📈 Probability Distributions

Common probability models that frequently appear in AMC problems.

🎯 Key Ideas

Binomial Distribution

Counts successes in $n$ independent trials with probability $p$ of success each. $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ $$\mathbb{E}[X] = np, \quad \text{Var}(X) = np(1-p)$$

Geometric Distribution

Counts trials until first success in independent trials with probability $p$ of success. $$P(X = k) = (1-p)^{k-1} p$$ $$\mathbb{E}[X] = \frac{1}{p}, \quad \text{Var}(X) = \frac{1-p}{p^2}$$

Hypergeometric Distribution

Counts successes in sampling without replacement from a finite population. $$P(X = k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$$ where $N$ = total population, $K$ = successes in population, $n$ = sample size.

💡 Micro-Examples

Binomial Distribution

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

Geometric Distribution

  • Problem: What’s the probability the first 6 appears on the 4th die roll?
  • Solution: $P(X = 4) = \left(\frac{5}{6}\right)^3 \cdot \frac{1}{6} = \frac{125}{1296}$

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: $P(X = 2) = \frac{\binom{10}{2}\binom{15}{3}}{\binom{25}{5}} = \frac{45 \cdot 455}{53130} = \frac{20475}{53130} = \frac{1365}{3542}$

⚠️ Common Traps & Fixes

Trap: Confusing binomial and hypergeometric

  • Wrong: “Drawing 5 cards with replacement” = Hypergeometric
  • Right: “Drawing 5 cards with replacement” = Binomial (with replacement)

Trap: Forgetting the “until first success” in geometric

  • Wrong: “What’s the probability of exactly 3 failures before success?” = Geometric
  • Right: “What’s the probability of exactly 3 failures before success?” = Geometric with $k = 4$

Trap: Misapplying hypergeometric formula

  • Wrong: “What’s the probability of exactly 2 aces in 5 cards?” = $\frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}}$
  • Right: This is actually correct! $N = 52$, $K = 4$, $n = 5$, $k = 2$.

🏆 AMC-Style Worked Example

Problem: A bag contains 6 red balls and 4 blue balls. You draw 3 balls without replacement. What’s the expected number of red balls?

Solution:

  • Method 1: Using hypergeometric distribution

    • $N = 10$, $K = 6$, $n = 3$
    • $\mathbb{E}[X] = n \cdot \frac{K}{N} = 3 \cdot \frac{6}{10} = \frac{18}{10} = 1.8$

  • Method 2: Using indicators

    • Let $I_i$ be 1 if ball $i$ is red, 0 otherwise
    • $\mathbb{E}[\text{red balls}] = \mathbb{E}[I_1 + I_2 + I_3] = 3 \cdot \frac{6}{10} = 1.8$

Key insight: Both methods give the same answer! The hypergeometric approach is often more direct.

🔗 Related Topics

Next: Symmetry & Invariance → Recurrences & GF