Joint distribution weird result 
We have 3x3 board   Each board cell has $0.5$ chance to be white
  (there is no dependency between different cell colors)  
Let X = number of white rows (a row with only white cells on it)
   Let
  N = total number of white cells in the whole board
Find $P\{X=0,N=3\}$
  Find $P\{X=0 |N=5\}$

What I tried to do is to calculate the $P\{X=0,N=3\}$ , I said that:
$P\{X=0,N=3\}=P(N=3)*P(X=0|N=3)$ is it true?
Because now I got that it equals to $\frac{21}{128}*\frac{1}{168}=\frac{1}{1024}$
Which just seems way too low and also I'm not sure if I would change my answer if the question was to find $P\{X=0|N=3\}$ instead of $P\{X=0,N=3\}$ which seems odd to me

First note that
$$P(X=0,N=3)=P(X=0\,|\,N=3) P(N=3)$$
We will find $P(X=0 \,|\,N=3)$ then $P(N=3)$ separately, then multiply them together.

Part 1: $P(X=0\,|\,N=3)$
With three white squares, there can be at most one white row. This means that if $N=3$, then $X=0$ or $X=1$ are the only possibilities. Therefore
$$P(X=0\,|\,N=3)=1-P(X=1 \,|\,N=3)$$
The total number of arrangements of the $9$ squares in the grid is $$\binom{9}{3}=\frac{9\cdot 8 \cdot 7}{3 \cdot 2} = 84$$
There are $3$ arrangements such that there is a white row.  That means
$$P(X=0\,|\,N=3)=1-\frac{3}{84} = 1-\frac{1}{28} = \frac{27}{28}$$

Part 2: $P(N=3)$
There are $\binom{9}{3}$ arrangements in which there are three white squares. Each of these arrangements has a $(1/2)^3(1/2)^6$ probability of occuring. Therefore
$$P(N=3)=\binom{9}{3}\left(\frac{1}{2}\right)^9=\frac{21\cdot 2^2}{2^9}=\frac{21}{128}$$

Our final answer is therefore
$$P(X=0,N=3)=\left(\frac{27}{28}\right)\left(\frac{21}{128}\right)=\boxed{\frac{81}{512}}$$
