Ok this might seem like a dumb question which I think it is but it is really bugging me.
Lets say you have a 1 in 5 chance at something ( doesn't matter what) and you get 2 attempts at it. By basic probability you can just take the inverse and multiply by 2 subtracting the result by 1.
$$1-(.80*.80) = .36$$
Or 36% which seems simple enough but now do the same problem incorporating the Binomial theorem
$${n \choose k}*p^k*(1-p)^{n-k}$$ or
$${2 \choose 1}*(.20)^1*(1-.20)^{2-1} = .32$$
$.32 \neq .36$ last time I checked. So will someone please tell me what foolish thing I am doing wrong?
The first number you computed, $0.36$, is the probability of at least one success, which is the opposite of all failures.
The second number, $0.32$, is the probability of exactly one success, which leaves out the case of two successes. If you add the probability of exactly 2 successes (namely, $(1/5)^2=0.04$) to this number, you obtain your first answer.