Title: "2024 Slot 2 Quant" Answer Key for Indian Game Problems
Problem Statement (Hypothetical Example):
In a popular Indian card game, players draw 3 cards from a deck of 52 cards. What is the probability that the hand contains exactly one pair (e.g., two kings and one 7) and no three-of-a-kind?
Solution:
Total Possible Hands:
( \binom{52}{3} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22,100 ).
Number of Valid Hands (One Pair):
Choose the rank for the pair: 13 options (e.g., kings, queens).
Choose 2 suits out of 4 for the pair: ( \binom{4}{2} = 6 ).
Choose a distinct rank for the third card: 12 remaining ranks.
Choose 1 suit out of 4 for the third card: 4 options.
Total Valid Hands: ( 13 \times 6 \times 12 \times 4 = 3,744 ).
Probability:
( \frac{3,744}{22,100} = \frac{24}{143} \approx 16.78% ).
Answer: ( \boxed{\frac{24}{143}} ).
Key Steps Explained:
Combinatorial Analysis: Used combinations to count valid hand configurations.
Exclusion Principle: Avoided counting three-of-a-kind by ensuring the third card is of a different rank.
Simplification: Reduced fractions to lowest terms for precision.
Common Indian Exam Tips:
Time Management: Allocate 2 minutes for combinatorial probability questions.
Check for Overcounting: Ensure pair ranks and suits are uniquely selected.
Practice Standard Problems: Drills on deck probability (e.g., flushes, straights) improve speed.
Similar Problems to Practice:
Probability of a flush in poker.
Expected value for dice games in Indian board exams.
Let me know if you need further clarification or additional problem examples!
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