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There are three bags P, Q, and R. Bag P contains 1 red and 2 green balls; bag Q contains 2 red and 1 green ball; bag R contains only 1 green ball. One ball is drawn from P and placed in Q; then one ball is drawn from Q and placed in R; finally one ball is drawn from R and placed back in P. After this sequence of operations, what is the probability that bag P contains 2 red and 1 green ball?

1. 1/4
2. 1/2
3. 1/3
4. 1/6

Correct answer: 1/6

Solution

Initial: P=1R2G, Q=2R1G, R=1G. Track the only path where P ends as 2R1G: (1) Draw green from P (p=2/3): P becomes 1R1G; Q becomes 2R2G. (2) Draw red from Q (p=1/2): Q becomes 1R2G; R becomes 1R1G. (3) Draw red from R (p=1/2): R becomes 1G; P gets red => P becomes 2R1G. Total probability = (2/3)*(1/2)*(1/2) = 1/6.

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