⚖️ At Least/At Most

Problems involving “at least” and “at most” constraints, solved using complement counting or inclusion-exclusion.

🎯 Recognition Cues

  • Keywords: “at least”, “at most”, “at most”, “no more than”, “at least one”, “exactly”
  • Setup: Counting with constraints on minimum or maximum occurrences
  • Constraints: Numerical bounds on counts or probabilities

📋 Solution Template

  1. Identify the constraint type:

    • At least: Use complement counting (total - “less than”)
    • At most: Use complement counting (total - “more than”)
    • Exactly: Direct counting or inclusion-exclusion

  2. Choose the appropriate method:

    • Simple complement: When “less than” or “more than” is easy to count
    • Inclusion-exclusion: When multiple constraints overlap
    • Casework: When complement is complex

  3. Apply the method:

    • Complement: $P(\text{at least } k) = 1 - P(\text{less than } k)$
    • Inclusion-exclusion: Add individual cases, subtract overlaps
    • Casework: Count each case separately

💡 Micro-Examples

At Least (Complement)

  • Problem: What’s the probability of at least one head in 3 coin flips?
  • Solution: $1 - P(\text{no heads}) = 1 - \left(\frac{1}{2}\right)^3 = 1 - \frac{1}{8} = \frac{7}{8}$

At Most (Complement)

  • Problem: How many 4-digit numbers have at most one 7?
  • Solution: Total - “At least two 7s” = $9000 - \text{count of numbers with 2+ sevens}$

Exactly (Direct)

  • Problem: What’s the probability of exactly 2 heads in 3 coin flips?
  • Solution: $\binom{3}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^1 = \frac{3}{8}$

⚠️ Common Pitfalls & Variants

Pitfall: Using complement when direct counting is easier

  • Wrong: “At least 2 heads in 3 flips” = $1 - P(\text{0 heads}) - P(\text{1 head})$
  • Right: $P(\text{2 heads}) + P(\text{3 heads}) = \binom{3}{2} \left(\frac{1}{2}\right)^3 + \binom{3}{3} \left(\frac{1}{2}\right)^3 = \frac{3}{8} + \frac{1}{8} = \frac{1}{2}$

Pitfall: Forgetting to account for overlaps in inclusion-exclusion

  • Wrong: “At least one of A, B, C” = $P(A) + P(B) + P(C)$
  • Right: $P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$

Pitfall: Confusing “at least” with “exactly”

  • Wrong: “At least 2 heads” = “Exactly 2 heads”
  • Right: “At least 2 heads” = “Exactly 2 heads” + “Exactly 3 heads”

🏆 AMC-Style Worked Example

Problem: What’s the probability that a random 5-card hand has at least one ace?

Solution:

  • Method 1: Complement counting

    • Total hands: $\binom{52}{5} = 2598960$
    • Hands with no aces: $\binom{48}{5} = 1712304$
    • Hands with at least one ace: $2598960 - 1712304 = 886656$
    • Probability: $\frac{886656}{2598960} = \frac{18472}{54145}$

  • Method 2: Direct counting

    • Exactly 1 ace: $\binom{4}{1}\binom{48}{4} = 4 \cdot 194580 = 778320$
    • Exactly 2 aces: $\binom{4}{2}\binom{48}{3} = 6 \cdot 17296 = 103776$
    • Exactly 3 aces: $\binom{4}{3}\binom{48}{2} = 4 \cdot 1128 = 4512$
    • Exactly 4 aces: $\binom{4}{4}\binom{48}{1} = 1 \cdot 48 = 48$
    • Total: $778320 + 103776 + 4512 + 48 = 886656$
    • Probability: $\frac{886656}{2598960} = \frac{18472}{54145}$

Key insight: Complement counting is often simpler than direct counting!

🔗 Related Topics

Next: Grid Paths → Urn, Coin & String Problems