Please help on this Probability problem 
A bag contains 5 red marbles and 7 green marbles. Two marbles are drawn randomly one at a time, and without replacement. Find the probability of picking a red and a green, without order.

This is how I attempted the question: I first go $P(\text{Red})= 5/12$ and $P(\text{Green})= 7/11$ and multiplied the two: $$\frac{7}{11}\times \frac{5}{12}= \frac{35}{132}$$
Then I got $P(\text{Green})= 7/12$ and $P(\text{Red})= 5/11$ $\implies$
$$\frac{5}{11} × \frac{7}{12}= \frac{35}{132}$$
So I decided that $$P(\text{G and R}) \;\text{ or }\; P(\text{R and G}) =\frac{35}{132} + \frac{35}{132} =\frac{35}{66}$$ Is this correct? 
 A: Very nice and successful attempt. You recognized that there are two ways once can draw a red and green marble, given two draws: Red then Green, or Green then Red. You took into account that the marbles are not replaced. And your computations are correct: you multiplied when you needed to multiply and added when you needed to add:
$$\left[P(\text{1. Red}) \times  P(\text{2. Green})\right]+ \left[P(\text{1. Green}) \times P(\text{2. Red})\right]$$
Your method and result are correct.
A: Your method is correct. For more complicated problems of the same general kind, one might take a slightly different approach.
Imagine that the $12$ marbles are distinct, they have different driver license numbers. There are $\binom{12}{2}$ equally likely ways to choose $2$ marbles from the $12$.  
There are $\binom{5}{1}$ ways to choose a red marble, and $\binom{7}{1}$ ways to choose a green marble. Thus there are $\binom{5}{1}\binom{7}{1}$ ways to choose a red and a green. It follows that our probability is 
$$\frac{\binom{5}{1}\binom{7}{1}}{\binom{12}{2}}.$$
Harder! However, suppose we have $25$ marbles, $10$ red and $15$ green. We choose (without replacement) $8$ marbles. What is the probability that we get $3$ red and $5$ green? The same analysis shows that the probability is 
$$\frac{\binom{10}{3}\binom{15}{5}}{\binom{25}{8}}.$$
