Question 759800
a box contains 12 balls of which 4 are green and 8 are red. if 7 balls are drawn from the box at random, one by one with replacement, find the probability that exactily 3 balls drawn are green. 
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First we will find the probability of drawing the 7 in this particular order:
GGGRRRR, where G = drawing a green ball and R = drawing a red ball.
That is,<br>
{{{(4/12)(4/12)(4/12)(8/12)(8/12)(8/12)(8/12)}}}{{{""=""}}}{{{(4/12)^3(8/12)^4}}}{{{""=""}}}{{{(1/3)^3(2/3)^4}}}{{{""=""}}}{{{(1/3^3)(2^4/3^4)}}}{{{""=""}}}{{{2^4/3^7}}}{{{""=""}}}{{{16/2187}}}<br>
But they do not have to be chosen in the particular order GGGBBBB.
For instance, the order BBGBGBG or GBGBGBB would do just as well. 
So we must multiply the probability of the particular order GGGBBBB 
by the number of all possible orders. That is the number of ways to
choose 3 of the 7 positions to draw the 3 green balls in or 
C(7,3) = 35. (The other 4 positions will of course be red balls).<br>
So the answer is {{{35}}}{{{""*""}}}{{{16/2187}}}{{{""=""}}}{{{560/2187}}}
<br>
Edwin</pre>