5 players of equal strength play one each with each other. $P(A) =$ probability that at least one player wins all matches he (they) plays. $P(B) =$ probability that at least one player loses all his (their) matches.
Then:
Step-by-Step Solution
Key Concept: In a round-robin tournament with 5 players of equal strength, each match has probability 1/2 for either outcome. Event A (at least one player wins all matches) and Event B (at least one player loses all matches) require counting favorable outcomes among 2^10 total possible match results, then applying inclusion-exclusion: P(A∪B) = P(A) + P(B) - P(A∩B).
Given $P(A) = \frac{^5C_2^6}{2^{10}} = \frac{5}{16}$ and $P(B) = \frac{^5C_2^2^5}{2^{10}} = \frac{5}{16}$, we compute $P(A ∩ B) = \frac{^5C_2 × 2 × 2^3}{2^{10}} = \frac{15}{32}$. Using inclusion-exclusion, $P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = \frac{5}{16} + \frac{5}{16} - \frac{15}{32} = \frac{15}{32}$.
Correct Answer: 1,3,4