Conditional Probability

What is Conditional Probability?

Conditional Probability: Probability of event A given that event B has occurred

Notation: P(A|B) (read: "probability of A given B")

Key insight: New information (B occurred) changes the probability of A

Conditional Probability Formula

$P(A|B) = \frac{P(A \cap B)}{P(B)}$

where P(B) > 0

Interpretation: Of all outcomes where B occurred, what fraction also have A?

Denominator P(B): Reduces sample space to just outcomes in B
Numerator P(A ∩ B): Outcomes in both A and B

Example 1: Two-Way Table

Survey of 100 students:

| | Male | Female | Total | |-----------|------|--------|-------| | Athlete | 25 | 15 | 40 | | Non-athlete| 35 | 25 | 60 | | Total | 60 | 40 | 100 |

Find P(Athlete|Male):

$P(\text{Athlete}|\text{Male}) = \frac{P(\text{Athlete and Male})}{P(\text{Male})} = \frac{25/100}{60/100} = \frac{25}{60} = \frac{5}{12}$

Interpretation: Of the 60 male students, 25 are athletes, so 25/60 = 5/12

Alternative approach: Restrict to males only (60 students), find fraction who are athletes (25/60)

Example 2: Cards

Draw one card from standard deck.

P(Ace|Red) = ?

• P(Red) = 26/52
• P(Ace and Red) = 2/52 (Ace of Hearts, Ace of Diamonds)
• P(Ace|Red) = (2/52)/(26/52) = 2/26 = 1/13

Interpretation: Of 26 red cards, 2 are aces

Rearranging the Formula

Multiplication Rule:

$P(A \cap B) = P(B) \times P(A|B)$

Also:

$P(A \cap B) = P(A) \times P(B|A)$

Use: Find probability of both events when you know conditional probability

Example: P(Draw 2 aces without replacement)

• P(First ace) = 4/52
• P(Second ace|First ace) = 3/51
• P(Both aces) = (4/52) × (3/51) = 12/2652 = 1/221

Independence Test

Events A and B are independent if:

$P(A|B) = P(A)$

Equivalently: P(B|A) = P(B)

Meaning: Knowing B occurred doesn't change probability of A

Example: Flip coin twice

• P(Second heads) = 1/2
• P(Second heads|First heads) = 1/2
• These are equal, so independent

Non-example: Cards without replacement

• P(Second ace) = 4/52 (before first draw)
• P(Second ace|First ace) = 3/51
• These differ, so NOT independent

Tree Diagrams for Conditional Probability

Example: Disease testing

• P(Disease) = 0.01
• P(Positive|Disease) = 0.95 (sensitivity)
• P(Positive|No disease) = 0.05 (false positive rate)

Find P(Positive):

Tree diagram:

• Branch 1: Disease (0.01) → Positive (0.95): 0.01 × 0.95 = 0.0095
• Branch 2: Disease (0.01) → Negative (0.05): 0.01 × 0.05 = 0.0005
• Branch 3: No disease (0.99) → Positive (0.05): 0.99 × 0.05 = 0.0495
• Branch 4: No disease (0.99) → Negative (0.95): 0.99 × 0.95 = 0.9405

P(Positive) = 0.0095 + 0.0495 = 0.059

Bayes' Theorem

Find P(B|A) when you know P(A|B):

$P(B|A) = \frac{P(A|B) \times P(B)}{P(A)}$

Example continued: Find P(Disease|Positive)

$P(\text{Disease}|\text{Positive}) = \frac{P(\text{Positive}|\text{Disease}) \times P(\text{Disease})}{P(\text{Positive})}$

$= \frac{0.95 \times 0.01}{0.059} = \frac{0.0095}{0.059} \approx 0.161$

Interpretation: Even with positive test, only 16.1% chance of having disease (because disease is rare!)

Common Two-Way Table Calculations

Given table with events A and B:

Joint probability: P(A and B) = (count in both)/(total)

Marginal probability: P(A) = (row/column total)/(grand total)

Conditional probability: P(A|B) = (count in both)/(count in B)

Conditional Probability Notation

P(A|B) ≠ P(B|A) (usually)

Example:

• P(Positive test|Disease) = 0.95 (sensitivity)
• P(Disease|Positive test) = 0.161 (very different!)

Always read carefully and identify which event is the condition!

Applications

Medical testing: P(Disease|Positive test)
Quality control: P(Defective|From certain machine)
Weather: P(Rain tomorrow|Rain today)
Sports: P(Win|Home game)

Common Mistakes

❌ Confusing P(A|B) with P(B|A)
❌ Assuming conditional independence means independence
❌ Forgetting to restrict to condition when calculating from table
❌ Using wrong denominator in formula

Practice Approach

1. Identify condition: What do we know occurred?
2. Restrict sample space: Consider only outcomes where condition is true
3. Find fraction: Of those outcomes, what fraction satisfies event?
4. Check: P(A|B) should be between 0 and 1

Quick Reference

Definition: $P(A|B) = \frac{P(A \cap B)}{P(B)}$

Multiplication Rule: $P(A \cap B) = P(B) \times P(A|B)$

Independence Test: P(A|B) = P(A)

Bayes' Theorem: $P(B|A) = \frac{P(A|B) \times P(B)}{P(A)}$

Remember: Conditional probability is about updating probabilities based on new information!