Conditional matching

Let $M$, $T$, $W$ be the events that Monday, Tuesday, Wednesday produce a matching pair, respectively. We want

1. $P(M^c)$. Total ways to choose $2$ socks from $10$: $\binom{10}{2}=45$. Matching pairs: $5$ (one per color). So $P(M)=5/45=1/9$, hence

2. $P(T^c \mid M^c)$. After a non‑matching Monday, two socks of different colors are removed. The remaining $8$ socks consist of $3$ colors with $2$ socks each and $2$ colors with $1$ sock each. Total pairs from $8$: $\binom{8}{2}=28$. Only the three colors with $2$ socks can produce a matching pair, so $3$ matching pairs. Thus $P(T\mid M^c)=3/28$ and

3. $P(W \mid M^c \cap T^c)$. After Monday and Tuesday are both non‑matching, the composition of the remaining $6$ socks depends on how Tuesday's non‑matching pair was formed. Starting from the Monday state ($3$ colors with $2$ socks, $2$ colors with $1$ sock), Tuesday's non‑matching pair can be of three types:

• Case A: Both socks from the $2$-sock colors. Number of ways: $\binom{3}{2}\cdot2\cdot2 = 12$. Resulting state: $1$ color with $2$ socks, $4$ colors with $1$ sock. Probability of a match on Wednesday: $\frac{1}{\binom{6}{2}} = \frac{1}{15}$.
• Case B: One sock from a $2$-sock color and one from a $1$-sock color. Number of ways: $3\cdot2\cdot2\cdot1 = 12$. Resulting state: $2$ colors with $2$ socks, $2$ colors with $1$ sock, $1$ color with $0$ socks. Probability of a match on Wednesday: $\frac{2}{15}$.
• Case C: Both socks from the two $1$-sock colors. Number of ways: $1$. Resulting state: $3$ colors with $2$ socks, $2$ colors with $0$ socks. Probability of a match on Wednesday: $\frac{3}{15} = \frac15$.

Total non‑matching pairs on Tuesday: $12+12+1 = 25$, so the conditional probabilities of the cases given $T^c$ are $\frac{12}{25},\frac{12}{25},\frac{1}{25}$. Hence

4. Multiply.

Thus the desired probability is $\boxed{\frac{26}{315}}$.

(One can verify by counting all sequences: total sequences $\binom{10}{2}\binom{8}{2}\binom{6}{2}=18900$, favorable sequences $5\cdot24\cdot13=1560$, giving the same fraction.)