From bag A containing 4 black and 8 blue marbles, 6 marbles are transferred into an empty bag B. Three marbles are now drawn from bag B, and each of them happens to be black.
Determine the probability that a fourth marble drawn from bag B will be black, given that:
(I) The first three marbles are returned to bag B.
(II) None of the first three marbles is returned to bag B.
In order to do Bayesian calculations, we need to know the probabilities that bag B will contain zero to four black balls:
p(n) = (4!/(4-n)! / (12!/(12-n)!)) * (8!/(2+n)! / ((12-n)!/6!)) * C(6,n)
Those values are:
0 1/33
1 8/33
2 5/11
3 8/33
4 1/33
The overall a priori probability that the first three drawn from bag B would be black is
(8/33) * (3*2*1/(6*5*4)) + (1/33) * (4*3*2*1/(6*5*4*3)) = 7/495
But from the relative contributions of the two terms (3 black balls transferred vs 4 black balls), the probability is 1/7 that four balls had been transferred and 6/7 that three had been.
Part 1:
The probability that, after replacement, an individual ball will be black is (6/7)*(3/6) + (1/7)*(4/6) = 22/42 = 11/21.
Part 2:
Without replacement, there are either zero or 1 black ball remaining of the last three balls:
(6/7)*(0/3) + (1/7)*(1/3) = 1/21.
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Posted by Charlie
on 2010-02-14 14:13:00 |
solution
