 

[matt grime]

let's do a simple example - 1 bit. so with parity check there are two bits sent.

they must by either 00 or 11. if one error happens we get 10 or 01, which we detect though we don't know what the error is. but if two errors occur we receive 11 or 00 resp. so we don't detect an error.

is it now clear why you need to work out the chances of 2,4,6, or 8 errors?

do you understand how binomial random variables work?

 

[KataKoniK]

Thanks matt grime for the detailed explanation. Everything is clear now :)