2. The only way to arrive at the latter alternative is to draw four different socks in the first drawing or to draw only one pair in the first drawing and then draw two different socks in the last drawing. We will call these probabilities respectively $p_{1}, p_{2}, p_{3}$. We calculate them as follows:  
$$ p_{1}=\frac{\binom{5}{4} 2^{4}}{\binom{10}{4}}=\frac{8}{21}, \quad p_{2}=\frac{5\binom{4}{2} 2^{2}}{\binom{10}{4}}=\frac{4}{7}, \quad p_{3}=\frac{4}{\binom{6}{2}}=\frac{4}{15} . $$  
We finally calculate the desired probability: $P=p_{1}+p_{2} p_{3}=\frac{8}{15}$.
2. (BEL 4) From a bag containing 5 pairs of socks, each pair a different color, a random sample of 4 single socks is drawn. Any complete pairs in the sample are discarded and replaced by a new pair draw from the bag. The process continues until the bag is empty or there are 4 socks of different colors held outside the bag. What is the probability of the latter alternative?