All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Black and Blue Transaction (Posted on 2010-02-14)
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.

 No Solution Yet Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 solution | Comment 1 of 2

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       1/331       8/332       5/113       8/334       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.

 Posted by Charlie on 2010-02-14 14:13:00

 Search: Search body:
Forums (0)