As Adriano pointed, we gonna split this problem in 3 sub-problems. What's the probability of drawing a 2 blue socks in the first 2 drawing. It's:
$$\frac{4}{14} \times \frac{3}{13} = \frac{12}{182}$$
Now what's the probability of drawing 2 red socks? It's:
$$\frac{7}{14} \times \frac{6}{13} = \frac{42}{182}$$
And the final sub-problem, what's the probability of drawing 2 yellow socks? It's:
$$\frac{3}{14} \times \frac{2}{13} = \frac{6}{182}$$
Now we add up this 3 fractions and we end up with:
$$\frac{12}{182} + \frac{42}{182} + \frac{6}{182} = \frac{60}{182} \approx 32.97 \%$$
There may be a more elegant way than this, but here's one way.
First, write down all of the color distributions that are not terminating, but could terminate on the next draw. This means that you need to have drawn at least one color twice. I'll designate the colors $1-4$ and not worry about order:
- $2$ balls: $11$
- $3$ balls: $112$
- $4$ balls: $1122, 1123$
- $5$ balls: $11223, 11234$
- $6$ balls: $112233, 112234$
- $7$ balls: $1122334$
- $8$ balls: $11223344$
Next, for each of these color distributions, calculate the number of ways to arrange the digits. This can be done with multinomials. For example, the number of ways to arrange $11223$ is
$$\frac{5!}{2!2!1!} = 30.$$
Next, observe that, for a given set of colors, the probability of drawing those colors is independent of the order. For example,
$$BBGGP = \frac{3}{20} \frac{2}{19} \frac{8}{18} \frac{7}{17} \frac{4}{16},$$
but
$$PGBGB = \frac{4}{20} \frac{8}{19} \frac{3}{18} \frac{7}{17} \frac{2}{16},$$
which is exactly the same value. All that happened was that the numerators got rearranged.
Now comes the somewhat tedious part, which is to calculate the individual probabilities for each substitution of colors. We can at least be a little smart about it, and calculate the numerator part for each color:
$$B = 3, BB = 6, P = 4, PP = 12, O = 5, OO=20, G = 8, GG = 56.$$
Let's do the case for drawing three blue balls, and drawing the blue on the fifth ball. We assign $BB$ to $11$ in the combination, and then write down all the others: $$BBPP, BBOO, BBGG, BBPO, BBPG, BBOG.$$
(The first three are $1122$ types, and the last three are $1123$ types.)
To calculate the probability of getting each one of these, we multiply the probability of getting them in the stated order with the number of ways to rearrange them. So,
$$P(BBPP) = \frac{6 \cdot 12}{20 \cdot 19 \cdot 18 \cdot 17}\cdot \frac{4!}{2!2!}.$$
Now, to find the probability of ending on the next turn with blue, given that you have two blues and two purples (drawn in any order), multiply by $1/16$.
Then it's just a lot of rinse/repeat. Do the same calculation for the rest of the four-ball colors. Then do all of the other numbers of balls, for all of the applicable color combinations each time.
And that's just the first part of a four-part problem!
You can possibly reuse some of the calculations you did before if you keep track.
Best Answer
Considering the fact that you picking the socks at the same time, the solution is as follow,
$1^{st}$ question, $$P_1=\frac{\binom{4}{2}}{\binom{10}{2}}=\frac{6}{45}=\frac{2}{15}$$
$2^{nd}$ question, $$P_2=\frac{\binom{4}{1}\binom{2}{1}}{\binom{10}{2}}=\frac{4\times2}{45}=\frac{8}{45}$$