Title: 2022 Slot 1 Quant: Analyzing an Indian Strategy Game
Content: English Solution and Explanation
Problem Statement (Hypothetical Example):
In a traditional Indian strategy game called Kho-Kho, two teams of 5 players each compete in a circular arena. Players take turns rolling a die to determine their movement. The team that captures all opposing players first wins. The die has numbers 1–6, and each roll moves the player clockwise by the number of positions. If a player lands on an opponent’s position, the opponent is captured.
Question:
If a player starts at position 0 and rolls a 4, how many unique positions could they capture an opponent in?
What is the probability that a player captures at least one opponent in two consecutive rolls?
Solution:
Question 1: Unique Capture Positions
Given:
Circular arena with positions numbered 0–9 (total 10 positions).
Player starts at position 0, rolls a 4.
Steps:
Calculate the new position after rolling 4:
[
(0 + 4) \mod 10 = 4
]
Determine opponent positions. Assume opponents are equally spaced. For simplicity, assume opponents are at positions 1, 3, 5, 7, 9 (odd positions).
Check if position 4 matches any opponent positions.
Opponent positions: 1, 3, 5, 7, 9.
Player lands on position 4 → No capture occurs.
Answer:
0 unique positions captured.
Question 2: Probability of Capturing in Two Rolls
Assumptions:
Opponent positions: 1, 3, 5, 7, 9 (5 positions).
Each die roll is independent.
Steps:
First Roll Probability:
To capture on the first roll, the player must land on an opponent’s position.
Possible rolls: 1, 3, 5, 7, 9 (since starting at 0).
Die has 6 outcomes. Probability of capturing on first roll:
[
P_1 = \frac{5}{6}
]
If captured on the first roll, the second roll is irrelevant.
Second Roll Probability (if not captured first):
Probability of not capturing on first roll:
[
P_{\text{not }1} = 1 - \frac{5}{6} = \frac{1}{6}
]
After not capturing, the player moves to a new position. For example, rolling a 2 would take them to position 2.
To capture on the second roll, the new position must land on an opponent’s position.
Assume the second roll is independent. The probability of capturing on the second roll is again ( \frac{5}{6} ).
Total Probability:
[
P_{\text{total}} = P_1 + P_{\text{not }1} \times P_2 = \frac{5}{6} + \frac{1}{6} \times \frac{5}{6} = \frac{5}{6} + \frac{5}{36} = \frac{35}{36}
]
Answer:
Probability = ( \frac{35}{36} ) (≈ 97.2%).
Key Concepts:
Circular Motion in Modular Arithmetic: Positions wrap around using modulo operations.
Probability of Compound Events: Use the law of total probability for sequential actions.
Assumption of Opponent Positions: Clarifying game rules is critical for accuracy.
Common Mistakes to Avoid:
Forgetting to account for modular wrap-around (e.g., position 10 ≡ 0).
Overlapping opponent positions (if not spaced uniquely).
Assuming independence without verifying game rules.
Final Note:
This solution assumes hypothetical rules for Kho-Kho. For precise answers, refer to the exam’s official problem statement. Let me know if you need further clarification!
Word Count: 298
Style: Formal, step-by-step, with mathematical notation and critical analysis.
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