 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.) My Thoughts | Comment 1 of 5
`Taking 5 marbles from 1st box(A) is completely independent of taking 30 marbles from second box(B).As they are independent events, P(A^B)=P(A)P(B){^: intersection}Let 1st box:2/3 red and 3x be no. of marbles.P(A1)=C(2x,4)*C(x,1),P(B1)=C(x,20)*C(2x,10)P(E1)=P(A1^B1)=P(A1)*P(B1)Similarly for the other case, 1st box: 2/3 whiteP(A2)=C(2x,1)*C(x,4),P(B2)=C(2x,20)*C(x,10)P(E2)=P(A2)*P(B2)Clearly x≥20 and for this P(E2)>P(E1)So, 2nd box is more likely to have majority of red marbles.`

 Posted by Praneeth on 2008-02-19 06:17:55 Please log in:

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