Marble Probability Problem (Not Putting Marbles Back In Bag) A friend of mine asked this question to me recently. It has been ages since I did probability, but it seems interesting mathematically regardless. Here it is:
Suppose you have $5$ yellow marbles and $135$ green marbles in a bag. $10$ marbles will be pulled out and not put back in the bag, i.e. one marble will be pulled out and not placed in the bag followed by another marble being pulled out and not placed in the bag and so forth. What is the probability of pulling a yellow marble?

Let $P$ be the probability of pulling at least one yellow marble, and let $P^*$ be the probability of pulling no yellow marbles. Then
$$P=1-P^*$$
The probability of pulling no yellow marbles is
$$p_1p_2...p_{10}$$
where $p_i$ is the probability of pulling a green marble on the $i$th draw. The probability on the first draw is
$$p_1=\frac{135}{140}$$
and on the second, it is
$$p_2=\frac{134}{139}$$
and so on, so that
$$p_n=\frac{136-n}{141-n}$$
and so their product is equal to
$$\frac{\frac{135!}{125!}}{\frac{140!}{130!}}$$
$$=\frac{135!130!}{125!140!}$$
Which is the final answer.
