🎯⭐ INTERACTIVE LESSON

Conditional Probability

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Conditional Probability - Complete Interactive Lesson

Part 1: Basic 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.

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

Probability	Meaning
$P(A) = 0$	Impossible — can never happen
$P(A) = 1$	Certain — always happens
$0 < P(A) < 1$

Key Vocabulary

Term	Symbol	Meaning
Sample space	$S$	Set of ALL possible outcomes
Event	$A$ , $B$ , etc.	A subset of the sample space
Complement	$A\text{'}$ or

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

Interpretation	Description
Classical	Equally likely outcomes (dice, coins, cards)
Relative frequency	Long-run proportion from many trials
Subjective	Personal belief based on experience

🔑 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 ✅

Part 2: Addition Rule

📊 Addition Rule

Part 2 of 7 — "Or" Probabilities

The General Addition Rule

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

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

Part 3: Multiplication Rule

🔢 Multiplication Rule

Part 3 of 7 — "And" Probabilities

The General Multiplication Rule

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

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

Part 4: Conditional Probability

📈 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:

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

Part 5: Independence

🧮 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:

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

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop

Part 6 of 7 — Combining Probability Rules

Strategy: Which Rule Do I Use?

Key Word / Phrase	Rule	Formula
"or", "either", "at least one of"	Addition	$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Part 7: Review & Applications

🏆 Review & Applications

Part 7 of 7 — Comprehensive Probability Review

Complete Formula Reference

Rule	Formula	When to Use
Complement	$P(A\text{' }) = 1 - P(A)$	"not", "fails", "none"
Addition (General)

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

Region	Represents	Probability
Left only	$A$ but not $B$	$P(A) - P(A \cap B)$
Overlap	$A$ and $B$	$P(A \cap B)$
Right only	$B$ but not $A$	$P(B) - P(A \cap B)$
Outside both	Neither $A$ nor $B$	$1 - P(A \cup B)$

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:

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

Rolling a 2 and rolling a 5 on one die → mutually exclusive
Being male and being female → 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:

	Freshman	Sophomore	Total
Male	50	30	80
Female	60	60	120
Total	110	90	200

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

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

Key Concepts 🔍

Exit Quiz — Addition Rule ✅

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

Scenario	Type	Why
Flip a coin, then roll a die	Independent	Coin doesn't affect die
Draw 2 cards WITH replacement	Independent	Deck is reset each time
Draw 2 cards WITHOUT replacement	Dependent	Deck shrinks after 1st draw
Select 2 people from a small group	Dependent	Pool changes after 1st selection

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 ✅

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

Reading Conditional Probability

Notation	Read As
$P(A	B)$
$P(B	A)$

⚠️ 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:

	Plays Sport	No Sport	Total
Male	120	80	200
Female	100	200	300
Total	220	280	500

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!

Bayes' Theorem (Tree Diagrams)

For problems where you know $P(B|A)$ but need $P(A|B)$ :

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

🔑 AP Tip: On the AP exam, you can use a tree diagram instead of memorizing Bayes' formula. Draw branches for $A$ and $A\text{' }$ , then for $B$ given each.

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 ✅

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

How to Check Independence

Method	Check	Independent If
Conditional probability	Compute $P(A	B)$
Multiplication check	Compute $P(A) \cdot P(B)$	$P(A) \cdot P(B) = P(A \cap B)$
Two-way table	Compare conditional proportions	Row/column proportions are equal

Independence vs Mutually Exclusive

These concepts are DIFFERENT and often confused:

Property	Independent	Mutually Exclusive
Can both occur?	Yes	No
$P(A \cap B)$	$P(A) \cdot P(B) > 0$	$0$
$P(A	B)$	$P(A)$
Knowing one affects the other?	No	Yes (if one occurs, the other can't)

🔑 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

	College Degree	No Degree	Total
Promoted	30	10	40
Not Promoted	120	40	160
Total	150	50	200

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

Sampling Method	Independent?
With replacement	Yes
Without replacement ( $n < 10\%$ of $N$ )	Approximately yes
Without replacement ( $n \geq 10\%$ of $N$ )	No — use other methods

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 ✅

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

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

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

Rule Selection 🔍

Exit Quiz — Problem-Solving Workshop ✅

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

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

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

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

Comprehensive Worked Example

A survey of 500 adults:

	Exercises Regularly	Does Not	Total
Healthy Weight	180	120	300
Not Healthy Weight	60	140	200
Total	240	260	500

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

Mistake	Correction
Using addition rule for "and"	"And" → multiplication rule
Forgetting $-P(A \cap B)$ in addition rule	Always subtract overlap unless ME
Treating dependent events as independent	Check if events affect each other
Confusing ME with independent	ME ≠ independent (opposite when $P > 0$ )
$P > 1$ in final answer	Probability is always between 0 and 1

Formula Identification 🎯

Mixed Calculations 🧮

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

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

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

Concept Review 🔍

Exit Quiz — Comprehensive Review ✅

Conditional Probability

Venn Diagram View

Mutually Exclusive Events

Worked Example 1: General Addition Rule

Worked Example 2: Two-Way Table

Independent Events (Special Case)

Independent vs Dependent

Worked Example 1: Independent Events

Worked Example 2: Dependent Events (Without Replacement)

Extending to Multiple Events

"At Least One" Using the Complement

Reading Conditional Probability

Worked Example 1: Formula

Worked Example 2: Two-Way Table

Connection to the Multiplication Rule

Bayes' Theorem (Tree Diagrams)

How to Check Independence

Independence vs Mutually Exclusive

Worked Example 1: Multiplication Check

Worked Example 2: Two-Way Table

Independence in Sampling

Worked Example 1: "At Least One" with Complement

Worked Example 2: Combining Multiple Rules

Worked Example 3: Sequential Events with Dependence

Decision Flowchart

Key Relationships

Comprehensive Worked Example

Common Mistakes to Avoid