Intuition for the formula of Binomial Distribution Let's take an example, let's say that 8% of men have Color Blindness, what is the probability that 2 men in a sample of 10 men have Color Blindness?
My question is:
It makes sense to apply Multiplication Theorem for say the probability of getting 2 Heads if we toss a coin twice. First we toss, 50% of the people get Heads, Then of that new sample space(50% of the original sample space) 50% of the people get Heads, so in total, 50% of 50% of Original Sample Space gets 2 heads, that is, 1/2 of 1/2. But how do we apply it here?
I can't understand, 92% of the given sample doesn't have Color Blindness, 92% of that 92% don't, and so on, until 92% of 92% of 92% of 92% of 92% of 92% of 92% of 92% of 8% of 8% has color Blindness, and why add 10 choose 2 in the front?
It's a bit confusing as to how we start with people who don't have Color Blindness (92% of Sample Space) and end with people with color Blindness by multiplying 92% 8 times then 8% 2 times.
 A: Adam, Bob, Charlie ... Jason are your sample.  Adam and Bob could be the two with colorblindness, or Adam and Charlie, or Charlie and Fred, or etc.
The number of pairs of men you could pick is the binomial coefficient.
Each particular combination of two men having colorblindness has the same probability: $p^2(1-p)^{10-2}$ -- Two with, eight without.
So you have to add together the $\binom{10}{2}$ different ways, each with individual probability $p^2(1-p)^{10-2}$ to get the overall probability $\binom{10}{2} p^2(1-p)^{10-2}$
In Gory Detail
First consider $N=1$.  You take a random person off the street, call him Bob, and assess whether he has colorblindness ("cb") or not ("non").  With probability $p$ he is, with probability $1-p$ he is isn't . Note that these probabilities add to $1$.
Later, you do an experiment with $N=2$.  You grab two random people.  There are 4 potential outcomes: non&non, non&cb, cb&non, cb&cb.  These have probabilities $(1-p)^2$, $(1-p)p$, $p(1-p)$, $p^2$.  These sum to 1 (exercise for the reader).  Note that two combinations, cb&non and non&cb, correspond to "1 out the two people sampled are colorblind".
For $N=3$ there are 8 states: non&non&non, non&non&cb, non&cb&non, non&cb&cb, cb&non&non, cb&non&cb, cb&cb&non, cb&cb&cb.  Each has a probability of the form $p^m(1-p)^{N-m}$.  They sum to 1.  Note that there are $3$ states corresponding to the statement $2$ out of the three people sampled are colorblind".
And so on...
A: It's a case of Bernoulli's trials , and is very similar to the case of successive throws of coins , say $n$ coins , and we can find the probability of , say $'r'$ heads in $'n'$ successive throws by finding the coefficient of $h^r$ in $(h+t)^n$ , where $h$ and $t$ are probabilty of heads and tails in each throw respectively , and $h+t=1$.
In your example , let's denote $'p'$ as probabilty of having color blindeness , and $'q'$ denoting not having it, and $p+q=1$ . Therefore , $p=\frac{8}{100}$ and $q=\frac{92}{100}$.
We have  to choose from a total of $10$ men , therefore $n=10$.
Therefore , the probability of $2$ men having color blindeness among $10$ men ,is equal to coefficient of $p^2$ in the expansion of $(q+p)^{10}$ , which is given by $${10 \choose 2}{q^{10-2}}{p^2}={10 \choose 2}{q^{8}}{p^2}$$, where $p=0.08$ , and $q=0.92$.
