Expected number of draws until the first good element is chosen
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 $$E[X] = \frac{B}{G+1} + 1 = \frac{B + G + 1}{G+1} = \frac{N+1}{G+1}.$$