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.
Best Answer
Since probability has been asked for, we can, without loss of generality, simplify the problem to two draws of two socks
With $4$ blue, $7$ red and $3$ yellow,
P(both are pairs) $= \dfrac{\binom42\binom22 + \binom72\binom52+2[\binom42\binom72+ \binom42\binom32+\binom72\binom32]}{\binom{14}2\binom{12}2}$
Can you now proceed to the second part of he problem. There will be a number of cases that you will need to count, and one warning: your computation for one pair towards the end is actually computation for at least one pair.