Conditional binomial expectactation and variance For example, we have standard case $X~B(n;p)$

How can we find $E(X|X&gt;=1)$ and $Var(X|X&gt;=1)$

I'd be glad to provide my work, but I'm  really stuck, I've only seen  problems, involving another variables. 

We can integrate the expression $pdf*x$ with the limits from 1 to infinity right?
But this value goes to infinity, and I don't know what to do, again :(
I give you the solutino for the expectation value, hope you can find the solution for the variance yourself:

$$E[X|x\geq 1]=\frac{\int_1^\infty p(x)\cdot x dx}{\int_1^\infty p(x)dx}$$

Why do we divide by $\int_0^\infty p(x)dx$?
Lets take the finite case with a dice, what is the expectation if we know, that we rolled higher than 2? All the outcomes have the same porbability, i.e. $\frac14$. I we didn't know anything about the outcome, each number had the same probability $\frac16$. We normalize, such that the total probability is again $1$.
I give you the solutino for the expectation value, hope you can find the solution for the variance yourself:

$$E[X|x\geq 1]=\frac{\int_1^\infty p(x)\cdot x dx}{\int_1^\infty p(x)dx}$$

Why do we divide by $\int_0^\infty p(x)dx$?
Lets take the finite case with a dice, what is the expectation if we know, that we rolled higher than 2? All the outcomes have the same porbability, i.e. $\frac14$. I we didn't know anything about the outcome, each number had the same probability $\frac16$. We normalize, such that the total probability is again $1$.