Probability
Probability
star_batch_jee_advanced_2025
Grade None
Question:
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:
P(A) = \frac{5}{16}
P(B) = \frac{7}{16}
P(A \cap B) = \frac{5}{32}
P(A \cup B) = \frac{15}{32}
Step-by-Step Solution
Key Concept: Combinatorial probability with binomial coefficients requires careful counting of favorable outcomes and applying inclusion-exclusion for union probabilities.
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