Probability of a draft without replacement There is an urn with $N_1$ balls of type $1$, $N_2$ of type $2$ and $N_3$ of type $3$. I want to show that the probability of picking a type $1$ ball before a type $2$ ball is $N_1/(N_1+N_2)$. (without replacement = when you pick a ball you don't put it back in the urn, you keep it and keep picking balls)
Can you help me ?

You can ignore the type 3 balls, as picking one leaves you with the same number of type 1 and 2 balls.  Take all the type 3 balls out and pick one ball.  It is type 1 with probability $\frac {N_1}{N_1+N_2}$ as you say.
