Here’s a structured solution to a typical Quantitative Ability (QA) problem from the CAT 2021 Slot 2 exam, formatted as per Indian standardized test conventions:
Problem Statement (CAT 2021 Slot 2):
A game consists of three rounds. In each round, a player can choose to either play or pass. If they play, they earn points equal to the round number (Round 1 = 1 point, Round 2 = 2 points, Round 3 = 3 points) multiplied by a random integer ( k ) (1 ≤ ( k ) ≤ 5). If they pass, they earn 0 points for that round. The player’s total score is the sum of points from all rounds. What is the probability that a player’s total score is exactly 10 points?
Solution:
Understanding the Game Mechanics:
Each round (1, 2, 3) offers a choice: Play (earn ( k \times \text{round number} )) or Pass (0 points).
( k ) is uniformly random: 1, 2, 3, 4, 5 (each with probability ( \frac{1}{5} )).
Total score = Sum of points from Rounds 1, 2, and 3.
Objective:
Calculate the probability that the total score = 10.
Approach:
Enumerate all possible combinations of play/pass decisions and ( k )-values that sum to 10. Divide by the total number of possible outcomes.
Total Possible Outcomes:
Each round has 2 choices (Play/Pass), and each play has 5 ( k )-values.
Total outcomes = ( 2^3 \times 5^3 = 8 \times 125 = 1000 ).
Valid Scenarios for Total = 10:
We need to find all sequences of Play/Pass and ( k )-values such that:
( a \times 1 + b \times 2 + c \times 3 = 10 ),
where ( a, b, c \in {0, 1, 2, 3, 4, 5} ) (since ( k ) can repeat across rounds).
Constraints:
( a, b, c ) are non-negative integers.
Each ( k )-value (1-5) is independent per round.
Enumerating Valid Combinations:
Let’s break down possible contributions from each round:
Case 1: Play all three rounds
( a + 2b + 3c = 10 ).
Possible ( (a, b, c) ):
( c = 2 ): ( a + 2b = 4 ). Solutions: (4,0,2), (2,1,2), (0,2,2).
( c = 1 ): ( a + 2b = 7 ). Solutions: (5,1,1), (3,2,1), (1,3,1).
( c = 0 ): ( a + 2b = 10 ). Solutions: (10,0,0) invalid (max ( a = 5 )), (8,1,0), (6,2,0), (4,3,0), (2,4,0), (0,5,0).
Case 2: Pass one round
Pass Round 1: ( 2b + 3c = 10 ). Solutions: (b=2, c=2), (b=5, c=0).
Pass Round 2: ( a + 3c = 10 ). Solutions: (a=1, c=3), (a=4, c=2), (a=7, c=1) invalid, (a=10, c=0) invalid.
Pass Round 3: ( a + 2b = 10 ). Solutions: (a=0, b=5), (a=2, b=4), (a=4, b=3), (a=6, b=2), (a=8, b=1), (a=10, b=0) invalid.
Case 3: Pass two rounds
Pass Rounds 1 & 2: ( 3c = 10 ) → No solution.
Pass Rounds 1 & 3: ( 2b = 10 ) → ( b=5 ).
Pass Rounds 2 & 3: ( a = 10 ) → Invalid.
Case 4: Pass all rounds → Total = 0 (invalid).
Valid Combinations Count:
After filtering valid ( (a, b, c) ) within ( 0 \leq a, b, c \leq 5 ):
Case 1: 6 (from c=2) + 3 (from c=1) + 5 (from c=0) = 14.
Case 2: 2 (Pass Round 1) + 3 (Pass Round 2) + 6 (Pass Round 3) = 11.
Case 3: 1 (Pass Rounds 1&3).
Total valid combinations = 14 + 11 + 1 = 26.
Probability Calculation:
Each valid combination corresponds to a unique sequence of play/pass and ( k )-values.
Probability = ( \frac{26}{1000} = 0.026 ).
Final Answer:
The probability is ( \boxed{\dfrac{13}{500}} ).
Key Takeaways for CAT备考:
Break Down Complex Problems: Split into cases (play/pass scenarios).
Constraints Matter: Ensure ( a, b, c \leq 5 ) (since ( k ) is per-round).
Efficient Enumeration: Use systematic iteration to avoid missing cases.
Probability Weighting: Each ( k )-value is equally likely, so total outcomes = ( 2^3 \times 5^3 ).
Let me know if you need further clarification!
|
