There is a bag of 15 marbles, 13 red and two blue. If you pick 9 marbles without replacement, what is the probability that:
a) One blue marble is left in the bag
b) Both blue marbles are still the bag
c) No blue marbles are left in the bag
This was an intro to probability question that I can't crack. I'm not even sure where to start - intuition doesn't help much. How can I solve these problems?
$\underline{An\; alternative\;method}$
You can solve all the 3 problems by considering only the blue marbles.
There are 6 "in bag" slots and 9 "out of bag" slots.
P(one blue marble in bag) $=\dfrac{\binom61\binom91}{\binom{15}2}$
You have a bag containing 13 red and 2 blue marbles, and on removing 9 marbles without replacement want to know the probabilities for counts of blue marbles in the bag.
This is the essentially same a taking a bag of 13 red and 2 blue marbles, and on removing 6 calculating the probabilities for counts of blue marbles removed. That is, the count has a Hypergeometric distribution.
a) One blue marble is left in the bag
Assuming this means "exactly" one, then it is the probability for leaving 1 from 2 blue marbles and 5 from 13 red marbles in the bag when leaving any 6 from the 15 marbles in the bag without bias or replacement of the 9 removed marbles.
$$\mathsf P(B=1)~=~ \dfrac{\dbinom 2 1\dbinom {13}5}{\dbinom{15}6}$$
The rest is left to you.