Probability of Selection with Mixed Groups
a. In how many ways can four people be selected from this group of twelve?
b. In how many ways can four Republicans be selected from the eight Republicans?
c. Find the probability that the selected group will consist of all Republicans.
What This Problem Teaches
- How to distinguish between combinations and permutations in selection problems
- Application of the combination formula C(n,r) = n!/(r!(n-r)!)
- Converting counting problems into probability calculations
- Understanding that probability equals favorable outcomes divided by total outcomes
- Working with multi-part problems that build toward a final probability
Solution: Method 1 — Direct Combination Counting
This is a classic combination problem where we need to count selections without regard to order. We'll work through each part systematically.
Step 1 — Calculate total ways to select 4 from 12 (Part a)
Since we're selecting people for a conference and order doesn't matter, we use the combination formula:
Computing this step by step:
Step 2 — Calculate ways to select 4 Republicans from 8 (Part b)
Now we focus only on the 8 Republicans and count ways to choose 4 of them:
Step 3 — Calculate the probability (Part c)
The probability of selecting all Republicans equals the favorable outcomes divided by total outcomes:
Step 4 — Simplify the fraction
We can reduce this fraction by finding the greatest common divisor of 70 and 495:
Solution: Method 2 — Sequential Probability Approach
Instead of counting combinations, we can think of this as drawing 4 people one at a time and multiplying the probability that each person drawn is a Republican.
Step 1 — Set up the sequential selection
For all 4 selected people to be Republicans, we need:
- 1st person: Republican (8 Republicans out of 12 total)
- 2nd person: Republican (7 Republicans out of 11 remaining)
- 3rd person: Republican (6 Republicans out of 10 remaining)
- 4th person: Republican (5 Republicans out of 9 remaining)
Step 2 — Calculate the sequential probabilities
Step 3 — Multiply the fractions
Step 4 — Simplify to match Method 1
Dividing both numerator and denominator by 24:
This matches our combination approach, confirming both methods are equivalent.
a) 495 ways to select 4 people from 12
b) 70 ways to select 4 Republicans from 8
c) P(all Republicans) = 14/99 ≈ 0.141 or about 14.1%
Verification
Let's verify our combination calculations:
Part a verification: C(12,4) = 12!/(4!×8!) = (12×11×10×9)/(4×3×2×1) = 11,880/24 = 495 ✓
Part b verification: C(8,4) = 8!/(4!×4!) = (8×7×6×5)/(4×3×2×1) = 1,680/24 = 70 ✓
Part c verification: Both methods give 70/495 = 14/99. Converting to decimal: 14 ÷ 99 ≈ 0.1414, which is about 14.1% ✓
Reasonableness check: Since Republicans make up 8/12 = 2/3 of the group, we'd expect the probability of selecting all Republicans to be less than (2/3)⁴ ≈ 0.20. Our answer of 0.141 is indeed smaller, which makes sense because we're selecting without replacement.
Where Students Go Wrong
✗ Using permutations instead of combinations
Some students calculate P(12,4) = 12!/(12-4)! = 11,880 instead of C(12,4) = 495. Remember: when order doesn't matter (selecting a committee), use combinations, not permutations.
✗ Forgetting that selection is without replacement
In Method 2, students sometimes write (8/12)⁴ = 0.296, treating each selection as independent. This ignores that each Republican selected reduces both the numerator and denominator for subsequent selections.
✗ Setting up the wrong probability fraction
Some students flip the fraction and calculate C(12,4)/C(8,4) = 495/70, which gives a probability greater than 1. Always remember: probability = favorable outcomes / total outcomes.
✗ Calculation errors with factorials
When computing C(8,4), students sometimes write 8×7×6×5 = 1,880 instead of 1,680. Double-check your arithmetic, especially when multiplying several numbers together.
The Pattern Behind This
This problem illustrates the fundamental probability principle for selection without replacement:
Where:
- n = total population size (12 people)
- k = size of desired subgroup (8 Republicans)
- r = number selected (4 people)
This formula works whenever you're calculating the probability that all selected items come from a specific subset of the population. The beauty is that it doesn't matter whether we think of it as combination counting or sequential probability—both approaches yield the same result.
More generally, for selecting exactly m items from subgroup k and (r-m) items from the remaining (n-k), the probability becomes:
Our problem is the special case where m = r = 4.
Real Applications
Quality Control: Manufacturing companies use this exact calculation when randomly sampling products. If 20% of items are defective, what's the probability that a random sample of 5 contains all defective items?
Medical Research: Clinical trials often analyze the probability that randomly selected patients all respond positively to treatment, given known response rates in different demographic groups.
Survey Sampling: Political pollsters calculate the likelihood that a random sample accidentally over-represents one party, helping them assess whether their results might be skewed by chance.
What If?
Total people = 8 + 4 = 12, selecting 4: C(12,4) = 495 (same as before)
Ways to select 4 Democrats from 8: C(8,4) = 8!/(4!×4!) = 1,680/24 = 70
P(all Democrats) = C(8,4)/C(12,4) = 70/495 = 14/99 ≈ 0.141
Notice this is exactly the same probability as selecting all Republicans in the original problem, since both scenarios involve selecting 4 from a subgroup of 8 within a total group of 12. Answer: 14/99 ≈ 14.1%
Ways to select 5 people from 12: C(12,5) = 12!/(5!×7!) = 95,040/120 = 792
Ways to select 5 Republicans from 8: C(8,5) = 8!/(5!×3!) = 40,320/720 = 56
P(all Republicans) = 56/792 = 7/99 ≈ 0.071
This probability is lower than selecting 4 Republicans (14/99), which makes sense—it's harder to select more people all from the same subgroup. Answer: 7/99 ≈ 7.1%
Total ways to select 4 from 12: C(12,4) = 495
Ways to select 2 Democrats from 4: C(4,2) = 6
Ways to select 2 Republicans from 8: C(8,2) = 28
Combined: 6 × 28 = 168
P(2D, 2R) = 168/495 = 56/165 ≈ 0.339
This is much more likely than all Republicans (14.1%) because we're allowing mixed groups. Answer: 56/165 ≈ 33.9%
Total people = 4 + 6 + 2 = 12, selecting 5: C(12,5) = 792
Ways to select 2 Democrats from 4: C(4,2) = 6
Ways to select 2 Republicans from 6: C(6,2) = 15
Ways to select 1 Independent from 2: C(2,1) = 2
Total favorable: 6 × 15 × 2 = 180
P(2D, 2R, 1I) = 180/792 = 45/198 = 15/66 ≈ 0.227
Answer: 15/66 ≈ 22.7%
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2026-07-18