Question

Consider N users who share a communication link. Suppose each

user transmits a fraction p of the time, i.e., at any time instant, each user is transmitting with probability p.

The probability that at any time instant, n users are transmitting is given by the Binomial distribution:

 C(N,n) = N!/[n!(N-n)!]

i.   What is the probability of more than 12 users transmitting 
     at the same time?  ( 10 pts)

ii.  What is the probability of less than 5 users transmitting 
     at the same time?  (20 pts)

iii. Suppose p = 0.15 and N = 35.  Calculate the numerical values 
     of the probabilities in parts i. and ii.  (20 pts)
     (You might need to write a small program to do this.) .

where C(N,n) = N!/[n!(N-n)!].  C(N,n) means the number of combinations of N items taken n at a time.  See  a tutorial on elementary probability  if you do not remember this from mathematics

Answer 1

Total number of users = N

(i) Now we want to calculate the probability that more than n = 12 users are transmitting at the same time.

Let P(i) be the probability that i users are transmitting at the same time.

P(i) =

So the probability that more than 12 users are transmitting at the same time =  

(ii) The probability that less than 5 users are transmitting at the same time =

(iii) Now, p = 0.15 and N = 35. Using MATLAB , we write the code for calculating the above probabilities

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% Matlab Code below

N = 35;
p = 0.15;

pAns = prob(p,N,4);
fprintf('The probability of less than 5 users transmitting is %f\n',pAns);
pAns = 1.0-prob(p,N,12);
fprintf('The probability of more than 12 users transmitting is %f\n',pAns);

function ans = prob(p,N,k)
pans = 0.0;
for i = 0:k
comb = nchoosek(N,i);
bernp = comb*(p.^i)*((1-p).^(N-i));
pans = pans + bernp;
end
ans = pans;
end

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Answer for part(i) which is probability of more than 12 users transmitting at the same time = 0.001098

Answer for part(ii) which is probability of less than 5 users transmitting at the same time = 0.380749