A bag contains 4 Black and 3 Red balls, 2 balls are drawn one by one without replacement, what is the probability of both being red? I  tried searching the internet regarding the third approach i did, but did not find any explanation for the same, here i have tried approaching this problem in 3 different ways.
approach 1:
since two balls are drawn without replacement, this can be treated as 
$P (\text{selecting $2$ out of $7$ balls such that both are red})$
$=\frac{^3C_2}{^{7}C_2}$ 
approach 2:
$P (\text{select 1 R out of 3R })\times$ $P (\text{select 1 out of 2 remaining R })$
$=\frac{^3C_1}{7C_1}\times$$\frac{^2C_1}{6C_1}$ 
approach 3:
$P (\text{selecting 1 out of 3 R and then 1 out of 2 R balls in the second draw })$
$=\frac{N(RR)}{{N(RR)+N(RB)+N(BR)+N(BB)}}$
$=\frac{^3C_1\times^2C_1}{^3C_1\times^2C_1+^3C_1\times^4C_1+^4C_1\times^3C_1+^4C_1\times^3C_1}$
all the above approaches give correct answer, but i am not sure how approach 1 and approach 2 are equivalent,
clearly this is a case of conditional probability without replacement, in the first case, i take 2 balls simultaneously out of 7 and compute probability of both being red, that is same as drawing 2 balls without replacement! 
is this the correct way to approach a problem?
in the approach 3 , i tried calculating the sample space where 2 balls are drawn without replacement, where there are four cases:


*

*Red in first and Red in second draw

*Red in first and Black in second draw

*Black in first and Red in second draw

*Black in first and Black in second draw


I am quite unsure about the validity of approach 3!
I am not sure if this is the right way to approach a problem.
 A: All $3$ approaches are okay.
Number the red balls by $1,2,3$ and number the black balls by $4,5,6,7$
Then you could say that approach1 is linked with a probability space where: $$\Omega=\{\{i,j\}\mid i,j\in\{1,2,3,4,5,6,7\}\text{ and }i\neq j\}$$
Here $|\Omega|=\binom72=21$ agreeing with the denominator in approach1.
Also you could say that approach3 is linked with a probability space where: $$\Omega=\{\langle i,j\rangle\}\mid i,j\in\{1,2,3,4,5,6,7\}\text{ and }i\neq j\}$$
Here $|\Omega|=7\times6=42$ agreeing with the denominator in approach3.
Concerning approach2 (my favorite) both spaces can be used but actually we do not meet any impact of a choice for a probability space. 
If $E_i$ denotes the event that the $i$-th ball is red for $i=1,2$ then we apply:$$P(\text{both balls are red})=P(E_1\cap E_2)=P(E_1)P(E_2\mid E_1)=\frac37\frac26$$
Calculating such probability we are not even aware of any underlying probability space.
A: Let $R$ represent red balls and $B$ black balls
$P(R)=3/7$
$P(B)=4/7$
Therefore $P(RR) = \frac37 \cdot \frac26$
Ans = $6/42$
