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Playing With Marbles II (Posted on 2014-08-29) Difficulty: 3 of 5
Contents of three identical looking bags are as follows:

Bag 1: Five black marbles, three red marbles and one white marble.
Bag 2: Five red marbles, three white marbles and one black marble.
Bag 3: Five white marbles, three black marbles and one red marble.

You choose one of the three identical looking bags at random.
After choosing a bag you draw two marbles out at random. You notice that the first one is black, and the second one is white.

You then put the two marbles back in the bag they came from and draw two marbles out of the same bag at random.

What is the probability that the first marble drawn is black and the second marble drawn is white?

No Solution Yet Submitted by K Sengupta    
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Solution solution | Comment 1 of 2
If you chose the first bag, the probability that you'd observe what you did would have been 5/9 * 1/8 = 5/72. The probability for getting these in either order would be twice as large, but that applies to all these probabilities, so it doesn't matter when we apply Bayes' rule below.

If you chose the second bag, the probability that you'd observe what you did would have been 1/9 * 3/8 = 3/72.

If you chose the first bag, the probability that you'd observe what you did would have been 3/9 * 5/8 = 15/72.




The probability that the bag chosen was indeed bag 1, by Bayes' rule, is ((5/72) / 3) / ((5/72) / 3 + (3/72) / 3 + (15/72) / 3)

The probability that the bag chosen was indeed bag 2, by Bayes' rule, is ((3/72) / 3) / ((5/72) / 3 + (3/72) / 3 + (15/72) / 3)

The probability that the bag chosen was indeed bag 1, by Bayes' rule, is ((15/72) / 3) / ((5/72) / 3 + (3/72) / 3 + (15/72) / 3)

These come out to 5/23, 3/23 and 15/23 respectively.



Then the probability that you'd observe that black-white sequence again would be 5/23 * 5/72 + 3/23 * 3/72 + 15/23 * 15/72 = 259/1656 ~=  .156400966183575.




  Posted by Charlie on 2014-08-29 11:39:58
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