Question 1203189
assuming this is a binomial distribution, i get the following:


n = 80,000
p = 1/20,000 = .00005
q = (1 - 1/20,000) = .99995
x = 0, 1, or 2


formula is p(x) = p^x * q^(n-x) * c(n,x)


we get:


p(0) = .00005^0 * .99995^(80,000 - 0) * c(80,000,0) = 0.018313807.


p(1) = .00005^1 * .99995^(80,000 - 1) * C(80,000,1) = 0.073258892.


P(2) = .00005^2 * .99995^(80,000 - 2) * C(80,000,2) = 0.146523279.


the combination formula is c(n,x) = n! / (x! * (n-x)!)


for example, when n = 80,000 and x = 2, this formula becomes:
c(80,000,2) = 80,000 / (2! * 79,998!) which becomes:
c(80,000,2) = (80,000 * 79,999 * 79,998!) / (2! * 79,998!) which becomes:
c(80,000,2) = (80,000 * 79,999) / 2 = 3,199,960,000.


P(2) = .00005^2 * .99995^(80,000 - 2) * C(80,000,2) = 0.146523279 becomes:
P(2) = .00005^2 * .99995^(80,000 - 2) * 3,199,960,000 = 0.146523279.
you can verify with your calculator that this is correct.