Hi there everyone,
Usually these standard draw x red marbles and y blue marbles from a box questions are no issue but here is a variant and friend asked me which I'm scratching my head about a bit:
You have a box with 10 blue marbles and 5 red marbles. When you draw a blue marble you keep it (without replacement) but when you draw a red marble you put it back in the box (replacement).
When you draw 3 blue marbles the game is over.
What is the average number of marbles expected to be drawn per game?
Now, obviously the drawing process for blue marbles would be a simple hypergeometric distribution if there were no red marbles. But the red marbles are going to skew this.
My initial thoughts were something like this:
- We know that the game is always going to end when we draw the 3rd blue marble so we will always have exactly 2 blue marbles and some number between 0 and infinity or red marbles (should be monitonically decreasing after a certain point).
- P[3 Blues|No Reds] = (10/15) * (9/14) * (8/13) = 0.2637
- P[3B|1R] = ((10C2 * 5C1(the 1 red))/(15C3) ) * (8/13) = 0.3043
- P[3B|2R] = ((10C2 * 6C2)/(16C4)) * (8/13) = 0.2282
- P[3B|3R] = ((10C2 * 7C3)/(17C5)) * (8/13) = 0.15663
- P[3B|4R] = ((10C2 * 8C4)/(18C6)) * (8/13) = 0.1044
Basically it is (combination without repetition -for blues) * (combination with repetition - for reds) / total combination) * (8/13).
Then I would muliply the number of marbles * probability to get expected number of total balls drawn per game.
But this way won't work as the probabilities (although diminishing) are going to total over 1.
Any one have any thoughts on this? Probably something very simple I'm overlooking. I'm stumped all the same!
Thanks very much for any and all help.