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Question
a box contain 3 black balls and 3 white balls.someone blindly takes 3 balls from the box and replaces it by 3 red balls.what is the probability to take 3 balls of different colors from the box.
Note that C(a,b) is "a choose b", and that is a!/(b!(a-b)!).
Lets look at taking 3 balls out of the box.
There are C(6,3) ways of doing this; that is 6!/(3!(6-3)!) = 720/(6*6) = 20.
There is 1 way of all white and 1 way of all black.
There are 9 ways of 2 whites and a black and 9 ways of 2 blacks and a white.
If the 3 balls left were all white or all black and those were replace with 3 red, there would be 2 chances of getting all of the balls as the same color out of 120 possible draws.
If the 3 left were of two different color, there would only be 1 way of getting 3 balls of the same color (and they would all be red) out of the 120 ways to draw them.
So of the 20 ways to draw out the balls, we would have 2 that gave the same color
and 18 that gave 2 different colors. That is 0.1 chance of the same color and 0.9 chance of two different colors. Of the balls with the 0.1 percent chance of being the same, there are 2 ways out of 120 ways of getting the same color, so that is 0.2/120 = 2/1200. Out of the balls of different colors, which has a chance of 0.9, there is only 1 way of drawing balls the same color.
That is, 0.9(1)/120 = 9/1200.
Adding the two together gives 11/1200, or a little less than 1%.
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Scott A Wilson
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