Question:

For part (A), using combinations: $\frac{^{11}C_5}{^{12}C_6} = \frac{1}{2}$. For part (B), let $E_1$ be the event that $S_3$ and $S_4$ are in the same group with $P(E_1) = \frac{1}{11}$, and $E_2$ be the event they are in different groups with $P(E_2) = \frac{10}{11}$. The probability that exactly one of $S_3$ and $S_4$ is among losers is $P(E) = \frac{1}{11} \times \frac{10}{2} + \frac{10}{11} \times (\frac{1}{2} + \frac{1}{2} - \frac{1}{4}) = \frac{6}{11}$. For part (C), $S_2$ and $S_4$ must be in different groups for both to be winners, giving required probability $= \frac{10}{11}(\frac{1}{2} \cdot \frac{1}{2}) = \frac{5}{22}$ or $^{10}C_6 = \frac{5}{22}$.