Coding theory - binary symmetric channel

coverband
Messages
170
Reaction score
1
Hi

Let us suppose we transmit the binary digit '1'. The probability of not receiving '1' is p. Thus the probability of receiving '1' is 1-p. Suppose we send a longer code of length n. The probability of this code being received correctly is (1-p)^n.

Now I don't understand this next statement: The probability that one error will occur in a specified position is p(1-p)^(n-1).

Taking an example let's say we transmit the code 000. The probability of receiving this code without error is (1-p)^3 [fine] + 3p(n-1)^2 [What !?]
 
"The probability that one error will occur in a specified position is p(1-p)^(n-1). "

This means that, for example, the probability that an error will occur in the first digit, and digits 2-n will be correct, is p(1-p)^(n-1). You have to specify the position in order for this formula to hold.
 
Last edited:

Similar threads

  • · Replies 1 ·
Replies
1
Views
10K