Probability
Probability
star_batch_jee_advanced_2025
Grade None
Question:
$8n$ players $P_1, P_2, \ldots, P_{8n}$ play a knock out tournament. It is known that all the players are of equal strength. The tournament is held in 3 rounds where the players are paired at random in each round. If it is given that $P_1$ wins in the third round. The probability that $P_2$ looses in the second round is:
n/(8n-1)
n/(8n+1)
2n/(4n-1)
None of these
Step-by-Step Solution
Key Concept: Use conditional probability with careful tracking of winning/losing conditions across multiple rounds and combination simplification.
Let $A$ = event of $P_1$ winning round 3, and $B$ = event of $P_2$ winning round 1 but losing round 2. Using conditional probability, $P(A) = \frac{\binom{8n-1}{8n-1}}{\binom{8n}{8n}} = \frac{1}{8}$. Through combinatorial analysis of the three rounds: $P(B \cap A) = \frac{n}{4(8n-1)}$, giving $P\left(\frac{B}{A}\right) = \frac{2n}{8n-1}$.
Correct Answer: 4