# Probability of getting the first red ball before the first blue ball

Consider an urn containing $N = N_1 + N_2 + N_3$ balls, more precisely, an urn containing $N_i \gt 0$ balls of colour $i$, $i = 1, 2, 3$. We draw the balls without replacement. Show that the probability of getting a ball of colour $1$ before a ball of colour $2$ is $\frac{N_1}{N_1 + N_2}$.

I've done a similar problem before, where we assumed we did a sequence of independant trials of results $R_1$, $R_2$ or $R_3$ with probability $p_i \gt 0$, $i = 1, 2, 3$ and $p_1 + p_2 + p_3 = 1$. The probability that the result $R_1$ happens before the result $R_2$ is $\frac{p_1}{p_1 + p_2}$.

It looks like the above problem. In that case, the hint was to consider first, for $0 \lt n_1 \lt n_2$, the probability of getting the first $R_1$ result at the $n_1$-th trial and the first $R_2$ result at the $n_2$-th trial.

I've tried to compute something similar for this problem, but it led to something quite complicated....

Disregarding the irrelevant colour-3 balls, an outcome of the experiment gives you a random permutation of the $N_1+N_2$ balls of colours 1 and 2. The event in question is that the first of these balls is colour $1$. Since each of the $N_1 + N_2$ balls is equally likely to be first, and $N_1$ of them are of colour $1$, the probability is $N_1/(N_1 + N_2)$.
There are $N_1+N_2$ total balls of types $1$ and $2$. The number of type $1$ balls is $N_1$, so you divide. The type $3$ balls aren't relevant if you notice pulling them doesn't affect the first in the group of type $1$ and type $2$.