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C
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Difficulty:
35%
(medium)
Question Stats:
68% (02:04) correct
32%
(02:19)
wrong
based on 60
sessions
History
Date
Time
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Not Attempted Yet
An English class consists of six girls and three boys. If three of the students are chosen for a trip to Ireland, approximately what percent of all of possible groups of three students consist of two girls and one boy?
A. 18%
B. 36%
C. 54 %
D. 71%
E. 89%
A. 18%
B. 36%
C. 54 %
D. 71%
E. 89%
12 Feb 2026, 13:12
Kudos
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This is a combinations problem testing your ability to calculate favorable outcomes versus total outcomes. The key is setting up the calculation correctly and not getting confused by the grouping.
**Step 1: Find total possible groups of 3 students**
Total students = 6 girls + 3 boys = 9 students
Total ways to choose 3 from 9: C(9,3) = 9!/(3!×6!) = (9×8×7)/(3×2×1) = 504/6 = 84
**Step 2: Find favorable groups (2 girls and 1 boy)**
We need to choose 2 girls from 6 AND 1 boy from 3.
Girls: C(6,2) = 6!/(2!×4!) = (6×5)/2 = 15
Boys: C(3,1) = 3
Favorable groups = 15 × 3 = 45
**Step 3: Calculate the percentage**
Percentage = (Favorable/Total) × 100% = (45/84) × 100%
Let me simplify: 45/84 = 15/28 ≈ 0.536 = 53.6%
**Answer: (C) 53%**
**Step-by-step calculation verification:**
- 45/84 = 45/84
- Divide both by 3: 15/28
- 15 ÷ 28 = 0.5357... ≈ 53.6%
**Common trap:** Students sometimes add combinations instead of multiplying them. When you need 2 girls AND 1 boy, you must multiply: C(6,2) × C(3,1). The word "and" in probability/combinations signals multiplication. Also, some students forget that choosing 3 from 9 gives you 84 total groups, not 27 (which would be 33).
**Takeaway:** In combination problems with multiple groups, use the formula C(n,r) = n!/(r!(n-r)!) for each selection, then multiply when you need items from different groups. Always calculate total possible outcomes first, then favorable outcomes, and express as a percentage.
**Step 1: Find total possible groups of 3 students**
Total students = 6 girls + 3 boys = 9 students
Total ways to choose 3 from 9: C(9,3) = 9!/(3!×6!) = (9×8×7)/(3×2×1) = 504/6 = 84
**Step 2: Find favorable groups (2 girls and 1 boy)**
We need to choose 2 girls from 6 AND 1 boy from 3.
Girls: C(6,2) = 6!/(2!×4!) = (6×5)/2 = 15
Boys: C(3,1) = 3
Favorable groups = 15 × 3 = 45
**Step 3: Calculate the percentage**
Percentage = (Favorable/Total) × 100% = (45/84) × 100%
Let me simplify: 45/84 = 15/28 ≈ 0.536 = 53.6%
**Answer: (C) 53%**
**Step-by-step calculation verification:**
- 45/84 = 45/84
- Divide both by 3: 15/28
- 15 ÷ 28 = 0.5357... ≈ 53.6%
**Common trap:** Students sometimes add combinations instead of multiplying them. When you need 2 girls AND 1 boy, you must multiply: C(6,2) × C(3,1). The word "and" in probability/combinations signals multiplication. Also, some students forget that choosing 3 from 9 gives you 84 total groups, not 27 (which would be 33).
**Takeaway:** In combination problems with multiple groups, use the formula C(n,r) = n!/(r!(n-r)!) for each selection, then multiply when you need items from different groups. Always calculate total possible outcomes first, then favorable outcomes, and express as a percentage.
