Q:

distributing n different balls into r numbered cells

We distribute 5 different balls into 3 numbered cells
All the different distributions have equal probability

I am trying to find all the possible probabilities of empty cells but I get some weird results
[the point in this exercise is to find Var(Y)]:
Let Y be the number of the empty cells.
$P(Y=0)=1-P(Y=1)-P(Y=2)-P(Y=3)$
$P(Y=1)=3*\frac{2^5}{3^5}$
$P(Y=2)=3*\frac{2^5}{3^5}$    [this part is probably wrong]
$P(Y=3)=0$
My problem is that the chance that all the cells are empty must be 0 I think because we distribute the balls into the cells so at least 1 cell got a ball
But I get weird results I need to calculate the variance of Y but with these chances I get that
$Var(Y)=0.722$
This is wrong.

If I look at this problem like a binomial distribution problem:
Lets remember that the chance of a given cell to be empty is $(\frac{2}{3})^5=0.1316$
$Y$~$B(3, (\frac{2}{3})^5)$
So $Var(Y)=np(1-p)=0.34$
This is wrong too AND it is different from the first part which is even worse :(
Any help appreciated :)

A:

Let $Y_{i}$ take value $1$ if cell $i$ stays empty and let it take
value $0$ otherwise. 
Then $Y=Y_{1}+Y_{2}+Y_{3}$. 
The $Y_{i}$ have
equal distribution but are not independent. 
With linearity of expectation
and symmetry we find:
$\mathbb{E}Y=3\mathbb{E}Y_{1}=3\mathbb{P}\left(Y_{1}=1\right)$
$\mathbb{E}Y^{2}=3\mathbb{E}Y_{1}^{2}+6\mathbb{E}Y_{1}Y_{2}=3\mathbb{P}\left(Y_{1}=1\right)+6\mathbb{P}\left(Y_{1}=1=Y_{2}\right)$
Can you take it from here?

