4/9
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To solve this problem, we need to consider the two scenarios in which bag A ends up with exactly 7 red beads after the process:
- Scenario 1: A red bead is taken from bag A and transferred to bag B, and then a red bead is taken from bag B and transferred back to bag A.
- Scenario 2: A yellow bead is taken from bag A and transferred to bag B, and then a yellow bead is taken from bag B and transferred back to bag A.
We need to calculate the probability of each scenario happening and then add these probabilities together to get the total probability that bag A still contains exactly 7 red beads.
Let's calculate the probability of each scenario:
Scenario 1: Red bead taken from A, then red bead taken from B
Probability of taking a red bead from bag A: Since there are 7 red beads and 2 yellow beads, the probability is7/9.
After transferring a red bead to bag B, bag B now has 2 red beads and 4 yellow beads. The probability of taking a red bead from bag B is 2/6 = 1/3.
The probability of this scenario, where a red bead is returned to bag A after one was taken out, is the product of the two probabilities:
(7/9)*(1/3) = 7/27
Scenario 2: Yellow bead taken from A, then yellow bead taken from B
Probability of taking a yellow bead from bag A, since there are 7 red beads and 2 yellow beads, the probability is 2/9.
After transferring a yellow bead to bag B, bag B now has 1 red bead and 5 yellow beads. The probability of taking a yellow bead from bag B is 5/6.
The probability of this scenario, where a yellow bead is returned to bag A after one was taken out, is the product of the two probabilities:
(2/9)*(5/6) = 5/27
Finally, we add the probabilities of the two scenarios together to get the total probability that bag A still contains exactly 7 red beads:
Total Probability = 7/27 + 5/27 = 12/27 = 4/9
Thus, the probability that bag A still contains exactly 7 red beads after the process is 4/9.
