Originally Posted by
Vinod Poisson distribution as a limiting case of binomial probability distribution under the following conditions
1)n= number of trials indefinitely large. 2)p= constant probability of success for each trial is indefinitely small 3)np=m is finite
In binomial distribution, the probability of x successes p(x)=n_C_x (p^x) q^(n-x),
Put now p=m/n and q=1-p=1-(m/n) we get
p(x)=n(n-1) (n-2)
(n-x+1)/x! (m/n)^x (1-(m/n)^(n-x)
=1 [1-(1/n)][1-(2/n)]
[1-{(x-1)/n}] m^x. [(1-(m/n))^n]
--
x! [(1-(m/n))^x]
Now,for fixed x,as n -->∞
[1-(1/n)][1-(2/n)]
[1-(x-1)/n] all tend to 1 and [1-(m/n)]^(-n) to. e^(-m)
Can anyone tell me how [1-(m/n)]^(-n) is = e^(-m)[/tex]