Binomial Distrinution - is my answer reasonable?

  • Thread starter KataKoniK
  • Start date
168
0
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

Science Advisor
Homework Helper
9,394
3
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
 
168
0
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

Science Advisor
Homework Helper
9,394
3
yes that is what you ned to do
 
168
0
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

Science Advisor
Homework Helper
9,394
3
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?
 
168
0
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

Science Advisor
Homework Helper
9,394
3
no no no!!!

you recieve 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
 
168
0
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

Science Advisor
Homework Helper
2,449
5
KataKoniK said:
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.
 
168
0
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

Science Advisor
Homework Helper
2,449
5
KataKoniK said:
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.
 
168
0
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

Science Advisor
Homework Helper
9,394
3
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?
 
168
0
Thanks matt grime for the detailed explanation. Everything is clear now :)
 

Related Threads for: Binomial Distrinution - is my answer reasonable?

Replies
1
Views
1K
  • Posted
Replies
2
Views
3K
Replies
5
Views
2K
Replies
4
Views
469
  • Posted
Replies
2
Views
2K
  • Posted
Replies
2
Views
1K
  • Posted
Replies
2
Views
1K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top