 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Counting One's Marbles (Posted on 2008-02-18) There are two closed boxes, filled with red and white marbles. One of the boxes is 2/3 red and the other is 2/3 white.

You're allowed to sample 5 marbles from the first box, and 30 from the second. As a result, you get 4 red and 1 white marble from the first box, and 20 red and 10 white from the second. Which is more likely to be the one with majority red marbles?

Assume the two boxes contain the same number of marbles.

 See The Solution Submitted by Charlie Rating: 1.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Solution | Comment 4 of 5 | First assume we have an infinite number of marbles or are sampling with replacement.

Probability of given event occurring with First Box being majority red: 5C1 * (2/3)^4 * (1/3) * 30C10 * (2/3)^10 * (1/3)^20 �

Probability of given event occurring with Second Box being majority red: 5C1 * (1/3)^4 * (2/3) * 30C10 * (1/3)^10 * (2/3)^20

The ratio of the two probabilities reduces to: 2^14:2^21 or 1:2^7

Therefore the probability of the second box being majority red is 128 times more likely than the first box being majority red.�

In the finite case without replacement this ratio only decreases as the number of marbles is reduced until there are fewer than 60 marbles in each box.

 Posted by Eric on 2008-02-21 03:39:38 Please log in:

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