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
