$4$ balls are randomly distributed into $3$ cells. What is the probability that there is a cell that contains exactly $2$ balls? $4$ balls are randomly distributed into $3$ cells ($3^4=81$ possibilities of equal probability).
What is the probability that there is a cell that contains exactly $2$ balls?
The correct answer is: $\frac{2}{3}$, but i know don't where was i mistaken.
Here was my idea:
Let's define: $\forall _{i=1,2,3}:A_i$ = The event that cell #$i$ contains exactly $2$ balls.
Then, according to the Inclusion–exclusion principle, the answer should be:
$P_{solution} = P(A_1 \cup A_2 \cup A_3) = P(A_1)+P(A_2)+P(A_3)-P(A_1\cap A_2)-P(A_1\cap A_3)-P(A_2\cap A_3)+P(A_1\cap A_2 \cap A_3)$
Where:
$$
\forall _{i=1,2,3}: \quad P(A_i)=\frac{{4 \choose 2}*2^2}{3^4}=\frac{8}{27}
$$
$$
\forall_{i \neq j}: \quad P(A_i \cap A_j)= \frac{{4 \choose 2}*2}{3^4}=\frac{4}{27}
$$
$$
P(A_1\cap A_2 \cap A_3)=0
$$
and so:
$$P_{solution}=3*\frac{8}{27}-3*\frac{4}{27} = \frac{4}{9}$$
I can see that IF my calculation of $P(A_i \cap A_j)$ was $\frac{{4 \choose 2}}{3^4}$ (without multiplying by $2$), then that would be correct, but i can't seem not to wonder why. I have to multiply by $2$. Suppose we look at cell #1 and cell #2: I need to choose $2$ balls out of $4$, that's $4 \choose 2$. Let's say i chose the the balls $\{1,3\}$ and $\{2,4\}$, then i must decide which cell will get the $\{1,3\}$ set and which will the $\{2,4\}$. That's $2$ options, so we multiply by $2$.
Any idea? Where was i mistaken? Can you show me your solutions?
 A: Choosing $\{1,3\}$ and bin $2$ is equivalent to choosing $\{2,4\}$ and choosing bin $1$.
Perhaps a cleaner way of finding numerator for $\Pr(A_1\cap A_2)$ is by treating ball1 as special.  Choose which bin ball1 goes into: $2$ choices.  Choose which other ball goes with ball1: $3$ choices.  The remaining balls go into the other bin.  This gives a numerator of $2\times 3 = 6$ as opposed to your $12$ you had in your attempt.
For an alternate approach to verify the answer of $\frac{2}{3}$, let us see which outcomes were "bad."  By pigeonhole principle every arrangement will have at least one bin with at least two balls in it.  We see then that the only bad outcomes are those where we have a bin with exactly three balls in it or a bin with exactly four.
For exactly three, pick which ball was not a part of the three.  Pick which bin it goes into.  Then, pick which bin the three go into.  For exactly four, pick which bin they all go into.  $\frac{4\cdot 3\cdot 2}{3^4}+\frac{3}{3^4} = \frac{1}{3}$ and so the probability we were originally after is $1-\frac{1}{3}=\frac{2}{3}$
A: In the approach you take, it should be,
$P =  \displaystyle 3 \cdot {4 \choose 2} \cdot \frac{2^2}{3^4} - 3 \cdot {4 \choose 2} \cdot \frac{1}{3^4}$
Please note that $ \displaystyle P(A_i \cap A_j) = {4 \choose 2} \cdot \frac{1}{3^4}$
Explanation: Once you choose two of the three cells for $2$ balls each, there are ${4 \choose 2}$ ways of choosing balls for the first of the selected two cells and the remaining two go to the second cell. In other words, they are already ordered and you should not multiply by $2$.
However for this question, instead of Principle of Inclusion Exclusion, you can choose direct counting too as there are only two cases.
