# Probability Question (Colored Socks)

In a drawer Sandy has 5 pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the 10 socks in the drawer. On Tuesday Sandy selects 2 of the remaining 8 socks at random and on Wednesday two of the remaining 6 socks at random. The probability that Wednesday is the first day Sandy selects matching socks is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

She has $5$ different colors: $a, b, c, d, e$.

The number of ways of selecting $2$ socks on Wednesday is:

$$\binom{6}{2}$$

But how do I go further?

There are three cases to consider:

1. Four socks, each of a different color, are selected on Monday and Tuesday, then a pair is selected on Wednesday from the one pair and four single socks that remain. The probability of this occurring is $$1 \cdot \frac{8}{9} \cdot \frac{6}{8} \cdot \frac{4}{7} \cdot \frac{\binom{2}{2}}{\binom{6}{2}}$$ since the first sock is guaranteed to be of a distinct color, $8$ of the remaining $9$ socks are a different color than that of the first sock that is selected, $6$ of the remaining $8$ socks are of a different color than those of both of the first two socks that have been selected, $4$ of the remaining $7$ socks are of a different color than those of the first three socks that have been selected, and we must choose both socks of the fifth color from the six remaining socks.
2. Four socks, of three different colors, are selected on Monday and Tuesday, then a pair is selected on Wednesday from the two remaining pairs and two remaining individual socks. The probability of this occurring is $$1 \cdot \frac{8}{9} \cdot \left(\frac{2}{8} \cdot \frac{6}{7} + \frac{6}{8} \cdot \frac{2}{7}\right) \cdot \frac{4}{6} \cdot {1}{5}$$ On Monday, the probabilities are the same as in the first case. On Tuesday, there are two possibilities. One is that a sock of one of the $2$ colors already selected on Monday is selected from one of the $8$ remaining socks, then a sock of one of the other $3$ colors is selected from the $7$ remaining socks. The other possibility is that one of the $6$ socks of one the $3$ remaining colors is selected from the $8$ socks that remain, then one of the $2$ socks of a color already selected on Monday is selected from one of the $7$ remaining socks. On Wednesday, there are $6$ socks left. One of the $4$ socks from the two remaining pairs must be selected first, then the only sock of that color that remains must be selected from the $5$ remaining socks.
3. Four socks, of two different colors, are selected on Monday and Tuesday, then a pair is selected on Wednesday from one of the three remaining pairs. The probability of this occurring is $$1 \cdot \frac{8}{9} \cdot \frac{2}{8} \cdot \frac{1}{7} \cdot 1 \cdot \frac{1}{5}$$ On Monday, the probabilities are the same as in the preceding cases. On Tuesday, the first sock selected must be one of the $2$ socks of a color already selected on Monday, then the other sock that is selected must match the other color selected on Monday. On Wednesday, only the three colors that have not already been used are available, so the first sock is guaranteed to be of a different color than those selected on Monday and Tuesday. Finally, we must select the only remaining sock that matches the first selection on Wednesday from the five remaining socks.

Since these cases are mutually exclusive, you can add the probabilities to determine the probability that Wednesday is the first day that Sandy selects a matching pair of socks.

Partial solution

Probability of getting matching socks on day 1 is $\frac19$.

If they were not matching, we have 3 pairs and 2 individual socks left. We must select a 'pairable' sock in our first pick, so $\frac68$ for that, and the other pick must be the other sock of the same pair, so $\frac17$. Therefore, probability of getting matching socks on day 2 (but not day 1) is $\frac89\times(\frac68\times\frac17)=\frac2{21}$

Probability of missing a match on day 1 as well as on day 2 should be $1-(\frac19+\frac2{21})=\frac{50}{63}$

I'm not sure how to get probability of getting a match on day 3, though.