Hello Hassan,
Since lambda is the mean value of the distribution, for small lambda there will not be a lot of traffic. If lambda is .1 and you take 100 draws with poissrnd(.1,1,100), then roughly speaking you would expect about 100*.1 = 10 ones, 90 zeros and the occasional draw of value two or greater. More precisely, for values n = 0,1,2,3 the expected number of draws are
lambda = .1;
n = 0:3;
Ndraws = 100;
expected_draws = Ndraws*(lambda.^n./factorial(n))*exp(-lambda)
expected_draws = 90.4837 9.0484 0.4524 0.0151
