My Math Forum Probability Question With Four Containers Of Marbles

 March 17th, 2014, 10:57 AM #1 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 575 Thanks: 79 Probability Question With Four Containers Of Marbles There are four containers of marbles. Each container has exactly 1 white marble and however many black marbles are necessary so that the containers have 2, 3, 4, and 5 total marbles. The object is to pick all the white marbles while picking as few black marbles as possible. You stop picking from a container once you get the white marble from it. The marbles are picked without replacement. 1. What is the expected number of black marbles picked before all 4 white marbles are picked? 2. What is the probability of each number of black marbles from 0 to 10 being picked before all 4 white marbles are picked? For 0 black marbles it would be (1/2)(1/3)(1/4)(1/5) = 1/120 and that should be the probability for 10 black marbles also. I think the probabilities should be the same because the probability of the white marble being picked first from each container should equal the probability of the white marble being picked last from each container. I want to calculate the probability of winning a game. For each number of black marbles, I know the probability of winning, so I want to know the probability of picking each number of black marbles.
 March 19th, 2014, 07:55 AM #2 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 575 Thanks: 79 Re: Probability Question With Four Containers Of Marbles I wrote out all the permutations and solved it myself.

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