- #1
coverband
- 171
- 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 probabilty of receiving this code without error is (1-p)^3 [fine] + 3p(n-1)^2 [What !?]
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 probabilty of receiving this code without error is (1-p)^3 [fine] + 3p(n-1)^2 [What !?]