Looking at a preparatory exam, I'm a little dumbfounded by a question on probability. There are $19$ balls in a box: $5$ red, $3$ white, and $11$ blue. The question is: what is the probability of getting a white ball AFTER first taking out $0, 1, 2, 3, 4$ OR $5$ red balls.
Let $r$ be the amount of red balls. My reasoning is that
for $r = 0$, it is $\frac 3 {19}$;
for $r = 1$, it is $\frac 5 {19} \frac 3 {18}$;
for $r = 2$, it is $\frac 5 {19} \frac 4 {18} \frac 3 {17}$
and so on...
Is there a way to express this in one equation? Since this is without replacement, I presume binomial distribution won't work.
Your reasoning is wrong: take $r=0$; then $\frac{3}{19}$ is the probability to get a white ball in the first draw. However, you might also get a blue ball first and then a white ball in the second draw, and still have $r=0$. So the probability for $r=0$ has to be larger than $\frac{3}{19}$.
A solution might look as follows: First, calculate the probability for getting exactly r red balls and exactly b blue balls (in any order) before the first white ball. Second, sum over all possible values of b.
First part: If we get exactly r red balls and exactly b blue balls (in any order) before the first white ball, this means two things: i) we get r red balls and b blue balls in the first (r+b) draws, ii) conditional on this, we get a white ball in the (r+b+1)-th draw. The probability for this is $$\frac{\binom{5}{r}\binom{11}{b}}{\binom{19}{r+b}}\cdot \frac{3}{19-r-b}. $$
Second part: Let's sum over all possible values of b:
$$P(\text{first white after r red balls})=\sum_{b=0}^{11}\frac{\binom{5}{r}\binom{11}{b}}{\binom{19}{r+b}}\cdot \frac{3}{19-r-b}= \sum_{b=0}^{11}\frac{3\cdot\binom{5}{r}\binom{11}{b}}{19\cdot\binom{18}{r+b}} $$
(Maybe there's a shorter way to write this last expression that I'm not aware of at the moment.)