i'm taking an introductory probability course and having a bit of trouble on this one. (a) A bag contains 8 White, 12 Green and 14 Red marbles. If I remove half of each color marble from the bag, how many ways are there to choose which marbles to keep. (b) If I randomly pour out half of the marbles from the original 34, what is the chance that exactly half of each color remain.
For part (a) After removing half of each color, there remain 4W 6G and 7R marbles. Does that make the number of ways to choose the remaining marbles = 8C4 * 12C6 * 14C7 OR is is 4! * 6! * 7!.
For part (b) I pour out 17 marbles, and want there to be exactly 4W 6G and 7R remaining. I'm a bit lost on this one. Choosing at random, we could figure out the probability of each color, but when we aren't replacing (i.e. pouring out half the bag) I am unsure of what to do. There are 34C17 ways of pouring out 17 balls. Do I just calculate the number of ways to arrange a correct setup, 8C4 * 12C6 * 14C7 and divide 34C17? That gave me an answer of 0.0951. Please advise
For part (a) it would indeed be $\binom{8}{4}\binom{12}{6}\binom{14}{7}$. The first term describes the number of ways in which we may select four of the eight white marbles to keep, the second term describes the number of ways in which we may select six of the twelve green marbles to keep, etc...
Your alternate attempt does not make sense as an answer to this question as $4!$ refers to the number of arrangements of four items, etc... but here we are not arranging anything.
As for part (b), yes you got it correct. The answer will indeed be $\dfrac{\binom{8}{4}\binom{12}{6}\binom{14}{7}}{\binom{34}{17}}$