$4$ balls are randomly distributed into $3$ cells. the probability that there is a cell that contains exactly $2$ balls

probabilityprobability distributions

$4$ 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?

Best Answer

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.

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