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so I have this question right here:

An urn has $12$ balls. $4$ in a row each pick a ball at random from the urn. After each pick and before the next pick, the ball is returned to the urn. What is the probability that all the balls picked are different?

So the way I thought about this is as follows:

We have $12$ balls, and we pick a ball and put it back with the other balls $4$ times in a row. So after each pick, before the next one, the ball that was picked is returned. So that means there's a chance that the ball that was picked may be picked again.

So first there are $12^4$ possibilities in the sample space. And then we use the formula $\frac{n!}{(n-r)!}$ where $n = 12$ and $r = 8$. So does that mean the answer is $\dfrac{\frac{12!}{8!}}{12^4}$?

If this is incorrect, I would love to have someone explain the process to me because I am very new to probability and counting.

Thanks

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  • $\begingroup$ Do you mind explaining why it is the case that it's correct? I kind of just randomly used the formula from class but I would appreciate a short explanation if possible. $\endgroup$
    – Stuy
    Jan 31, 2018 at 21:16
  • $\begingroup$ @Stuy As I said (presciently!) in the final paragraph of my wrong answer, it's right and proper to be worried about whether and why you've got this kind of question right. Counting is not trivial. $\endgroup$ Jan 31, 2018 at 21:19
  • $\begingroup$ @PatrickStevens so is my solution correct then? $\endgroup$
    – Stuy
    Jan 31, 2018 at 21:21
  • $\begingroup$ You might look up the Birthday Problem, if you haven't seen it before. It is exactly the same as this, combinatorially, although it is usually phrased as finding the probability that at least two are the same (the complement of your problem here). $\endgroup$
    – Ned
    Jan 31, 2018 at 22:59

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Your answer is correct. However, you made an error in your explanation. You meant $r = 4$. That gives you $$P(12, 4) = 12 \cdot 11 \cdot 10 \cdot 9 = \frac{12!}{(12 - 4)!} = \frac{12!}{8!}$$ favorable cases, corresponding to the number of ways of selecting four different objects from $12$ objects when the order of selection matters.

Here is another approach that justifies your correct answer: The first ball that is selected is guaranteed to be distinct. Since there are $12$ balls and the selected ball is replaced after each draw, the probability that the next selected ball is distinct from the first is $11/12$. The probability that the third selected ball is different from the first two is $10/12$. The probability that the fourth selected ball is different from the first three is $9/12$. Hence, the desired probability is $$1 \cdot \frac{11}{12} \cdot \frac{10}{12} \cdot \frac{9}{12} = \frac{12 \cdot 11 \cdot 10 \cdot 9}{12^4} = \frac{\frac{12!}{8!}}{12^4}$$

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    $\begingroup$ Since we're only picking 4 balls, that's why we start off with the first one is guaranteed to be distinct. But since we place the ball pack before the next pick, that means the 2nd one that's picked only has 11/12 of a chance to be distinct because the probability of the same one being picked is 1/12? So since we picked 1 ball, returned it to set of balls, we pick the 2nd one, and then the third, and when we pick the 4th, it has a 9/12 probability of it being distinct? $\endgroup$
    – Stuy
    Jan 31, 2018 at 21:28
  • $\begingroup$ Another thing I want to ask is, how do we know if order matters or not? $\endgroup$
    – Stuy
    Jan 31, 2018 at 21:32
  • $\begingroup$ You have correctly understood my argument. In this case, the order matters since we must compare the balls we select with the previously selected balls. $\endgroup$ Jan 31, 2018 at 21:45
  • $\begingroup$ Note also that each possible outcome is a sequence such as blue, green, blue, red, so order matters. $\endgroup$ Jan 31, 2018 at 22:13

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