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2014-09-19T01:00:38+05:30
There are 18 C_{4} ways to selecting 4 balls from 5 black, 6 white, 7 red balls
we can select x from 6 white balls and 4-x from remaining balls in (6 C_{x} )(12 C_{4-x} ) ways. hence
f(x) = (6 C_{x} )(12 C_{4-x} ) / 18 C_{4}   x = 0,1,2,3,4
I hope you can understood. if not please ask.
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2014-09-19T05:58:37+05:30

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There are four while balls (W) and 12 other balls (O) in the bag. Four balls are drawn one after another, let us say. The drawing of 4 balls are independent  events and mutually exclusive. So probabilities can be multiplied.

X = event that X number of the four drawn balls are white balls.

P(X=0) = all drawn balls are non-white.

P(X=0) = \frac{12}{18} * \frac{11}{17} * \frac{10}{16} * \frac{9}{15} = 0.1617 \\ \\ P(X=1) = 4 * \frac {6* 12*11*10} {18*17*16*15} = 0.4314 \\ \\ P(X=2) = 6 * \frac{6*5* 12*11}{18*17*16*15} = 0.3235 \\ \\ P(X=3) = 4 * \frac{6*5*4*12}{18*17*16*15} = 0.0784 \\ \\ P(X=4) = \frac{6*5*4*3}{18*17*16*15} = 0.0049 \\



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