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
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 $$1 \cdot \frac{11}{12} \cdot \frac{10}{12} \cdot \frac{9}{12} = \frac{12 \cdot 11 \cdot 10 \cdot 9}{12^4} = \frac{\frac{12!}{8!}}{12^4}$$