Stats1_Week8_Conditional_Probability

Stats 1 Week 8: Conditional Probability & Bayes’ Theorem

0. Prerequisites

NOTE

What you need to know:

• Intersection ($P(A\cap B)$): Probability of A AND B.
• Partitions: Splitting a sample space into non-overlapping chunks.

Quick Refresher

• Conditional Probability $P(A|B)$: Prob of A given B happened.
• Formula: $P(A|B)=\frac{P(A\cap B)}{P(B)}$.
• Bayes’ Theorem: Flipping the condition. Finding $P(B|A)$ using $P(A|B)$.

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1. Core Concepts

1.1 Conditional Probability

• Concept: The sample space shrinks. If we know B happened, B becomes the new “Universe”.
• Multiplication Rule: $P(A\cap B)=P(A|B)\times P(B)$.

1.2 Independence (Formal Definition)

• Two events A and B are independent IF AND ONLY IF:
  1. $P(A|B)=P(A)$ (Knowing B tells you nothing about A).
  2. $P(A\cap B)=P(A)\times P(B)$.

1.3 Bayes’ Theorem

• Scenario: You know $P(\text{Effect}|\text{Cause})$. You want $P(\text{Cause}|\text{Effect})$.
• Formula: $P(A|B)=\frac{P(B|A)\cdot P(A)}{P(B)}$
• Total Probability (Denominator): $P(B)=P(B|A)P(A)+P(B|A^c)P(A^c)$.

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2. Pattern Analysis & Goated Solutions

Pattern 1: The “Given” Keyword (Basic Conditional)

Context: “Roll 2 dice. Sum is 8. Prob that one die is 5?”

TIP

Mental Algorithm:

1. Identify Condition (B): “Sum is 8”. List outcomes.
   • $B=\{(2,6),(3,5),(4,4),(5,3),(6,2)\}$. Count = 5.
2. Identify Target (A): “One die is 5”.
3. Find Overlap ($A\cap B$): Which outcomes in B have a 5?
   • $\{(3,5),(5,3)\}$. Count = 2.
4. Divide: $\frac{\text{Overlap Count}}{\text{Condition Count}}=2/5$.

Pattern 2: Bayes’ Theorem (Tree Diagram Method)

Context: “Disease D (1% of pop). Test T is 99% accurate. You test Positive. Prob you have Disease?”

TIP

Mental Algorithm:

1. Draw Tree:
   • Branch 1: Disease ($0.01$) $\to$ Pos ($0.99$) / Neg ($0.01$).
   • Branch 2: No Disease ($0.99$) $\to$ Pos ($0.01$) / Neg ($0.99$).
2. Identify Goal: $P(D|Pos)$.
3. Numerator: Path “Disease AND Pos”.
   • $0.01\times 0.99=0.0099$.
4. Denominator: All “Pos” paths.
   • (Disease AND Pos) + (No Disease AND Pos).
   • $0.0099+(0.99\times 0.01)=0.0099+0.0099=0.0198$.
5. Divide: $0.0099/0.0198=0.5$.
   • Result: Only 50%! (Counter-intuitive).

Pattern 3: Independence Check

Context: “$P(A)=0.4,P(B)=0.5,P(A\cup B)=0.7$. Are A and B independent?”

TIP

Mental Algorithm:

1. Find Intersection: Use Addition Rule.
   • $0.7=0.4+0.5-P(A\cap B)⟹P(A\cap B)=0.2$.
2. Check Product: Calculate $P(A)\times P(B)$.
   • $0.4\times 0.5=0.2$.
3. Compare: Does Intersection = Product?
   • $0.2=0.2$. Yes.
   • Result: Independent.

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3. Practice Exercises

1. Conditional: $P(A\cap B)=0.1,P(B)=0.5$. Find $P(A|B)$.
   • Hint: $0.1/0.5=0.2$.
2. Independence: If A, B independent. $P(A)=0.5$. What is $P(A|B)$?
   • Hint: It’s just $P(A)=0.5$. B doesn’t matter.
3. Bayes: 2 Urns. U1(2R, 3B), U2(3R, 2B). Pick Urn (0.5), Pick Red. Prob it was U1?
   • Hint: Num: $0.5\times (2/5)$. Denom: $0.5(2/5)+0.5(3/5)$. Ans: $2/5$.

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🧠 Level Up: Advanced Practice

Question 1: Bayes’ Theorem (Medical Test)

Problem: Disease prevalence 1%. Test sensitivity 99% (True Positive), Specificity 95% (True Negative). If test is positive, prob of disease? Logic:

1. Events: $D$ (Disease), $T$ (Test Positive).
2. Priors: $P(D)=0.01,P(D^{\prime })=0.99$.
3. Conditionals: $P(T|D)=0.99,P(T|D^{\prime })=0.05$ (False Positive).
4. Bayes: $P(D|T)=\frac{P(T|D)P(D)}{P(T|D)P(D)+P(T|D^{\prime })P(D^{\prime })}$.
5. Calc:
   • Num: $0.99\times 0.01=0.0099$.
   • Denom: $0.0099+(0.05\times 0.99)=0.0099+0.0495=0.0594$.
   • Result: $0.0099/0.0594=1/6\approx 16.6\%$. Answer: ~16.7%. (Counter-intuitive! Even with positive test, mostly likely healthy).

Question 2: Independence Check

Problem: Two dice. A: Sum is 7. B: First die is 4. Are A and B independent? Logic:

1. P(A): Sum 7 pairs (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Total 6/36 = 1/6.
2. P(B): First die 4. Pairs (4,1)…(4,6). Total 6/36 = 1/6.
3. Intersection (A and B): First is 4 AND Sum is 7. Only (4,3). Prob 1/36.
4. Check: $P(A)\times P(B)=1/6\times 1/6=1/36$.
5. Result: $P(A\cap B)=P(A)P(B)$. Answer: Yes, Independent.

Question 3: Conditional Probability Trap

Problem: Family has 2 children. Given at least one is a girl, prob that both are girls? Logic:

1. Sample Space: BB, BG, GB, GG. (Prob 1/4 each).
2. Condition: “At least one girl” → {BG, GB, GG}. (3 cases).
3. Event: “Both girls” → {GG}. (1 case).
4. Result: 1/3. (Not 1/2!). Answer: 1/3.