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 ?
 A: 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.
A: Consider an fish tank with 2 black fishes and 1 white fish.  We need to find the probability that we pick white fish before black fsh is the same as white fish is the last to be picked. A simple enumeration would be 
$B_1B_2W_1$
$B_1W_1B_2$
$W_1B_1B_2$
$B_2B_1W_1$
$B_2W_1B_1$
$W_1B_2B_1$
In the above enumeration,of the six possible outcomes, there are two outcomes where white is the last.  The probability that White is the last type to be removed is $\frac{1.2!}{3!} = \frac{1}{3}$.  Extending the same for urn with $N_1$ type 1 , $N_2$ type2, and $N_3$ type 3 balls where $N_3$is irrelevant    the total number of outcomes $= (N_1+N_2)!$ with $N_3 $ interspersed among the types  Number of ways the last ball  is Type 1 is  $N_1.(N_1+N_2-1)!$.  Thus the probability is $\frac{N_1.(N_1+N_2-1)!}{(N_1+N_2)!} = \frac{N_1}{N_1+N_2}$.
