🎯 Hypergeo vs Binomial

Recognizing when to use hypergeometric vs binomial distributions and applying the correct formulas.

🎯 Recognition Cues

  • Keywords: “with replacement”, “without replacement”, “drawing”, “sampling”
  • Setup: Drawing objects from a population
  • Constraints: Replacement vs. no-replacement, population size

📋 Solution Template

  1. Identify the sampling type:

    • With replacement: Use binomial distribution
    • Without replacement: Use hypergeometric distribution
    • Large population: Binomial approximation may apply

  2. Apply the appropriate formula:

    • Binomial: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
    • Hypergeometric: $P(X = k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$

  3. Check for approximations:

    • Large population: Hypergeometric ≈ Binomial
    • Small sample: Use exact formulas

💡 Micro-Examples

Binomial Distribution

  • Problem: What’s the probability of exactly 3 heads in 5 coin flips?
  • Solution: $P(X = 3) = \binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^2 = \frac{10}{32} = \frac{5}{16}$

Hypergeometric Distribution

  • Problem: What’s the probability of exactly 2 red balls in 5 draws from an urn with 10 red and 15 blue balls?
  • Solution: $P(X = 2) = \frac{\binom{10}{2}\binom{15}{3}}{\binom{25}{5}} = \frac{45 \cdot 455}{53130} = \frac{1365}{3542}$

Approximation

  • Problem: What’s the probability of exactly 2 red balls in 5 draws from an urn with 1000 red and 1500 blue balls?
  • Solution: Hypergeometric ≈ Binomial with $p = \frac{1000}{2500} = \frac{2}{5}$

⚠️ Common Pitfalls & Variants

Pitfall: Using binomial when hypergeometric is needed

  • Wrong: “Drawing 5 cards without replacement” = Binomial distribution
  • Right: Use hypergeometric distribution for sampling without replacement.

Pitfall: Using hypergeometric when binomial is needed

  • Wrong: “Drawing 5 cards with replacement” = Hypergeometric distribution
  • Right: Use binomial distribution for sampling with replacement.

Pitfall: Forgetting to check population size

  • Wrong: “Large population” = Always use binomial
  • Right: Check if the population is large enough for the approximation to be valid.

🏆 AMC-Style Worked Example

Problem: An urn contains 6 red balls and 4 blue balls. You draw 3 balls without replacement. What’s the probability of getting exactly 2 red balls?

Solution:

  • Method 1: Hypergeometric distribution

    • $N = 10$, $K = 6$, $n = 3$, $k = 2$
    • $P(X = 2) = \frac{\binom{6}{2}\binom{4}{1}}{\binom{10}{3}} = \frac{15 \cdot 4}{120} = \frac{60}{120} = \frac{1}{2}$

  • Method 2: Direct counting

    • Total ways to draw 3 balls: $\binom{10}{3} = 120$
    • Ways to draw 2 red and 1 blue: $\binom{6}{2} \cdot \binom{4}{1} = 15 \cdot 4 = 60$
    • Probability: $\frac{60}{120} = \frac{1}{2}$

Key insight: Both methods give the same answer! Choose the one that’s easier for the specific problem.

🔗 Related Topics

Next: Digit Counting → Formulas