Probability of drawing no red balls from 20 draws without replacement given finite sample I understand this to be a binomial distribution: There are 100 balls in a bucket. 10 are red, 90 are blue. I select a ball at random and then replace it in the bucket, and I do this 20 times. I then calculate the probability that none of the selected balls were red.
But what if I don't put the ball back in the bucket? The probability then changes with each trial. Can anyone point me in the right direction of how to calculate the probability in this case?
 A: Let $B$ denotes blue balls, $R$ denotes red balls, then you may apply the formula for hypergeometric distribution:
$$P(B = 20, R = 0) = \frac{\binom{10}{0}\binom{90}{20}}{\binom{100}{20}} = \frac{\binom{90}{20}}{\binom{100}{20}}$$
The last term exactly matches the @Macro's answer, but hypergeometric formula is more general. The idea beyond the formula is simple: get the number of ways to draw $20$ $B$ of $90$, number of ways to draw $0$ $R$ from $10$ (there is only one possibility) and divide their product by the number or ways to draw any $20$ balls from $100$. Hope this was not your homework ;)
A: Well, on the first try, you have a $90/100$ probability of not drawing a red ball; if the first was not a red ball, then on the second try there are still 10 red balls left, but only 99 to choose from, so you have a $89/99$ chance of not drawing a red ball. Similarly, on the third draw, if the second draw was also not a red ball, then you have a $88/98$ chance of picking a red ball, and so on. In general, if you attempt $k$ times independently without replacement, the probability you seek is 
$$ \prod_{i=1}^{k} \frac{ 90-i+1 }{100-i+1} $$ 
One important thing to note is that this probability actually doesn't arise from a binomial distribution. You are not conducting independent trials with equal probability and counting the number of "successes". The trials are not independent because the success probability of a future trial depends on whether a past trial was a success, making it fundamentally different from the binomial distribution. If there was replacement, then you'd be correct in saying the number of success follows a binomial distribution. 
