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The last one probe for today:
There are 8 balls in a box; among them white, red, blue, and black ones. The rest four are colorless. Four balls are taken at random without repetition. What is the probability of having all the colored balls? :yak
There are 8 balls in a box; among them white, red, blue, and black ones. The rest four are colorless. Four balls are taken at random without repetition. What is the probability of having all the colored balls? :yak
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15 Jul 2003, 07:26
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I assume that there are four colorless balls in the box besides the four having any color. Thus, the question is an example of hypergeometric distribution.
P=1C1*1C1*1C1*1C1/8C4=1/70
P=1C1*1C1*1C1*1C1/8C4=1/70
16 Jul 2003, 01:39
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stolyar
I assume that there are four colorless balls in the box besides the four having any color.
How can you make this assumption based on the wording of the question? No where does it state that there is only one ball of each of the four colors.
The probability will certainly change according to the numbers of each type of ball in the bag. Unless, of course, you are asserting that the average probability, after considering the plethora of possible combinations are colored balls given that at least one of each color is in the bag, will reduce to that of having 4 colorless balls. But that cannot be the case, because any time you repeat the number of balls, you increase the number of ways you can pick out 4 different colored balls. So clearly, 1/70 is the smallest possible probability and hence cannot possible be the average.
It is *possible* to solve this problem given no information other than that this is at least one ball of each color and that the other balls each have an equal change of being one of the 5 colors, the four or a neutral one. But it will be very difficult and timeconsuming to do so because it involve summing up a large number of weighted probabilities based on the various possible distributions in the bag.
