Question

4- A binary communication channel introduces a bit error in a transmission with 10 pts probability 0.01. Let X be the number of errors in 10 independent transmissions. (a) Find the pmf of X if X takes on value in the set Sx-(1,2,3,4,53. (b) Find the mean, the mean square, and the variance of X. 10 t.

Answer 1

(a)

X can be any number between 0 and 10 and X will follow binomial distribution with n = 10 and p = 0.01

If it is given that X can takes only the values 1, 2, 3, 4, 5 then,

PMF of X = P(X = x | X = 1, 2, 3, 4, 5) = P(X = x) / P(X = 1, 2, 3,4, 5)

Now, P(X = 1, 2, 3,4, 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

= ¹⁰C₁ * 0.01¹ * (1 - 0.01)^10-1 + ¹⁰C₂ * 0.01² * (1 - 0.01)^10-2 + ¹⁰C₃ * 0.01³ * (1 - 0.01)^10-3 + ¹⁰C₄ * 0.01⁴ * (1 - 0.01)^10-4 + ¹⁰C₅ * 0.01⁵ * (1 - 0.01)^10-5

= 0.09135172 + 0.004152351 + 0.0001118478 + 0.000001977108 + 0.00000002396495

= 0.09561792

PMF of X is,

P(X = 1) = P(X = 1 | X = 1, 2, 3, 4, 5) =  0.09135172 / 0.09561792 = 0.9553828

P(X = 2) = P(X = 2 | X = 1, 2, 3, 4, 5) =  0.004152351 / 0.09561792 = 0.04342649

P(X = 3) = P(X = 3 | X = 1, 2, 3, 4, 5) =  0.0001118478 / 0.09561792 = 0.001169737

P(X = 4) = P(X = 4 | X = 1, 2, 3, 4, 5) =  0.000001977108 / 0.09561792 = 0.00002067717

P(X = 5) = P(X = 5 | X = 1, 2, 3, 4, 5) =  0.00000002396495 / 0.09561792 = 0.0000002506324

(b)

Mean of X is,

E(X) = 1 * P(X = 1) + 2 * P(X = 2) + 3 * P(X = 3 ) + 4 * P(X = 4) + 5 * P(X = 5)

= 1 * 0.9553828 + 2 * 0.04342649 + 3 * 0.001169737 + 4 * 0.00002067717 + 5 * 0.0000002506324

= 1.045829

Mean square of X is,

E(X²) = 1² * P(X = 1) + 2² * P(X = 2) + 3² * P(X = 3 ) + 4² * P(X = 4) + 5² * P(X = 5)

= 1 * 0.9553828 + 4 * 0.04342649 + 9 * 0.001169737 + 16 * 0.00002067717 + 25 * 0.0000002506324

= 1.139953

Variance of X is,

E(X²) - E(X)²   = 1.139953 - 1.045829² = 0.0461947