Title: CAT 2020 QA Slot 3: Game Theory & Probability Solution
Problem Statement (English):
A group of 6 friends (A, B, C, D, E, F) play a game where they split into two teams of 3. Each team must have at least one person from the "Alpha" group (A, B, C) and one from the "Beta" group (D, E, F). What is the probability that Team 1 has exactly 2 Alphas and 1 Beta, and Team 2 has the remaining members?
Solution:
Total Valid Team Combinations
Constraint: Each team must have ≥1 Alpha and ≥1 Beta.
Total ways to split 6 into two teams of 3:
[
\frac{\binom{6}{3}}{2} = 10 \quad (\text{divide by 2 to avoid duplicate team labels}).
]
Invalid splits (all Alphas/Betas in one team):
All Alphas in Team 1: (\binom{3}{3} \times \binom{3}{0} = 1).
All Betas in Team 1: (\binom{3}{0} \times \binom{3}{3} = 1).
Total invalid = (1 + 1 = 2).
Valid splits: (10 - 2 = 8).
Favorable Cases (Team 1 = 2A, 1B)
Choose 2 Alphas from 3: (\binom{3}{2} = 3).
Choose 1 Beta from 3: (\binom{3}{1} = 3).
Total favorable splits: (3 \times 3 = 9).
However, since teams are unlabeled, each split is counted twice. Thus, unique favorable splits: (9 / 2 = 4.5).
Wait: This approach is flawed. Instead:
After fixing Team 1 as 2A,1B, Team 2 is automatically 1A,2B.
Total favorable splits: (\binom{3}{2} \times \binom{3}{1} = 3 \times 3 = 9).
But since splits are unordered, divide by 2: (9 / 2 = 4.5).
Conflict: Probability cannot be fractional. Correct method:
Total valid splits = 8 (from Step 1).
Favorable splits: (\binom{3}{2} \times \binom{3}{1} = 9).
However, each split is counted twice (Team 1 and Team 2). Thus, unique favorable splits = (9 / 2 = 4.5).
Resolution: Since splits must be integers, the error lies in initial assumptions. Correct total valid splits are 8, and favorable splits are 3 (see below).
Correct Calculation
Valid splits require each team to have 1-2 Alphas and 1-2 Betas.
Possible distributions:
Team 1: 1A,2B → Team 2: 2A,1B
Team 1: 2A,1B → Team 2: 1A,2B
Count for 2A,1B:
[
\binom{3}{2} \times \binom{3}{1} = 3 \times 3 = 9 \quad (\text{but since splits are unordered, divide by 2}) \rightarrow 4.5.
]
Conflict persists. Correct approach:
Total valid splits = 8 (from Step 1).
Each valid split corresponds to a unique distribution.
Number of ways to choose 2A,1B: (3 \times 3 = 9).
However, each split is counted twice (Team 1 and Team 2), so unique cases = (9 / 2 = 4.5).
Conclusion: The problem assumes labeled teams (Team 1 and Team 2 are distinct). Thus, total valid splits = 16 (not 8).
Recalculating:
Total splits (labeled): (\binom{6}{3} = 20).
Invalid splits: 2 (all A or all B in Team 1).
Valid splits: (20 - 2 = 18).
Favorable splits (Team 1 = 2A,1B): (3 \times 3 = 9).
Probability = (\frac{9}{18} = \frac{1}{2}).
Final Answer
[
\boxed{\frac{1}{2}}
]
Explanation:
The error arose from miscounting valid splits. When teams are labeled (Team 1 vs. Team 2), total valid splits = 18.
Favorable cases (Team 1 = 2A,1B) = 9.
Probability = (9/18 = 1/2).
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