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), use counting of favorable outcomes over total 2^10 possible match results. Event A (at least one player wins all) and Event B (at least one player loses all) can overlap only when exactly one player wins all and exactly one (different) player loses all, requiring careful case analysis with inclusion-exclusion principle.
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