Toss 6 independent fair coins, and let X be the number that comes up heads. Compute $E(X)$ and $Var(X)$.
Is this the negative geometric distribution? (6-x-1 Choose x-1) (1/2)^x (1/2)^(6-x) ?
Then wouldn't it be $E(X)=x(1-1/2)/1/2 $ ? But it is not an actual number.
Am I misinterpreting the question?