A couple in a dark room pulls 2 socks each from a shared drawer. What's the probablity of them getting exactly one & two pairs without trading? A couple in a dark room pulls 2 socks each from a shared drawer. The drawer contains 4 blue, 7 red, and 3 yellow socks. The couple cannot trade socks. What is the probability of exactly one pair (of the same colour) being drawn? And what is the probability of both of them each getting a pair?
Similarly to this question I know that the probability of drawing a pair when only one person is drawing 2 socks from this drawer is
$$\frac{\binom{4}{2}}{\binom{14}{2}} + \frac{\binom{7}{2}}{\binom{14}{2}} + \frac{\binom{3}{2}}{\binom{14}{2}} = \frac{6}{91} + \frac{21}{91} + \frac{3}{91} = \frac{30}{91}$$
but I'm completely lost when it comes to drawing two pairs at once.
Additionally I know that the total amount of ways to draw these socks is $\binom{14}{2} \times \binom{12}{2} = 91\times66=6006$ since 2 socks are first drawn from 14 and then 2 socks are drawn from the remaining 12. This also means that the first person getting a pair and the second person getting any other 2 socks happen in $30\times\binom{12}{2}=1980$ out of the $6006$ ways (which is the same probability as $\frac{30}{91}$). But I'm not sure if this is useful for solving the problem.
I really need some insight for how to approach this problem as it's not the same thing as getting two pairs when drawing 4 or more socks/cards etc.

Since true blue anil has answered the second question, I will answer the first question.  What is the probability that they get exactly one pair?
The probability is
$$\frac{N}{\binom{14}{2} \times \binom{12}{2}} = \frac{N}{6006}. \tag1 $$
Assume that the people are A and B, and that A pulls two socks out first.
In computing the denominator in (1) above, for convenience, order is construed to be relevant.  That is, A pulling a pair of blue socks, then B pulling a pair of red socks is construed to be distinct from A pulling a pair of red socks, then B pulling a pair of blue socks.
Therefore, the numerator, $(N)$ must be computed in a consistent manner.
Further, by symmetry, the number of combinations where A pulls a pair and B does not must equal the number of combinations where A does not pull a pair and B does pull a pair.
Therefore, I will compute twice the number of combinations where A pulls a pair and B does not.
In all of the possibilities below, I will use the syntax: 
b-r to denote the number of ways that B can pull a blue and red. 
b-y to denote the number of ways that B can pull a blue and yellow. 
r-y to denote the number of ways that B can pull a red and yellow.

*

*A pulls a blue pair. 
$\binom{4}{2} = 6~$ ways. 
Then, there are 2 blue, 7 red, 3 yellow remaining.
Therefore, 
b-r + b-y + r-y equals 
$(2 \times 7) + (2 \times 3) + (7 \times 3) = 41.$ 
Overall partial sum is
$6 \times 41 = 246.$


*A pulls a red pair. 
$\binom{7}{2} = 21~$ ways. 
Then, there are 4 blue, 5 red, 3 yellow remaining.
Therefore, 
b-r + b-y + r-y equals 
$(4 \times 5) + (4 \times 3) + (5 \times 3) = 47.$ 
Overall partial sum is
$21 \times 47 = 987.$


*A pulls a yellow pair. 
$\binom{3}{2} = 3~$ ways. 
Then, there are 4 blue, 7 red, 1 yellow remaining.
Therefore, 
b-r + b-y + r-y equals 
$(4 \times 7) + (4 \times 1) + (7 \times 1) = 39.$ 
Overall partial sum is
$3 \times 39 = 117.$
Therefore,
$$N = 2 \times \left(246 + 987 + 117\right) = 2700.$$
Therefore, the probability of this happening is
$$\frac{2700}{6006} = \frac{450}{1001}.$$
