Probability of picking some color combinations of socks A box contains $3$ yellow socks, $4$ blue socks, $1$ orange sock, and $2$ green socks. What is the probability of picking $2$ blue socks at the same time? What is the probability of picking $1$ green and $1$ blue sock at the same time?
I was thinking it is:
$\large\frac{4}{10} \cdot \frac39$ for the $1^{st}$ question,  
$\large\frac{2}{10} \cdot \frac49$ for the $2^{nd}$ question.
 A: 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,
$$P_2=\frac{\binom{4}{1}\binom{2}{1}}{\binom{10}{2}}=\frac{4\times2}{45}=\frac{8}{45}$$
A: You are correct for Q1, your solution for Q2 is incorrect because you must multiply by 2 since you can either choose the blue or the green sock first, accounting for combinations. 
Therefore, you must multiply by 2 in your final answer to account for that.
A: Your solution to the first one is correct, but I think the second solution is incorrect.
Another way to do this is through combinations.  There are ${10 \choose 2} = 45$ different pairs of socks you can pull out.  There are ${4 \choose 2} = 6$ ways to get a pair of blue socks, so the probability for the first question is $2/15$ as you have it.
But for the green/blue pair, there are ${4 \choose 1} \cdot {2 \choose 1} = 8$ distinct pairs.  So your probability is $8/45$, not $8/90$ as you had.
Where the difference lies is that you only took account of half of the possibilities.  You counted the possibilities where you pulled a green sock, then a blue sock.  You also need to count the possibilities where you pull a blue sock, then a green sock.  There's your factor of $2$ that you were missing.
Why did you get the right answer with the two green socks, though?  You did it exactly the same way.  Well, you essentially counted the green pairs twice, but also counted all of the pairs twice, so those factors of two canceled.
A: I think the second answer is not correct.
If you pick one sock after another the probility for a green sock (fist pick) and a blue sock (second pick) is:
$P(gb)=\frac{2}{10} \cdot \frac{4}{9}$
OR you pick first a blue sock and then a green sock:
$P(bg)=\frac{4}{10} \cdot \frac{2}{9}$
Now sum this two probabilities.
In principle it doesn´t matter, if you handle this problem as a successive-pick problem or a simulaneous-pick problem. 
