The expected value of a binomial distribution, $B(k, n, p)$ is $np$.
How do I go about calculating the conditional expectation, given that $k>0$, $E(B(k, n, p) | k>0)$?
I assume the answer is very simple but I do not know where to start.
ATTEMPT AT A SOLUTION
The expected value of a binomial distribution is given by:
$\mathrm{E}(X)=\sum_{k=0}^{n} k\left(\begin{array}{l} n \\ k \end{array}\right) p^{k} q^{n-k}$
I would assume that, in order to get $k>0$, one would simply start the sum from $k=1$. However, this term in the sum is already zero; and is part of the derivation for the expected value of the mean in any case.
I am aware of a relationship:
$\begin{aligned} \mathrm{E}(X \mid A) &=\sum_{x} x P(X=x \mid A) \\ &=\sum_{x} x \frac{P(\{X=x\} \cap A)}{P(A)} \end{aligned}$
Yet I am not sure how it applies to my problem.
Probability for $k\gt 0$ is $1-q^n$ so $E(X|k\gt 0)=\frac{np}{1-q^n}$.