How many balls to take from an urn so that the probability of drawing at least $2$ red balls is $0.9$? Suppose we have a urn with $24$ red balls and $8$ black balls. We randomly draw $n$ balls from the urn without putting them back. How big must $n$ be such that the probability of drawing at least $2$ red balls is $0.9$?
The probability is the same as $1$ minus the probability of drawing at most $1$ red ball, which is perhaps easier to find. But how do I find it?

Not so elegant, but I think this is the fastest way to answer this question (that is, obtain a minimal value of $n$ such that the probability of having at least $2$ balls is more than or equal to $0.9$)
By constraints of the question we have $0 \leq n \leq 32$. Furthermore there are $8$ blue balls, and $24$ red balls.
So it is obvious that P( at least 2 are red ) increases as $n$ increases. In fact, by the Pigeonhole Principle, for $ 10 \leq n \leq 32 $ the probability of having at least $2$ red balls is $1$.
For $n=0$ or $n=1$, it is impossible to have at least $2$ red balls. The probability is $0$.
For $n=2$, there is only the case $2$ out of $2$ of the balls being red to consider. So we get:
$\dfrac{(24 \cdot 23)}{(32 \cdot 31)} \approx 0.55645$
For $n = 3$, we need to consider $2$ out of $3$, as well as $3$ out of $3$ of the balls being red. So we get:
$\dfrac{(8)(24 \cdot 23) {3 \choose 2} }{( 32 \cdot 31 \cdot 30 )} + \dfrac{(24 \cdot 23 \cdot 22)}{32 \cdot 31 \cdot 30} \approx 0.85323$
Now for $n=4$, we need to consider $2$ out of $4$, $3$ out of $4$, as well as $2$ out of $4$ of the balls being red. So we get:
$\dfrac{(8 \cdot 7 )(24 \cdot 23) {4 \choose 2} }{( 32 \cdot 31 \cdot 30 \cdot 29 )} + \dfrac{(8)(24 \cdot 23 \cdot 22) {4 \choose 3}}{( 32 \cdot 31 \cdot 30 \cdot 29)} + \dfrac{(24 \cdot 23 \cdot 22 \cdot 21)}{( 32 \cdot 31 \cdot 30 \cdot 29)} \approx 0.96068 $
Hence the answer is $n=4$.
