Probability
Probability
Allen Star Batch
Grade 12
Question:
$2^n$ players of equal strength are playing a knock out tournament. If they are paired randomly in all rounds, the probability that out of two particular players $S_1$ and $S_2$ exactly one will reach in semi final is $(n \in N, n \geq 2)$:
$\frac{8(2^n - 4)}{2^n(2^n - 1)}$
$\frac{(2^n - 4)}{2^n(2^n - 1)}$
$\frac{(2^{n-1} - 4)}{2^n(2^n - 1)}$
None of these
Step-by-Step Solution
Key Concept: Probability requires counting favorable selections from equal-strength players using combinations.
Since 4 players reach the semi-final from $2^n$ equal-strength players, the total number of ways to select them is $^{2^n}C_4$. The favorable outcomes where exactly 2 come from each group of $2^{n-2}$ players is $^{2^{n-2}}C_2 imes ^{2^{n-2}}C_2$. The required probability is $ rac{^{2^{n-2}}C_2 imes 2}{^{2^n}C_4} = rac{8(2^n - 4)}{2^n(2^n - 1)}$.
Correct Answer: 1