đź”— Conditional Chains

Complex probability problems involving sequences of conditional events and Bayes’ theorem.

🎯 Recognition Cues

  • Keywords: “given that”, “if”, “conditional”, “total probability”, “Bayes”
  • Setup: Multiple conditional events in sequence
  • Constraints: Dependencies between events, information updates

đź“‹ Solution Template

  1. Identify the chain structure:

    • Sequential: Events happen in order
    • Parallel: Events happen simultaneously
    • Mixed: Combination of sequential and parallel

  2. Apply the appropriate method:

    • Sequential: Use multiplication principle
    • Parallel: Use law of total probability
    • Mixed: Combine both methods

  3. Check for Bayes’ theorem:

    • Forward: Given cause, find effect probability
    • Backward: Given effect, find cause probability

đź’ˇ Micro-Examples

Sequential Chain

  • Problem: What’s the probability of getting heads, then tails, then heads in 3 coin flips?
  • Solution: $P(H) \cdot P(T \mid H) \cdot P(H \mid H,T) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$

Total Probability

  • Problem: What’s the probability of rain given that 60% of days are cloudy and it rains on 30% of cloudy days?
  • Solution: $P(\text{rain}) = P(\text{rain} \mid \text{cloudy}) \cdot P(\text{cloudy}) + P(\text{rain} \mid \text{not cloudy}) \cdot P(\text{not cloudy}) = 0.3 \cdot 0.6 + 0.1 \cdot 0.4 = 0.22$

Bayes’ Theorem

  • Problem: Given that it’s raining, what’s the probability it was cloudy?
  • Solution: $P(\text{cloudy} \mid \text{rain}) = \frac{P(\text{rain} \mid \text{cloudy}) \cdot P(\text{cloudy})}{P(\text{rain})} = \frac{0.3 \cdot 0.6}{0.22} = \frac{0.18}{0.22} = \frac{9}{11}$

⚠️ Common Pitfalls & Variants

Pitfall: Confusing conditional and joint probability

  • Wrong: “Given A, what’s the probability of B?” = $P(A \cap B)$
  • Right: $P(B \mid A) = \frac{P(A \cap B)}{P(A)}$

Pitfall: Forgetting to normalize in Bayes’ theorem

  • Wrong: “Given rain, probability of cloudy” = $P(\text{rain} \mid \text{cloudy}) \cdot P(\text{cloudy})$
  • Right: $\frac{P(\text{rain} \mid \text{cloudy}) \cdot P(\text{cloudy})}{P(\text{rain})}$

Pitfall: Misapplying total probability

  • Wrong: “Total probability of rain” = $P(\text{rain} \mid \text{cloudy}) + P(\text{rain} \mid \text{not cloudy})$
  • Right: $P(\text{rain} \mid \text{cloudy}) \cdot P(\text{cloudy}) + P(\text{rain} \mid \text{not cloudy}) \cdot P(\text{not cloudy})$

🏆 AMC-Style Worked Example

Problem: A bag contains 3 red balls and 2 blue balls. You draw 2 balls without replacement. Given that the first ball is red, what’s the probability that the second ball is also red?

Solution:

  • Step 1: After drawing 1 red ball, the bag has 2 red and 2 blue balls
  • Step 2: $P(\text{second red} \mid \text{first red}) = \frac{2}{4} = \frac{1}{2}$

Alternative approach using Bayes’ theorem:

  • Step 1: $P(\text{first red}) = \frac{3}{5}$
  • Step 2: $P(\text{both red}) = \frac{3}{5} \cdot \frac{2}{4} = \frac{3}{10}$
  • Step 3: $P(\text{second red} \mid \text{first red}) = \frac{P(\text{both red})}{P(\text{first red})} = \frac{3/10}{3/5} = \frac{1}{2}$

Key insight: Sometimes the direct approach is simpler than using Bayes’ theorem!

đź”— Related Topics

Next: Derangements → Circular Permutations