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Each of Alice and Bob has an identical bag containing 6 balls numbered 1, 2, 3, 4, 5, and 6. Alice randomly selects one ball from her bag and places it in Bob’s bag, then Bob randomly selects one ball from his bag and places it in Alice’s bag. What is the probability that after this process the content in two bags remains unchanged?

I would have thought that the probability would be $\frac{1}{6} + \frac{2}{7}$ as Alice picks 1 out of balls, now Bob has 7 balls and 2 contain the same number. Or should I have multiplied them in this scenario? The balls are indistinguishable.

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    $\begingroup$ The balls should be distinguishable because they are numbered. I think you mean balls with the same number are indistinguishable. Anyway, think about the $1/6$ for Alice. Think about what that probability represents. It is the probability of her choosing a specific ball. Does it matter which ball she puts in Bob's bag? $\endgroup$ – InterstellarProbe Jul 31 at 13:25
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    $\begingroup$ Yes, that is what I mean. It doesn't really matter as long Bob puts the same number back. No matter what ball she puts into his bag, Bob will have 2/7 chance of selecting that ball. $\endgroup$ – Libor Zachoval Jul 31 at 13:28
  • $\begingroup$ That is the correct answer :) $\endgroup$ – InterstellarProbe Jul 31 at 13:28
  • $\begingroup$ I see. That makes sense. $\endgroup$ – Libor Zachoval Jul 31 at 13:29
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    $\begingroup$ Anyway, you basically had your own problem answered. You should post the answer and mark it as correct. $\endgroup$ – InterstellarProbe Jul 31 at 13:30
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After the action of Alice there are exactly $7$ balls in the bag of Bob, and of exactly $2$ of them it is true that placing it back in the bag of Alice will restore the original situation.

So the probability of this event is $\frac27$.

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The choice of Alice's ball does not matter and so the probability that the content of two bags remains unchanged is $2/7$.

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