Question

In a multiuser wireless network the data frames are being transmitted when the probability of collision during the transmission is 0.75. The transmitters try to send the collision-free frames using up to 6 independent trials (transmission) as needed. The transmitters indefinitely discard the data frames after 6 failures. Calculate the probability that a data frame can be successfully transmitted (not be discarded). Let ? denote the random number of trails until the frame is transmitted without collision, find ?(? ≤ 5) =?

Answer 1

mean = np

n= 6

where n = total number of repetitions experiment is executed

p = success probability = 0.75 q = 1-0.75 = 0.25

mean = 6 * 0.75

= 4.5

variance = npq

where n = total number of repetitions experiment is executed p = success probability q = failure probability

variance = 6 * 0.75 * 0.25

= 1.125

standard deviation = sqrt( variance )

= sqrt(1.125)

=1.060 Let ? denote the random number of trails until the frame is transmitted without collision, the probability that a data frame can be successfully transmitted ?(? ≤ 5) = P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)    = ( 6 5 ) * 0.75^5 * ( 1- 0.75 ) ^1 + ( 6 4 ) * 0.75^4 * ( 1- 0.75 ) ^2 + ( 6 3 ) * 0.75^3 * ( 1- 0.75 ) ^3 + ( 6 2 ) * 0.75^2 * ( 1- 0.75 ) ^4 + ( 6 1 ) * 0.75^1 * ( 1- 0.75 ) ^5 + ( 6 0 ) * 0.75^0 * ( 1- 0.75 ) ^6

= 0.8220