Basic probability question, confusion with basic concept There is a bag with $95$ green and $5$ red balls. If $3$ balls are drawn, what is the probability that $2$ of them are red and $1$ green?
Solution 1: 
Total number of ways is $100\choose3$
Favorable number of ways = $\binom{5} {2} * \binom{95}{1}$
So required probability = $\frac{\binom{5}{2} \binom{95}{1}}{\binom{100}{3}}$
Solution 2:
   However since the favorable number of ways is obtained by multiplying $5\choose2$ and $95\choose1$, it takes the order between two red and one green in to account while the total number of ways does not take order in to account. 
Should I therefore divide by $2$ to get the accurate favorable number of ways?
so required probability will be $\frac{\binom{5}{2} \binom{95}{1}}{\binom{100}{3}*2}$
Solution 3:
Total number of ways = $\binom{100}{3} * 3!$ (counting in way to account for order)
Favorable number of ways = $\left(\binom{5}{2} * 2!\right) * \left(\binom{95}{1} * 1!\right)$ (to also account for order like in denominator)
So required probability = $\frac{\binom{5}{2} * \binom{95}{1}}{\binom{100}{3} *3}$)
I understand that as long as we are consistent in taking in to account the order when counting favorable and total ways, we should get the same probability.
However I am finding it hard to reconcile solution 1 vs solution 2 vs solution 3.
 A: Solution $1$ is correct.
Your statement under solution $2$ that multiplying $\binom{5}{2}$ by $\binom{95}{1}$ "takes order into account" is erroneous.  I don't understand why you say this.
In solution $3$, you've taken order into account properly in the denominator, but your attempt to do so in the numerator is flawed.  In order to take order into account, we say there are $3$ possible orders, RRG, RGR, GRR.  Each of these can occur in $5\cdot4\cdot95$ ways, so we get $$\frac{3\cdot5\cdot4\cdot95}{100\cdot99\cdot98}$$ for the probability, the same answer as in $1$.
The way you did it in solution $3$, you took the order in which the red balls are drawn into account, but not the order of the green ball in relation to the red balls.
A: ${^{100}\mathrm C_{3}}$ counts ways to select three from one hundred balls.   This does not count ways to arrange those selected items.
${^{5}\mathrm C_{2}}\cdot{^{95}\mathrm C_{1}}$ counts ways to select two from five red balls and one from ninety-five green balls.   Likewise this does not count ways to arrange those selected items.
Therefore the probability you seek is: $$\dfrac{{^{5}\mathrm C_{2}}\cdot{^{95}\mathrm C_{1}}}{{^{100}\mathrm C_{3}}}$$

Alternatively, consider the task as lining up the one hundred balls and selecting the first three.   We shall then count ways to select places for the red balls.
${^{3}\mathrm C_{2}}\cdot{^{97}\mathrm C_{1}}$ counts ways to select two among the first three places and one among the latter ninety-seven places, while ${^{100}\mathrm C_{5}}$ counts ways to select five among the one-hundred places . 
Thus the probability we seek is: $$\dfrac{{^{3}\mathrm C_{2}}\cdot{^{97}\mathrm C_{3}}}{{^{100}\mathrm C_{5}}}~~=~~\dfrac{\tfrac{3!}{2!~1!}\cdot\tfrac{97!}{3!~94!}}{\tfrac{100!}{5!~95!}}~~=~~\dfrac{\tfrac{5!}{2!~3!}\cdot\tfrac{95!}{1!~94!}}{\tfrac{100!}{3!~97!}}~~=~~\dfrac{{^{5}\mathrm C_{2}}\cdot{^{95}\mathrm C_{1}}}{{^{100}\mathrm C_{3}}}$$
