Probability help please!

Question

In random access communication protocols a population of users is contending for the use of a common channel to transmit their packets, without the control and supervision of a master station. In the sequel we assume that time is divided into timeslots. The communication channel can handle at most two packets in a timeslot. Because all users transmit packets when they themselves find appropriate, it will be possible of course that multiple users transmit a packet during the same time. If one or two users transmit a packet over the channel in a certain timeslot, then these transmissions are successfull. If three users transmit a packet over the channel in the timeslot, a small collision occurs and only one of the three packets is transmitted successfully. If four or more users transmit a packet in the timeslot, then a big collision occurs and all the packets transmitted in the timeslot are lost. 1 Let the probability mass function of the number of users, K, that transmit a packet over the communication channel in a certain timeslot be given by

PK(k) =

0.1, for k = 0,

0.1, for k = 1,

0.3, for k = 2,

0.3, for k = 3,

0.1, for k = 4,

0.04, for k = 5,

0.03 for k = 6,

0.02, for k = 7,

0.01, for k = 8,

0, otherwise.

Compute the expectation and the variance of K.

Calculate the expectation and the variance of L, the number of successfully transmitted packets in a timeslot.

What is the probability that in a period of 10 successive timeslots there will be 3 big collisions and 2 small collisions?

Answer 1

E{K) = ∑k P(k) = 2.6 E(k²) = ∑k²P(k) = 9.3 var(k) = E(k²) - E{k)²= 9.3 - 2.6^2 = 2.54

P(L=0) = P(k = 0) + P(k >=4) = 1 - P(1) - P(2) - P(3) = 1 - .1 - .3 - .3 = .3

P(L=1) = P(k = 1) + P(k = 3) = .1 + .3 = .4

P(L = 2) = P(k = 2) = .3

E{L) = ∑l P(l) = 1; E(L²) = ∑l²P(l) = 1.6; var(L) = 1.6 - 1² = .6

P(small collision) = P(k = 3) = .3

P(big collision) = P(k >= 4) = 1 - .1 - .1-.3 - .3 = .2

P(no collision) = 1 - .3 - .2 = .5

P(3 big collisions and 2 small collisions) = 10!/(3!2!5!).2³.3².5⁵ = 2520 * .008 * .09 * .03125 = .0567