Comments on: 50/50 Marbles http://www.laranevans.com/2007/10/31/5050-marbles/ I develop software people love to use. Sat, 07 Aug 2010 01:17:10 +0000 hourly 1 By: John Brook http://www.laranevans.com/2007/10/31/5050-marbles/comment-page-1/#comment-90 John Brook Sat, 05 Jul 2008 12:43:07 +0000 http://www.laranevans.com/2007/10/31/5050-marbles/#comment-90 This solution assumes that once a jar is chosen at random, a marble is chosen in a nonrandom fashion. I disagree with this solution because I think that, judging from the context of the problem (we're asked to compute "exact odds"), we should assume that selection of the marble from the chosen jar is random as well. In that case, you should put 1 red marble and 0 blue marbles in the first jar, and you should put 49 red marbles and 50 blue marbles in the second jar. .50 + .50*(49/99) = .7474747474 There is a 74.747474% chance that a red marble will be picked. This solution assumes that once a jar is chosen at random, a marble is chosen in a nonrandom fashion. I disagree with this solution because I think that, judging from the context of the problem (we’re asked to compute “exact odds”), we should assume that selection of the marble from the chosen jar is random as well.

In that case, you should put 1 red marble and 0 blue marbles in the first jar, and you should put 49 red marbles and 50 blue marbles in the second jar.

.50 + .50*(49/99) = .7474747474

There is a 74.747474% chance that a red marble will be picked.

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