Probability
Classical Probability
Grade 12
Question:
<p>\(n\) teams (\(n>5\)), each plays every other once, all equally likely to win each game. \(P(T_1, T_2, T_3\) finish in top 3, any order\() =\) <em>[JEE Advanced 2015]</em></p>
Step-by-Step Solution
Key Concept: By symmetry, any 3 of the n teams are equally likely to occupy the top 3 positions. P(specific 3 teams in top 3) = C(3,3)/C(n,3) = 6/[n(n-1)(n-2)].
<p>By symmetry, each set of 3 teams is equally likely to occupy positions 1,2,3.</p><p>Number of ways to choose top 3 from n: $\binom{n}{3}$.</p><p>P(T_1, T_2, T_3 are top 3 in any order) $= \dfrac{1}{\binom{n}{3}} = \dfrac{6}{n(n-1)(n-2)}$. Answer: B ✓</p>
Correct Answer: B