A sequence of characters is transmitted over a channel that
Solution
a) you must calculate the probability there will be n error free characters between two erroneous characters.
 Probability of an error free character: 1-p= 0.99
 Probability of n error free characters followed by an erroneous characters: P(N=n) = (0.99)^n*0.01
 which is a geometric distribution with p=0.01
 
 b) E(N) = (1-p)/p in a geometric distribution
 E(N) = 0.99/0.01 = 99
 That\'s to be expected. if you have a 1% chance of an erroneous character, you will have a mean of 1 erroneous character every 100 characters, and thus a mean of 99 error free characters between erroneous characters.
 
 c) We want P(N1000) = 0.99
 P(N=n) = p(1-p)^n
 P(N1000) =1-P(N1000) = 1 - (i=0 to 1000)P(N=i) = 1 - p(i=0 to 1000)(1-p)^i
 P(N1000) = 1 - p*(1- (1-p)^1001)/(1-(1-p))
 P(N1000) = (1-p)^1001
 since P(N1000) = 0.99,
 (1-p)^1001  0.99
 ln(1-p)  ln(0.99)/1001
 p  1 - e^(ln0.99/1001)
 p  0.00001

