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4/3/05 2:33 PM
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tower1
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Edited: 03-Apr-05
Member Since: 02/24/2003
Posts: 381
 
suppose that a die has been altered so the faces are 1,1,2,3,3,3. If the die is tossed 9 times, what is the probability that the numbers recorded are 1,1,2,2,2,3,3,3,3 in any order?
4/4/05 1:31 AM
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tower1
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Edited: 04-Apr-05
Member Since: 02/24/2003
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I'm pretty sure it follows this formula, but I'm struggling with the "# of arrangements." P(1)*P(2)*P(3)*(# of arrangements) the # of arrangements could equal "2*3*4" in which case P(1,1,2,2,2,3,3,3,3 in any order) would equal 2/3. or the # of arrangements could equal "(2!*3!*4!)/9!)" in which case P(1,1,2,2,2,3,3,3,3 in any order) would equal 1/45360. I'm losing hair over this one... Can you help me out? Please :)
4/4/05 10:00 PM
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tower1
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Edited: 04-Apr-05
Member Since: 02/24/2003
Posts: 383
actually, I think the answer is: (2/6)^2 * (1/6)^3 * (3/6)^4 * (9!/(2!*3!*4!)) which equals something... If some smart person could come along and tell me I'm correct that'd be great. Thanks
4/5/05 3:23 PM
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Andrew Yao
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Edited: 05-Apr-05
Member Since: 01/01/2001
Posts: 2748
It's pretty tough. I couldn't do it.
4/5/05 3:43 PM
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Andrew Yao
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Edited: 05-Apr-05 03:55 PM
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Okay, your answer is correct, I think. Let's pretend for a second that all the numbers were different, like this : 1 2 3 4 5 6 7 8 9. There are 9! ways to permute 9 different numbers, if order matters and without replacement. But in our case, we have some replacements: 1 1 2 2 2 3 3 3 3 We still get the same number 9! of ways to scramble them if we assume that the first 1 is different from the second 1, and if we swapped the first and second 1, it counts as a different way. We have to remove those collisions, because in the final solution, if you switch the first 1 and the second 1, it doesn't really count as a different way, it's still the same thing. For any given permutation, the 1's can collide 2! = 2 ways. Either the first one comes first, or the second one comes first. The twos can collide 3! ways. and the 3s can collide 4! ways. For each possible permutation, remove the collisions. So now, to remove all of these from the original count of permutations, we will divide by the number of collisions. For example, to remove the collisions of 1s, we will divide the original number of permutations by 2. This is because we only want to keep one permutation out of each n collisions, because they really count as the same thing. So the number of possible ways to scramble 1 1 2 2 2 3 3 3 3 = 9!/(2!*3!*4!) Each of these has the same probability of occuring, which is (2/6)^2 * (1/6)^3 * (3/6)^4, so the probability of any of them occuring is the sum of the individual probabilities, = (2/6)^2 * (1/6)^3 * (3/6)^4 * (9!/(2!*3!*4!))

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