Balls are drawn one by one at random from a box containing n…

Balls are drawn one by one at random from a box containing $n$ white, $m$ black and $l$ red balls: $1^{\circ}$ without replacement; $2^{\circ}$ with replacement after each draw. Determine in both cases the probability that a white ball is drawn before a black ball.

Step 1: $1^{\circ}$ — without replacement

All orders of the balls are equally likely. Ignore the red balls and look only at the order of the $n+m$ white and black balls: the first of them is equally likely to be any one of these $n+m$ balls. A white ball comes before any black ball iff this first ball is white:

\[ P_1=\frac{n}{n+m}. \]

Step 2: $2^{\circ}$ — with replacement

Each draw is independent, with probabilities $w=\frac{n}{N}$, $b=\frac{m}{N}$, $r=\frac{l}{N}$ ($N=n+m+l$) of white, black, red. White comes first iff the first $k-1$ draws are red and draw $k$ is white, for some $k\ge1$:

\[ P_2=\sum_{k=1}^{\infty}r^{k-1}w=\frac{w}{1-r}=\frac{w}{w+b}=\frac{n}{n+m}. \]

Step 3: The answer

\[ \boxed{P_1=P_2=\frac{n}{n+m}.} \]

Surprisingly, the result is the same in both cases and does not depend on $l$.

Step 4: Why the red balls do not matter

A red ball never decides the question; it only postpones the decision. Whenever the decision is made (the first white-or-black ball), the chance that it is white is $\frac{n}{n+m}$ — with replacement because the box is always the same, without replacement by symmetry.

Step 5: Example

$n=2$ white, $m=3$ black, $l=5$ red: $P=\frac25$ in both cases.