🎲 Urn, Coin & String Problems

Classic probability problems involving urns, coins, and strings with replacement and no-replacement scenarios.

🎯 Recognition Cues

• Keywords: “urn”, “bag”, “drawing”, “with replacement”, “without replacement”, “coin”, “flip”, “run”
• Setup: Drawing objects from containers or flipping coins
• Constraints: Replacement vs. no-replacement, specific sequences, runs

📋 Solution Template

1. Identify the scenario type:

   • With replacement: Each draw is independent
   • Without replacement: Each draw affects the next
   • Sequential: Order matters

2. Apply the appropriate method:

   • With replacement: Use independence and basic probability
   • Without replacement: Use conditional probability
   • Sequential: Use multiplication principle

3. Check for common patterns:

   • Binomial distribution (with replacement)
   • Hypergeometric distribution (without replacement)
   • Geometric distribution (until first success)

💡 Micro-Examples

With Replacement

• Problem: What’s the probability of drawing 2 red balls from an urn with 3 red and 2 blue balls, with replacement?
• Solution: $P(\text{red}) \cdot P(\text{red}) = \frac{3}{5} \cdot \frac{3}{5} = \frac{9}{25}$

Without Replacement

• Problem: What’s the probability of drawing 2 red balls from an urn with 3 red and 2 blue balls, without replacement?
• Solution: $P(\text{first red}) \cdot P(\text{second red} \mid \text{first red}) = \frac{3}{5} \cdot \frac{2}{4} = \frac{3}{10}$

Coin Runs

• Problem: What’s the probability of getting exactly 2 heads in a row in 4 coin flips?
• Solution: Count sequences: HHTT, THHT, TTHH, HHTH, THHH, HHHH. Probability: $\frac{6}{16} = \frac{3}{8}$

⚠️ Common Pitfalls & Variants

Pitfall: Confusing with and without replacement

• Wrong: “Drawing 2 red balls” = Always use the same probability
• Right: Check if replacement is mentioned. Without replacement changes the probabilities.

Pitfall: Forgetting about order in sequential problems

• Wrong: “Getting heads then tails” = “Getting tails then heads”
• Right: These are different events with potentially different probabilities.

Pitfall: Misapplying distribution formulas

• Wrong: “Drawing 2 red balls without replacement” = Binomial distribution
• Right: Use hypergeometric distribution for sampling without replacement.

🏆 AMC-Style Worked Example

Problem: An urn contains 5 red balls and 3 blue balls. You draw 3 balls without replacement. What’s the probability of getting exactly 2 red balls?

Solution:

• Method 1: Direct counting

  • Total ways to draw 3 balls: $\binom{8}{3} = 56$
  • Ways to draw 2 red and 1 blue: $\binom{5}{2} \cdot \binom{3}{1} = 10 \cdot 3 = 30$
  • Probability: $\frac{30}{56} = \frac{15}{28}$

• Method 2: Sequential probability

  • R-R-B: $\frac{5}{8} \cdot \frac{4}{7} \cdot \frac{3}{6} = \frac{60}{336} = \frac{5}{28}$
  • R-B-R: $\frac{5}{8} \cdot \frac{3}{7} \cdot \frac{4}{6} = \frac{60}{336} = \frac{5}{28}$
  • B-R-R: $\frac{3}{8} \cdot \frac{5}{7} \cdot \frac{4}{6} = \frac{60}{336} = \frac{5}{28}$
  • Total: $\frac{5}{28} + \frac{5}{28} + \frac{5}{28} = \frac{15}{28}$

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

🔗 Related Topics

• Probability Basics - Foundation for these problems
• Distributions - Binomial and hypergeometric distributions
• Conditional Chains - Advanced sequential problems

Next: Expected Counts → Conditional Chains