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Ramsey_KJ
Member since Sep-23-04
Oct-30-04, 06:02 PM (EST)   "An Urn and 2 Balls"

 Here is a probability problem that is ancient but a little demanding that I got off the MAA web sight. Suppose that there is an urn with two balls inside, both of which may be either white or black. On each draw, a ball is removed at random and then put back into the urn. If on the first two draws, a white ball is drawn, what is the probability that on the third draw a white ball is also drawn. Hint, it isn't 50/50.Have a Good DayKJ Ramsey

Subject     Author     Message Date     ID An Urn and 2 Balls Ramsey_KJ Oct-30-04 TOP RE: An Urn and 2 Balls TheSmith Aug-29-05 1 RE: An Urn and 2 Balls Graham C Sep-15-05 2 RE: An Urn and 2 Balls iliaden Apr-02-06 4 RE: An Urn and 2 Balls iliaden Apr-02-06 3 RE: An Urn and 2 Balls mr_homm Apr-04-06 5 RE: An Urn and 2 Balls iliaden Apr-06-06 6 RE: An Urn and 2 Balls mr_homm Apr-07-06 7

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TheSmith guest
Aug-29-05, 04:49 PM (EST)

1. "RE: An Urn and 2 Balls"
In response to message #0

 I get 2/3.There's four ways that the urn could have been set up: BB, BW, WB, WW. (This is a little sticky, but I imagined that was marble was slightly smaller than the other, thus making them distinct... much like making the two dice different colors to see that there's two 7's using a 3 and a 4.)Since a white was drawn at all, we know it's not BB, so that gives us 3 possible urns, each with equal probability of having happened. So there are essentially 6 marbles that could be drawn all with equal probability and 4 of them are white. 4/6 = 2/3. Graham C
Member since Feb-5-03
Sep-15-05, 10:17 AM (EST)    2. "RE: An Urn and 2 Balls"
In response to message #1

 >I get 2/3. >>There's four ways that the urn could have been set up: BB, >BW, WB, WW. (This is a little sticky, but I imagined that >was marble was slightly smaller than the other, thus making >them distinct... much like making the two dice different >colors to see that there's two 7's using a 3 and a 4.) >>Since a white was drawn at all, we know it's not BB, so that >gives us 3 possible urns, each with equal probability of >having happened. But that's not true, surely? If you drew two white balls in succession, the probability of the urn originally containing WW is higher than BW or WB.I get at first look* that the probability of the initial urn being WW is 2/3 and the probability of WB or BW (added) is 1/3. If it is WW then the probability of the next being white is 1; if it is BW or WB the probability is 1/2.So the total probability of the next ball being white is (1*2/3 plus 1/2*1/3) or (2/3 plus 1/6) or 5/6.The longer you go on drawing white balls the greater the probability the urn only contains white balls. (You've partially acknowledged this by allowing for the probability of BB to be zero, rather than the 1/4 that would have been assumed before no balls had been drawn.)> So there are essentially 6 marbles that could be drawn all with > equal probability and 4 of them are white. 4/6 = 2/3. * Using Bayes. Writing P(ww) for the a priori probability (1/4) that the initial distribution was both white, and P(WW) for the a priori probability (3/8) of drawing two white balls in a row you getP(ww|WW) = P(WW|ww)*P(ww)/P(WW) = (1*1/4)/(3/8) Member since Aug-14-05
Apr-02-06, 10:23 PM (EST)    4. "RE: An Urn and 2 Balls"
In response to message #1

 >I get 2/3. >>There's four ways that the urn could have been set up: BB, >BW, WB, WW. (This is a little sticky, but I imagined that >was marble was slightly smaller than the other, thus making >them distinct... much like making the two dice different >colors to see that there's two 7's using a 3 and a 4.) Up 'till here, I aggree, but it's the following part that has a problem.>>Since a white was drawn at all, we know it's not BB, so that >gives us 3 possible urns, each with equal probability of >having happened. This is wrong.Since you already know that ONE of the two balls is white, you must consider that there are two possibilities that are thrown away: the BB and the BW OR the BB and the WB, depending which ball you consider being drawn: the first one or the second one. You cannot keep both BW and WB because then we are unsure about the color of both balls, yet we are certain that one of them is white.My answer is 1/4 and it is explained in another posting.Ilia

Member since Aug-14-05
Apr-02-06, 10:23 PM (EST)    3. "RE: An Urn and 2 Balls"
In response to message #0

 I get 3/4I simply consider the probability of getting a black ball.Since one of the two balls is white, there is a 50% probability of the second one being black.Meanwhile, there is also a 50% probability of the second ball being drawn, otherwise, it will be the first ball, which is white.thus the probability of a black ball being drawn is 50%*50%=25%, or 1/4the probability of a white ball being drawn is thus 1-(1/4)=3/4.Ilia mr_homm
Member since May-22-05
Apr-04-06, 10:26 AM (EST)    5. "RE: An Urn and 2 Balls"
In response to message #3 Member since Aug-14-05
Apr-06-06, 05:36 PM (EST)    6. "RE: An Urn and 2 Balls"
In response to message #5

 Hi Stuart,>Your reasoning is correct, but you are finding the >probability that the SECOND ball you draw is white. The >problem asks for the probability that the THIRD ball you >draw is white. This is my mistake, I haven't read the problem carefully.>So when you draw that second ball, 3/4 of the time it is >white, which is correct. Now, which of these white balls >come from WW and which come from WB? Thinking about the >case of drawing a black ball, you get the black ball half >the time when you have BW and none of the time when you have >WW. Therefore, when the second ball drawn is white, that >ball comes from all the WW cases and half the BW cases. >Therefore, after seeing that the second ball is white, it is >twice as likely that you are in a WW case than in a BW case. > Therefore, 2/3 of the time you will have WW and 1/3 of the >time you will have BW. Thus here you agree that one of the BW and WB probabilities must be eliminated. I have encountered several people trying to proove me that I should keep both of them after the first draw. Right now I am confused. I can give you my explanation of why we SHOULD eliminate one of them, yet I am not certain about it. There goes:When you draw the first ball, you assign a number to it. Let's say ball #1. we know that the ball #1 is White, while the color of the other one is unknown. In our question, the ball #2 is unknown, so it has a 50% probability of being white, and 50% prob. of being black. This means that there should be two variantes left: the WW and the WB. The BW is eliminated. Am I the one who did the mistake, or were it the others?And another question: can you proove that if you keep drawing white balls, the odds of having a WW variation will go up?Thank you,Ilia Denotkine mr_homm
Member since May-22-05
Apr-07-06, 08:22 PM (EST)    7. "RE: An Urn and 2 Balls"
In response to message #6

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