In a tournament there are twelve players $S_1, S_2, \ldots, S_{12}$ and divided into six pairs at random. From each game a winner is decided on the basis of a game played between the two players of the pair. Assuming all the pairs are of equal strength, then match the following.
Step-by-Step Solution
Key Concept: The problem requires calculating conditional probabilities in random pairing scenarios. Key insight: when 12 players are randomly divided into 6 pairs, the probability that any specific player is paired with another specific player is 1/11, and the probability they're in different pairs is 10/11. This uses the principle that once one player is selected, there are 11 remaining positions, exactly 1 of which pairs with the chosen player.
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 $^{10}C_6 = \frac{5}{22}$.
Correct Answer: [A-r][B-s][C-p][D-q]