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")

Meaning: Adjust probability based on new information

Formula:

P(A|B) = P(A and B) / P(B) where P(B) > 0

Understanding P(A|B)

P(A|B) ≠ P(A) (usually)

B happening changes the sample space

Focus only on outcomes where B occurred

Example 1: Card Draw

Draw card from standard deck

A: Card is ace B: Card is spade

Find P(A|B):

P(A) = 4/52 (without condition)

But given B (card is spade):

• Sample space reduced to 13 spades
• Only 1 ace of spades

P(A|B) = 1/13

Example 2: Two Dice

Roll two dice

A: Sum is 8 B: First die shows 3

Find P(A|B):

Given first die is 3, need second die to be 5 for sum of 8

P(A|B) = 1/6 (second die must be 5)

Using the Formula

P(A|B) = P(A and B) / P(B)

Example: Family with Two Children

Family has 2 children, at least one is a girl (B)

What's probability both are girls (A)?

Sample space: (GG, GB, BG, BB)

Event B (at least one girl): (GG, GB, BG) P(B) = 3/4

Event (A and B) (both girls): (GG) P(A and B) = 1/4

Calculate:

P(A|B) = (1/4) / (3/4) = 1/3

Answer: 1/3

(Not 1/2 as might be expected!)

Multiplication Rule

Rearrange conditional probability formula:

P(A and B) = P(B) · P(A|B)

Or: P(A and B) = P(A) · P(B|A)

Use: Find probability of both events

Example: Drawing Without Replacement

Bag: 3 red, 2 blue marbles

Draw 2 without replacement

P(both red)?

First draw: P(red) = 3/5

Second draw given first red: P(red|first red) = 2/4

P(both red):

= P(first red) · P(second red | first red)
= (3/5) · (2/4)
= 6/20
= 3/10

Answer: 3/10

Independent Events

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

B doesn't affect probability of A

Test for independence:

A and B independent if and only if: P(A and B) = P(A) · P(B)

Example 1: Coin Flips

Flip coin twice

A: First flip heads B: Second flip heads

P(A) = 1/2 P(B) = 1/2 P(A and B) = 1/4

Check: (1/2)(1/2) = 1/4 ✓

Independent

Example 2: Cards With Replacement

Draw card, replace, draw again

A: First is ace B: Second is ace

P(A) = 4/52 P(B) = 4/52 (replaced, so same) P(A and B) = (4/52)² = 16/2704

Check: (4/52)(4/52) = 16/2704 ✓

Independent

Example 3: NOT Independent

Draw 2 cards without replacement

A: First is ace B: Second is ace

P(A) = 4/52 P(B|A) = 3/51 (only 3 aces left) P(A and B) = (4/52)(3/51)

Check independence: (4/52)(4/52) ≠ (4/52)(3/51)

NOT independent (dependent)

Tree Diagrams for Conditional Probability

Visual tool showing all paths and probabilities

Example: Medical Test

Disease: 1% of population Test: 95% accurate if disease, 90% accurate if no disease

P(D) = 0.01 (disease) P(no D) = 0.99

Tree:

       P(+|D)=0.95 ─ D and + : 0.01(0.95) = 0.0095
D=0.01
       P(-|D)=0.05 ─ D and - : 0.01(0.05) = 0.0005

           P(+|no D)=0.10 ─ no D and + : 0.99(0.10) = 0.099
no D=0.99
           P(-|no D)=0.90 ─ no D and - : 0.99(0.90) = 0.891

P(positive test) = 0.0095 + 0.099 = 0.1085

Bayes' Theorem

Reverse conditional probability

Formula:

P(A|B) = [P(B|A) · P(A)] / P(B)

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

Example: Disease Test (continued)

Person tests positive. Probability they have disease?

Want: P(D|+)

Know:

• P(+|D) = 0.95
• P(D) = 0.01
• P(+) = 0.1085

Bayes' Theorem:

P(D|+) = [P(+|D) · P(D)] / P(+)
       = (0.95 · 0.01) / 0.1085
       = 0.0095 / 0.1085
       ≈ 0.0876
       ≈ 8.8%

Only 8.8% chance of having disease despite positive test!

(Because disease is rare)

False Positives and False Negatives

False positive: Test positive, but no disease

False negative: Test negative, but have disease

From example:

• P(false positive) = P(+|no D) = 0.10
• P(false negative) = P(-|D) = 0.05

Law of Total Probability

If events B₁, B₂, ..., Bₙ partition sample space:

P(A) = P(A|B₁)P(B₁) + P(A|B₂)P(B₂) + ... + P(A|Bₙ)P(Bₙ)

Use: Find P(A) by considering all cases

Example: Jar Selection

Jar 1: 2 red, 3 blue Jar 2: 4 red, 1 blue

Pick jar randomly, then pick marble

P(red)?

P(Jar 1) = 1/2, P(Jar 2) = 1/2

P(red|Jar 1) = 2/5 P(red|Jar 2) = 4/5

Total:

P(red) = P(red|Jar1)·P(Jar1) + P(red|Jar2)·P(Jar2)
       = (2/5)(1/2) + (4/5)(1/2)
       = 1/5 + 2/5
       = 3/5

Answer: 3/5

Real-World Applications

Medical testing: Interpreting test results

Quality control: Defect detection

Machine learning: Spam filters, classification

Insurance: Risk assessment

Criminal justice: DNA evidence interpretation

Example: Spam Filter

P(spam) = 0.3 (30% of emails are spam) P(word "free" | spam) = 0.8 P(word "free" | not spam) = 0.1

Email contains "free". Probability it's spam?

First, find P("free"):

P("free") = 0.8(0.3) + 0.1(0.7)
          = 0.24 + 0.07
          = 0.31

Bayes' theorem:

P(spam | "free") = [P("free"|spam) · P(spam)] / P("free")
                 = (0.8 · 0.3) / 0.31
                 = 0.24 / 0.31
                 ≈ 0.774
                 ≈ 77.4%

77.4% chance it's spam

Common Mistakes

Don't confuse P(A|B) with P(B|A)

Example: P(rain|clouds) ≠ P(clouds|rain)

Don't assume independence without checking

Remember to reduce sample space for conditional probability

Practice