For the answer to the question above the probability of exactly x successes is
P(X=x)=b(x;n,p) = (nCx)(p^x)((1-p)^(n-x))
where nCx is number of combinations of n things taken x at a time, and "^" means exponentiation.
n = 10
p = 0.2
P(X=7) = 10C7*(0.2^7)*(0.8^3) = 0.00079
P(X=8) = 10C8*(0.2^8)*(0.8^2) = 0.00007
P(X=9) = 10C9*(0.2^9)*(0.8^1) = 0.00000
P(X=10) = 10C10*(0.2^10)*(0.8^0) = 0.00000
.00079 + .00007 + 0 + 0 = .00086 = .0009 rounded
