How is Poisson Distribution simply discerned? How is it related to the Binomial distribution? There is this question which I thought I had understood, until taking a look at the answers:
Let a floor tile be composed of different four tiles: a black one of size $1\times1$, a red $3\times 3$ one, and two yellow ones: $1\times 3$ and $3\times 1$. Let a big floor be composed of these floor tiles. Stones are being scattered one by one and independently over the tiles. The number of stones scattered is the number of colored tiles, that is $4n$, if the floor is made of $n$ floor tiles. No stone prevents other stones from being in the tile it is in. How does the number($X$) of stones in a specific black tile distribute? 
(There is a photo near the question that shows it. Does it matter? If it is, I will provide that if it were to be a matrix, the black would be $a_{11}$, the yellow ones would be $a_{12},a_{13},a_{14}$, and $a_{21},a_{31},a_{41}$, and the red one would be the rest, a $3\times 3$ sub-matrix.)
What I thought is that the distribution works that way: $P(X=k)={4n\choose k}({1\over 16})^k({12\over 16})^{n-k}$. I was, as it turns out, utterly wrong. The answers say: $X\sim Bin(4n,{1\over 9n})$ (where did this $9$ came from?) which is approximately $Pois({4n\over 9n})=Pois({4\over 9})$. I would really appreciate it if you could help me understand why I didn't arrive at ths aforementioned binomial distribution, and how it ever relates the Poisson distribution. 
 A: You seem a little confused about terminology in general and the binomial
and Poisson distributions in particular. The cure for that is 
to go back in your text or lecture notes to where these ideas
are explained. (You will rarely find an exact formula that
is 'tailor made' to answer a particular probability problem
so that you can plug into the formula without understanding it.)
In my view the Comments here have been scattered, somewhat above your level, and unhelpful, Also, there may be some confusion between
what I would call 'large tiles' and the nine 'small squares' within
each large tile.
so I will write some things that I hope will guide you as you go back and study these topics.
Each of $4n$ stones has probability $1/9n$ of falling on a particular little black square. The fraction of each large tile that is black is $\frac{1}{9},$
and there are $n$ of the large tiles. So the exact distribution
of the number of stones on a particular small black square is $X \sim Binom(4n, 1/9n)$. as stated.
A binomial distribution $Binom(m, p)$ with $m$ independent trials, and $P(\text{Success}) = p\,$ on each trial, has mean $\mu = mp.$ In your
case $m = 4n$ and $p = 1/9n$, so $\mu = mp = 4n/9n = 4/9.$
If $m$ is large and $p$ is not too near 1, a Binomial distribution
can be approximated by a Poisson distribution with a mean (often
designated as $\lambda$, 'lambda') matching the binomial mean.
That is why your answer says $Pois(4/9)$ is a reasonable
approximation. (Very roughly speaking, the approximation depends
on not having many small black squares with double hits.)
An illustration: Suppose $n = 100,$ so that there are $m = 400$ stones thrown each with probability $p = 1/900$ of hitting a particular small black square. Then there will be on average
$\mu = 4/9$ hits per small black square. (Maybe think of that
as an average of 4 hits for a particular 9 small black squares.)
Then the binomial distribution says $P(X = 0) = 0.6410$ and $P(X = 1) = 0.2852$. By contrast, the Poisson approximation says
$P(X = 0) = 0.6412$ and $P(X = 1) = 0.2850.$ I imagine you
will be able to find binomial and Poisson formulas in your book
to verify these four numbers. Can you find the binomial and Poisson values for $P(X = 3)?$
