I'm having a difficult time wrapping my head around how to solve this, and would really appreciate some help. Here's the problem:
I have a bag of 100 distinct marbles. I randomly choose 16 of them. If you were to randomly choose 12 marbles from that same bag of 100, what is the probability that all of your 12 were also marbles that I chose? Order does not matter, and there are no replacements.
Thank you in advance!
On
Basically you have selected and marked $16$ from the $100$ marbles and then returned them.
So you now want the probability for someone selecting $12$ from these $16$ marked marbles when selecting $12$ from $100$ marbles; selecting without bias nor replacement.
Does that help you wrap your head around the problem?
PS: You should know that $\binom n k$ counts the ways to select $k$ distinct elements from a set of $n$ such.
If I'm only drawing one marble, the answer is $16/100$.
How about two? The first one I draw needs to be one you've chosen. As remarked earlier, the probability of that is $16/100$. Since I'm not putting the marble back before I draw my second, there are in the bag $15$ marbles you've chosen, and $99$ marbles total. It seems reasonable to suppose that my second draw's probability is just based on the relative proportions of the marbles.
How about three? And so on..