Date: Wed, 16 Sep 1998 17:59:28 -0600 (CST) Subject: Probability Question
Name: Chris
Question: Hi Chris,
There are at least two different ways to interpret this question. The question might be interpreted to mean that we are to place 6 identical marbles in 3 boxes labeled A, B and C. Here, for each marble there are three choices for its placement, either in box A, B or C. Thus there are ways to place the 6 marbles in the 3 boxes. If you know about permutations and combinations then you see that there are ways to select 2 balls for box A, and for each selection there are ways to select 2 of the remaining balls for box B, and finally ways to select the last 2 balls for box C. Thus there are ways to put two balls in each box.
If we suppose the problem means 6 identical marbles into 3 identical boxes
we can answer in the following manner. We will use to represent a
marble. If we line up 6 marbles the define 7 spaces 5 between the marbles and 1 at each end. Thus if we wanted to divide the marbles into only two boxes all we would have to do is place a divider in one of the 7 spots, 7 ways to do this. For example if we use a red bar as the divider we might see
or or these would represent a division of 2 marbles in the first box and 4 in the second, 5 marbles in the first box and 1 in the second, and 0 marbles in the first box and 6 in the second. Our problem is more complicated in that we have three boxes which means we need two dividers, a red bar and a blue bar. In this notation an arrangement such as
for example means 2 marbles in the first box, 2 in the second and 2 in the third while
means 2 in the first, 1 in the second and 3 in the third, and
would mean 4 in the first, none in the second and 2 in the third. Our problem is then, how many ways can we put in the two dividers? Again if you know about permutations and combinations there is an easy way to see the answer is 28 but let's just do it step by step. First place one of the dividers, say the red bar as we did above. Where can we place the second? Let's have another look at the situation after the red bar has beeen placed -- we now have 7 objects lined up, the 6 marbles and the divider. These determine 8 spaces, 1 between any pair of adjacent objects and 1 at each end. For example
Notice that there is a space on either side of the red bar. So where can we place the blue bar? In any of the 8 places! For example
giving us a division of 2, 1 and 3 marbles. It looks like there are 7 x 8 = 56 possiblities but we must be careful! Look at the pairs of solutions
they each give 2 marbles in the first box, 1 in the second and 3 in the third, so we've counted this answer twice -- this is because just switching the red and the blue bars doesn't give us a new answer. This means all told there are (7*8)/2 = 28 different possiblities of which the situation with 2 marbles in the first box, 2 in the second and 2 in the third only appears once. Therefore, with this interpretation of the problem the probability is 1/28 that each box gets exactly 2 marbles.
Cheers, |

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