Probability
Probability
star_batch_jee_advanced_2025
Grade None
Question:
Sixteen players $S_1, S_2, \ldots, S_{16}$ play in a tournament. They are divided into eight pairs at random. From each pair, a winner is decided on the basis of a game played between the two players of the pair. Assume that all the players are of equal strength then:
(A) The probability that the player $S_1$ is among the eight winners is $1/2$
(B) The probability that the player $S_1$ is among the eight winners is $1/4$
(C) The probability that exactly one of the two players $S_1$ and $S_2$ is among the eight winners is $8/15$
(D) The probability that exactly one of the two players $S_1$ and $S_2$ is among the eight winners is $7/15$
The probability that the player $S_1$ is among the eight winners is $1/2$
The probability that the player $S_1$ is among the eight winners is $1/4$
The probability that exactly one of the two players $S_1$ and $S_2$ is among the eight winners is $8/15$
The probability that exactly one of the two players $S_1$ and $S_2$ is among the eight winners is $7/15$
Step-by-Step Solution
Key Concept: By symmetry, each player wins their match independently with probability $1/2$, and the probability two specific players are paired together is $1/15$.
Since all players are equally strong, $S_1$ wins their paired match with probability $1/2$ regardless of whom they're paired with. For $S_1$ and $S_2$ to have exactly one winner, they must either: (1) be paired together (probability $1/15$, then exactly one wins with probability $1$), or (2) be in different pairs (probability $14/15$, then exactly one wins with probability $1/2 \times 1/2 + 1/2 \times 1/2 = 1/2$). Thus $P(\text{exactly one of } S_1, S_2 \text{ wins}) = \frac{1}{15} \cdot 1 + \frac{14}{15} \cdot \frac{1}{2} = \frac{1}{15} + \frac{7}{15} = \frac{8}{15}$.
Correct Answer: 1,3