a three digit message is transmitted in communication system over a noisy channal having a probability of error Pe=2/5 per digit,

The probability distribution for errors is the binomial distribution B(3, 2/5), the probability distribution function (pdf) is

P(k)= (3 C k) (2/5)

^{k}(3/5)^{3-k},(3 C k) is the binomial coefficient.

The pdf has 4 values:

P(0)= (3/5)

^{3}= .216P(1)= 3 (2/5)(3/5)

^{2}= .432P(2)= 3 (2/5)

^{2}(3/5) = .288P(3)= (2/5)

^{3}= .064The cumulative distribution function (cdf) is the sum of these:

C(0)= .216

C(1)= .216+.432= .648

C(2)= .216+.432+.288 = .936

C(3)= .216+.432+.288+.064 = 1

You can use a TI-84 calculator to get these answers directly.