Conditional Probability

📐 Basic Probability

Part 1 of 7 — Foundations of Probability

What Is Probability?

Probability measures how likely an event is to occur, expressed as a number between 0 and 1.

$P(A) = \frac{\text{number of favorable outcomes}}{\text{total number of equally likely outcomes}}$

Key Vocabulary

Complement Rule

$P(A\text{' }) = 1 - P(A)$

🔑 Key Insight: An event and its complement always sum to 1. This is often the fastest way to solve "at least one" problems.

Worked Example 1: Rolling a Die

Find $P(\text{even})$ on a fair six-sided die.

• Sample space: $S = \{1, 2, 3, 4, 5, 6\}$ → 6 outcomes
• Favorable outcomes: $\{2, 4, 6\}$ → 3 outcomes

$P(\text{even}) = \frac{3}{6} = 0.5$

Worked Example 2: Using the Complement

A bag has 10 marbles: 3 red, 7 blue. Find $P(\text{not red})$.

$P(\text{not red}) = 1 - P(\text{red}) = 1 - \frac{3}{10} = \frac{7}{10} = 0.7$

Interpreting Probability

🔑 Law of Large Numbers: As the number of trials increases, the relative frequency approaches the true probability.

Basic Probability Concepts 🎯

Calculating Probabilities 🧮

1) $P(\text{heads})$ on a fair coin:

2) A bag has 5 red and 15 blue marbles. $P(\text{red}) =$ ? (as decimal)

3) Using the complement: $P(\text{not red}) =$ ?

Probability Vocabulary 🔍

Exit Quiz — Basic Probability ✅

📊 Addition Rule

Part 2 of 7 — "Or" Probabilities

The General Addition Rule

The probability that event $A$ or event $B$ (or both) occurs:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

🔑 Why subtract $P(A \cap B)$? Because the overlap is counted once in $P(A)$ and again in $P(B)$. Subtracting avoids double-counting.

Venn Diagram View

Mutually Exclusive Events

Two events are mutually exclusive (disjoint) if they cannot both occur: $P(A \cap B) = 0$.

For mutually exclusive events, the addition rule simplifies to:

$P(A \cup B) = P(A) + P(B)$

Examples:

• Rolling a 2 and rolling a 5 on one die → mutually exclusive
• Choosing one student who is a freshman and choosing one who is a senior → mutually exclusive
• Drawing a heart and drawing a spade → mutually exclusive

NOT mutually exclusive:

• Drawing a heart and drawing a king (king of hearts exists!)

Worked Example 1: General Addition Rule

$P(A) = 0.4$, $P(B) = 0.3$, $P(A \cap B) = 0.1$. Find $P(A \cup B)$.

$P(A \cup B) = 0.4 + 0.3 - 0.1 = 0.6$

Worked Example 2: Two-Way Table

A class of 200 students:

Find $P(\text{Male OR Freshman})$:

$P(M \cup F) = P(M) + P(F) - P(M \cap F) = \frac{80}{200} + \frac{110}{200} - \frac{50}{200} = \frac{140}{200} = 0.70$

🔑 Check: Count directly: 80 males + 60 female freshmen = 140 out of 200 = 0.70 ✓

Addition Rule Concepts 🎯

Addition Rule Calculations 🧮

1) $P(A) = 0.4$, $P(B) = 0.3$, $P(A \cap B) = 0.1$. $P(A \cup B) =$ ?

2) $P(A) = 0.6$, $P(B) = 0.4$, $P(A \cap B) = 0.2$. $P(A \cup B) =$ ?

3) $P(A \cup B) = 0.9$, $P(A) = 0.5$, $P(B) = 0.6$. $P(A \cap B) =$ ?

Key Concepts 🔍

Exit Quiz — Addition Rule ✅

🔢 Multiplication Rule

Part 3 of 7 — "And" Probabilities

The General Multiplication Rule

The probability that both $A$ and $B$ occur:

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

where $P(B | A)$ means "the probability of $B$ given that $A$ has occurred."

Independent Events (Special Case)

If events are independent (one doesn't affect the other):

$P(A \cap B) = P(A) \cdot P(B)$

🔑 Key Insight: For independent events, you just multiply. For dependent events, you must account for how the first event changes the second.

Independent vs Dependent

Worked Example 1: Independent Events

Two coin flips. Find $P(HH)$.

Flips are independent: $P(HH) = P(H) \cdot P(H) = 0.5 \times 0.5 = 0.25$

Worked Example 2: Dependent Events (Without Replacement)

A bag has 4 red and 6 blue marbles. Draw 2 without replacement. Find $P(\text{both red})$.

$P(R_1 \cap R_2) = P(R_1) \cdot P(R_2 | R_1) = \frac{4}{10} \cdot \frac{3}{9} = \frac{12}{90} = \frac{2}{15} \approx 0.133$

After removing one red marble, only 3 red remain out of 9 total.

Extending to Multiple Events

For three independent events: $P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C)$

Example: Probability of 3 heads in a row: $P(HHH) = 0.5^3 = 0.125$

"At Least One" Using the Complement

$P(\text{at least one}) = 1 - P(\text{none})$

Example: Roll a die 3 times. $P(\text{at least one 6})$: $P(\text{at least one 6}) = 1 - P(\text{no 6s}) = 1 - \left(\frac{5}{6}\right)^3 = 1 - \frac{125}{216} \approx 0.421$

Multiplication Rule Concepts 🎯

Computing "And" Probabilities 🧮

For independent events:

1) $P(A) = 0.5$, $P(B) = 0.5$. $P(A \cap B) =$ ?

2) $P(A) = 0.2$, $P(B) = 0.5$. $P(A \cap B) =$ ?

3) $P(\text{at least one head in 2 flips}) = 1 - P(TT) =$ ?

Independent vs Dependent 🔍

Exit Quiz — Multiplication Rule ✅

📈 Conditional Probability

Part 4 of 7 — "Given That" Probabilities

What Is Conditional Probability?

Conditional probability is the probability of event $A$ occurring, given that event $B$ has already occurred:

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

🔑 Key Insight: Knowing $B$ occurred restricts the sample space to only outcomes where $B$ happened.

Reading Conditional Probability

⚠️ Warning: $P(A|B) \neq P(B|A)$ in general! The order matters.

Worked Example 1: Formula

$P(A \cap B) = 0.12$, $P(B) = 0.4$. Find $P(A|B)$.

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.12}{0.4} = 0.3$

Worked Example 2: Two-Way Table

Survey of 500 students:

Find $P(\text{Sport} | \text{Male})$:

$P(\text{Sport}|\text{Male}) = \frac{120}{200} = 0.60$

We restrict to males only (200), then count how many play a sport (120).

Find $P(\text{Male} | \text{Sport})$:

$P(\text{Male}|\text{Sport}) = \frac{120}{220} \approx 0.545$

We restrict to sport players (220), then count how many are male (120).

Notice: $P(\text{Sport}|\text{Male}) \neq P(\text{Male}|\text{Sport})$!

Connection to the Multiplication Rule

Rearranging the conditional probability formula:

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

This IS the general multiplication rule!

Reversing the Condition with a Tree Diagram

Sometimes you know $P(B|A)$ but need $P(A|B)$. Draw a tree, then use the conditional probability formula.

Example: 2% of people have a disease. A test is positive for 95% of people with the disease and for 10% of people without it.

$P(\text{disease}|\text{positive}) = \frac{P(\text{disease} \cap \text{positive})}{P(\text{positive})} = \frac{0.019}{0.117} \approx 0.162$

🔑 Only about 16% of positive tests come from people with the disease, because the disease is rare. The tree plus $P(A|B) = P(A \cap B)/P(B)$ is all you need; no separate formula is required.

Conditional Probability Concepts 🎯

Computing Conditional Probabilities 🧮

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

1) $P(A \cap B) = 0.12$, $P(B) = 0.4$. $P(A|B) =$ ?

2) $P(A \cap B) = 0.06$, $P(B) = 0.2$. $P(A|B) =$ ?

3) From a table: 80 males, 50 play a sport. $P(\text{Sport}|\text{Male}) =$ ? (as decimal)

Conditional vs Joint Probability 🔍

Exit Quiz — Conditional Probability ✅

🧮 Independence

Part 5 of 7 — Testing for Independence

What Does Independence Mean?

Events $A$ and $B$ are independent if knowing one occurred does NOT change the probability of the other.

Two equivalent tests:

$P(A|B) = P(A) \qquad \text{or} \qquad P(A \cap B) = P(A) \cdot P(B)$

If either equation holds, events are independent. If not, they are dependent.

How to Check Independence

Independence vs Mutually Exclusive

These concepts are DIFFERENT and often confused:

🔑 Critical Fact: If $A$ and $B$ are mutually exclusive (and both have $P > 0$), they are ALWAYS dependent. If $A$ happened, $B$ definitely didn't!

Worked Example 1: Multiplication Check

$P(A) = 0.4$, $P(B) = 0.5$, $P(A \cap B) = 0.2$. Independent?

$P(A) \cdot P(B) = 0.4 \times 0.5 = 0.2 = P(A \cap B) \checkmark$

Yes, independent! The product equals the intersection.

Worked Example 2: Two-Way Table

200 employees:

Are promotion and degree independent?

$P(\text{Promoted}|\text{Degree}) = \frac{30}{150} = 0.20$ $P(\text{Promoted}|\text{No Degree}) = \frac{10}{50} = 0.20$

Since $P(\text{Promoted}|\text{Degree}) = P(\text{Promoted}|\text{No Degree}) = 0.20$, the events are independent. Having a degree doesn't affect the promotion rate.

Independence in Sampling

Independence Concepts 🎯

Testing Independence 🧮

Compute $P(A) \times P(B)$ and compare with $P(A \cap B)$:

1) $P(A) = 0.4$, $P(B) = 0.5$. $P(A) \times P(B) =$ ?

2) $P(A) = 0.3$, $P(B) = 0.6$. $P(A) \times P(B) =$ ?

3) $P(A) = 0.5$, $P(B) = 0.8$. $P(A) \times P(B) =$ ?

Key Distinctions 🔍

Exit Quiz — Independence ✅

🛠️ Problem-Solving Workshop

Part 6 of 7 — Combining Probability Rules

Strategy: Which Rule Do I Use?

Worked Example 1: "At Least One" with Complement

A factory produces items with a 5% defect rate. In a batch of 3 independent items, what is $P(\text{at least one defective})$?

Step 1: Find $P(\text{not defective}) = 1 - 0.05 = 0.95$

Step 2: $P(\text{none defective in 3}) = 0.95^3 = 0.857375$

Step 3: $P(\text{at least one}) = 1 - 0.857375 = 0.1426$

$\boxed{P(\text{at least one defective}) \approx 0.143}$

Worked Example 2: Combining Multiple Rules

In a class: 60% study math, 40% study science, 20% study both. A student is chosen. Given they study math, what is $P(\text{science})$?

Step 1: Identify: This is $P(\text{Science}|\text{Math})$ → conditional probability

Step 2: Apply formula: $P(\text{Sci}|\text{Math}) = \frac{P(\text{Sci} \cap \text{Math})}{P(\text{Math})} = \frac{0.20}{0.60} = 0.333$

Worked Example 3: Sequential Events with Dependence

A bag has 5 red and 3 blue marbles. Draw 2 without replacement. $P(\text{both red})$?

Step 1: $P(\text{1st red}) = \frac{5}{8}$

Step 2: After removing a red: $P(\text{2nd red}|\text{1st red}) = \frac{4}{7}$

Step 3: Multiply (general rule): $P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14} \approx 0.357$

Decision Flowchart

$\text{Read the problem} \to \begin{cases} \text{"or" / "either"} &\to \text{Addition Rule} \\ \text{"and" / "both"} &\to \text{Multiplication Rule} \\ \text{"given" / "if"} &\to \text{Conditional} \\ \text{"at least one"} &\to \text{Complement} \end{cases}$

🎯 Pro tip: "At least one" problems are almost ALWAYS easier with the complement: $1 - P(\text{none})$.

Choosing the Right Rule 🎯

Multi-Rule Calculations 🧮

1) $P(\text{defect}) = 0.1$, items independent. $P(\text{at least one defect in 3 items}) =$ ? (Use: $1 - P(\text{none}) = 1 - 0.9^3$. Round to 3 decimal places.)

2) Bag: 4 red, 6 blue. Draw 2 without replacement. $P(\text{both blue}) =$ ? (Round to 3 decimal places.)

3) $P(A) = 0.5$, $P(B) = 0.3$, $P(A \cap B) = 0.15$. $P(A|B) =$ ?

Rule Selection 🔍

Exit Quiz — Problem-Solving Workshop ✅

🏆 Review & Applications

Part 7 of 7 — Comprehensive Probability Review

Complete Formula Reference

Key Relationships

$\text{Mutually Exclusive:} \quad P(A \cap B) = 0$ $\text{Independent:} \quad P(A \cap B) = P(A) \cdot P(B)$ $\text{Complementary:} \quad P(A) + P(A\text{' }) = 1$

🔑 Remember: Mutually exclusive events with $P > 0$ are NEVER independent!

Comprehensive Worked Example

A survey of 500 adults:

a) $P(\text{Healthy Weight})$ $P(\text{HW}) = \frac{300}{500} = 0.60$

b) $P(\text{Exercises AND Healthy Weight})$ $P(\text{E} \cap \text{HW}) = \frac{180}{500} = 0.36$

c) $P(\text{Exercises OR Healthy Weight})$ $P(\text{E} \cup \text{HW}) = P(\text{E}) + P(\text{HW}) - P(\text{E} \cap \text{HW}) = 0.48 + 0.60 - 0.36 = 0.72$

d) $P(\text{Healthy Weight} | \text{Exercises})$ $P(\text{HW}|\text{E}) = \frac{P(\text{HW} \cap \text{E})}{P(\text{E})} = \frac{0.36}{0.48} = 0.75$

e) Are Exercise and Healthy Weight independent? $P(\text{HW}) = 0.60 \neq 0.75 = P(\text{HW}|\text{E})$ No — exercising is associated with higher rate of healthy weight.

Common Mistakes to Avoid

Formula Identification 🎯

Mixed Calculations 🧮

1) $P(A) = 0.4$, $P(B) = 0.3$, independent. $P(A \cap B) =$ ?

2) $P(A) = 0.6$, $P(B) = 0.5$, $P(A \cap B) = 0.3$. $P(A \cup B) =$ ?

3) $P(\text{pass}) = 0.9$, 2 independent attempts. $P(\text{at least one pass}) =$ ?

Concept Review 🔍

Exit Quiz — Comprehensive Review ✅