I recently ran into problems understanding when to use which combinatorical counting principle concerning "balls and bins" problems. I am not 100% (maybe 80%) sure if I have understood it correctly, so I will try to summarize my thoughts on that. I would really appreciate it if someone can confirm that I have understood it properly. I know there are already dozens of balls and bins questions but I haven't found any which explicitly adresses my issue. Maybe those questions are closest but do not fully answer my problem:
Equal probabilities in each count in stars and bars
Sorting undistinguishable balls - $l$-many in one slot - check my proof (was asked by me and made me aware that I had some misconceptions of counting principles and probabilities)
Let's say we have $5$ balls and $3$ numbered bins. We now look at different situations where we want to calculate the probability of a certain outcome.
1.) The balls are indistinguishable and thrown successively at the bins. It is equally probable to hit any bin. What is the probability of the two events
$A= (bin_1: 1~ ball,bin_2: 2~ balls,bin_3: 2~balls)$ and $B= (bin_1: 5~ balls,bin_2: 0~ balls,bin_3: 0~balls)$ ?
My answer: The sample space is $\Omega =\{(\omega_1, \omega_2,...,\omega_5)| \omega_i$ is the bin of the $i$-th throw $\}$. So the cardinality of the corresponding sample space $\Omega$ is: $3^5$. Hence, $P(B) = \frac{1}{3^5}$. I am not quite sure what $P(A)$ is, but it must be definitely greater than $P(B)$!
2.) The balls are indistinguishable and will simultanously appear in any bin. It's like pouring balls over the bins and every bin is equally probable for each ball. Again, what is the probability of the two events
$A= (bin_1: 1~ ball,bin_2: 2~ balls,bin_3: 2~balls)$ and $B= (bin_1: 5~ balls,bin_2: 0~ balls,bin_3: 0~balls)$?
My answer: Due to the specific setting we must change the sample space $\Omega$. It becomes: $\Omega =\{(\omega_1, \omega_2,\omega_3)| \omega_i$ is the number of balls in the $i$-th bin$\}$. The cardinality of $\Omega$ is: ${3+5-1 \choose 5}$ and the probabilities of the two events are $P(A)= P(B)=\frac{1}{{3+5-1 \choose 5}}$.
3.)Now the balls are distinguishable and thrown successively at the bins. It is equally probable to hit any bin. What is the probability of the two events $A$ and $B$?
My answer: the same as in 1.)
What do you think of my answers?