Each of the three urns contains six balls with an urn containing five white balls and one black ball, while another urn contains four white and two black balls and, the remaining urn contains three white and three black balls.
One white ball is drawn at random from one of the three urns and, one black ball is drawn without replacement from one of the remaining two urns. (The balls are drawn blindly, without knowing which the identities are of the urns from which they are drawn.)
Determine the probability of drawing a white ball from the remaining urn.
There are 6 possible orders of urns from which the balls can have been sampled. The probabilities of drawing first a white, then a black, given each of the 6 orders, are:
W B
1,2,3: (5/6)(1/3)
1,3,2: (5/6)(1/2)
2,1,3: (2/3)(1/6)
2,3,1: (2/3)(1/2)
3,1,2: (1/2)(1/6)
3,2,1: (1/2)(1/3)
Adding these up and dividing by 6 gives an overall a priori probability that the observed conditions would exist as 25/108.
The probabilities of drawing in sequence white, black, white are:
W B W
1,2,3: (5/6)(1/3)(1/2)
1,3,2: (5/6)(1/2)(2/3)
2,1,3: (2/3)(1/6)(1/2)
2,3,1: (2/3)(1/2)(5/6)
3,1,2: (1/2)(1/6)(2/3)
3,2,1: (1/2)(1/3)(5/6)
These average 17/108, and again, the average is the overall probability as each order is equally likely a priori.
Thus the sought probability, that the last ball drawn would be white, is 17/25 -- the ratio that the observed conditions would happen and the desired outcome would also happen, divided by the a priori probability that the observed conditions would happen at all (Bayes' rule).
list
10 A=(5//6)*(1//3)
11 B=(5//6)*(1//2)
12 C=(2//3)*(1//6)
13 D=(2//3)*(1//2)
14 E=(1//2)*(1//6)
15 F=(1//2)*(1//3)
20 print (A+B+C+D+E+F)//6
30
41 A=(5//6)*(1//3)*(1//2)
42 B=(5//6)*(1//2)*(2//3)
43 C=(2//3)*(1//6)*(1//2)
44 D=(2//3)*(1//2)*(5//6)
45 E=(1//2)*(1//6)*(2//3)
46 F=(1//2)*(1//3)*(5//6)
50 print (A+B+C+D+E+F)//6
OK
run
25//108
17//108
OK
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Posted by Charlie
on 2009-12-23 14:49:27 |
solution
