Expected number of draws until the first good element is chosen A population has $G$ good and $B$ bad elements, $G+B=N$. Elements are drawn one by one at random without replacement. Suppose the first good element appears on draw number $X$. Find a simple formula, not involving any summation from $1$ to $N$, for $E(X)$.
Hint: Write $X-1$ as sum of $B$ indicators.
Ok, so we know in the first $X-1$ draws we only get bad elements. Let $I_j$ be $1$ if $j$th draw gives bad ball and $0$ otherwise. $X-1=I_1+I_2+ \cdots +I_{x-1}$, all of which have value $1$.  $E(I_1)= \cdots=E(I_n)=B/N=(N-G)/N$. Now we can write $E(X)=E((X-1)+1)$ $=E(X-1)+E(1)=E(X-1)+1.$
I'm not exactly sure on the formula for $E(X-1)$. Hopefully, I'm on the right track. Thanks!

Your current attempt doesn't express $X-1$ as the sum of $B$ indicators.  Instead, number the bad balls $1$ through $B$.  Try $I_j = 1$ if bad ball $j$ is chosen before any of the good balls and $0$ otherwise.  Then you have $X-1 = \sum_{j=1}^B I_j$.  Now, can you finish off the problem by finding $E[I_j]$?
(Added, for completeness): We have $E[I_j] = \frac{1}{G+1}$, the probability that bad ball $j$ is chosen before any of the good balls.  Thus $$E[X] = \frac{B}{G+1} + 1 = \frac{B + G + 1}{G+1} = \frac{N+1}{G+1}.$$
