Chapter 7: Probability: The Basics
Learning Objectives
- Define sample spaces, outcomes, and events; calculate basic probabilities
- Apply complement, addition, and multiplication rules correctly
- Identify mutually exclusive events and use the special addition rule
- Calculate and interpret conditional probability using $P(B \mid A) = P(A \cap B)/P(A)$
- Determine whether two events are independent using the multiplication rule
- Use tree diagrams and two-way tables to organize and calculate probabilities
7.1 Sample Spaces and Basic Probability
Probability is the long-run relative frequency of an outcome. Before calculating any probability, we identify the experiment, the possible outcomes, and the sample space.
Definition: Key Vocabulary
- Experiment: Any process with an uncertain result (rolling a die, drawing a card, selecting a student).
- Outcome: A single possible result of an experiment.
- Sample space $S$: The set of all possible outcomes. For a fair die: $S = \{1, 2, 3, 4, 5, 6\}$.
- Event: Any collection of outcomes (a subset of $S$). Events are usually denoted by capital letters.
For equally likely outcomes: $$P(A) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes in } S}$$
Basic Properties of Probability
- $0 \le P(A) \le 1$ for any event $A$
- $P(S) = 1$ (something in the sample space must occur)
- $P(\emptyset) = 0$ (the impossible event has probability 0)
- Complement Rule: $P(A^c) = 1 - P(A)$, where $A^c$ is the event that $A$ does not occur
Theorem: Law of Large Numbers
As the number of trials of a random experiment increases, the relative frequency of an event approaches its theoretical probability. The law guarantees convergence in the long run — it says nothing about short-run behavior.
Example: Flip a fair coin 10 times and get 7 heads. The proportion 7/10 = 0.70 may be far from 0.50. But after 10,000 flips, the proportion will be very close to 0.50.
Example 7.1 — Rolling a Fair Die
Roll a fair six-sided die. Sample space: $S = \{1, 2, 3, 4, 5, 6\}$, each equally likely.
- $P(\text{even}) = P(\{2,4,6\}) = \dfrac{3}{6} = \dfrac{1}{2}$
- $P(\text{greater than 4}) = P(\{5,6\}) = \dfrac{2}{6} = \dfrac{1}{3}$
- $P(\text{not 3}) = 1 - P(3) = 1 - \dfrac{1}{6} = \dfrac{5}{6}$
Example 7.2 — Marbles in a Bag
A bag contains 4 red, 3 blue, and 2 green marbles (9 total). One marble is drawn at random.
- $P(\text{red}) = \dfrac{4}{9}$
- $P(\text{not blue}) = 1 - P(\text{blue}) = 1 - \dfrac{3}{9} = \dfrac{6}{9} = \dfrac{2}{3}$
Note: $P(\text{not blue}) = P(\text{red or green}) = \dfrac{4+2}{9} = \dfrac{6}{9}$ — same answer via direct count.
A student guesses on a 5-choice multiple-choice question (options A–E). What is the probability of getting it wrong?
Show Answer
Sample space for a fair die — even outcomes (2, 4, 6) shown in green, odd outcomes (1, 3, 5) in gray. $P(\text{even}) = 1/2$.
Figure 7.1 — Sample Space: Rolling a Fair Die
7.2 Addition Rule and Mutually Exclusive Events
When we want the probability of event $A$ or event $B$ occurring (i.e., their union $A \cup B$), we use the addition rule.
Theorem: General Addition Rule
For any two events $A$ and $B$: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
We subtract $P(A \cap B)$ because outcomes in both $A$ and $B$ are counted twice when we add $P(A) + P(B)$.
Special Case — Mutually Exclusive (Disjoint) Events: If $A$ and $B$ cannot both occur, then $P(A \cap B) = 0$, so: $$P(A \cup B) = P(A) + P(B)$$
The intersection $A \cap B$ (read "A and B") is the event that both $A$ and $B$ occur. Two events are mutually exclusive (disjoint) if they share no outcomes: $A \cap B = \emptyset$.
Example 7.3 — Activities Survey
A student is selected at random from a school. Let $A$ = plays sports and $B$ = plays a musical instrument.
$P(A) = 0.45$, $\quad P(B) = 0.30$, $\quad P(A \cap B) = 0.15$ (both sports and music).
Find $P(A \cup B)$:
$P(A \cup B) = 0.45 + 0.30 - 0.15 = \mathbf{0.60}$
60% of students participate in sports or music (or both).
Example 7.4 — Drawing a Card
One card is drawn from a standard 52-card deck. Let $K$ = king and $Q$ = queen.
- $P(K) = \dfrac{4}{52}$, $\quad P(Q) = \dfrac{4}{52}$
- Are $K$ and $Q$ mutually exclusive? Yes — a card cannot be both a king and a queen, so $P(K \cap Q) = 0$.
- $P(K \cup Q) = \dfrac{4}{52} + \dfrac{4}{52} = \dfrac{8}{52} = \dfrac{2}{13}$
$P(A) = 0.4$, $P(B) = 0.3$, $P(A \cap B) = 0.12$. Find $P(A \cup B)$. Are $A$ and $B$ mutually exclusive?
Show Answer
AP Exam Tip: On the AP exam, "or" always means union ($A \cup B$). Always apply the general addition rule and subtract the intersection — unless the problem states or you can verify the events are mutually exclusive.
Venn diagram: two overlapping circles representing events $A$ and $B$. The overlap is $A \cap B$. $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Figure 7.2 — Venn Diagram: Addition Rule for $P(A \cup B)$
7.3 Conditional Probability and Independence
Sometimes the probability of an event changes when we know that another event has occurred. This is conditional probability.
Definition: Conditional Probability
The conditional probability of event $B$ given that event $A$ has occurred is: $$P(B \mid A) = \frac{P(A \cap B)}{P(A)}, \quad P(A) > 0$$
Read "$P(B \mid A)$" as "the probability of $B$ given $A$." Knowing $A$ occurred restricts the sample space to outcomes in $A$.
Independence
Events $A$ and $B$ are independent if knowing that $A$ occurred does not change the probability of $B$:
- $P(B \mid A) = P(B)$ — equivalent definition
- $P(A \cap B) = P(A) \cdot P(B)$ — the multiplication rule for independent events
Theorem: Multiplication Rule
General form (always valid): $$P(A \cap B) = P(A) \cdot P(B \mid A)$$
Special case — independent events: $$P(A \cap B) = P(A) \cdot P(B)$$
Testing independence: $A$ and $B$ are independent if and only if $P(A \cap B) = P(A) \cdot P(B)$.
Example 7.5 — Athletes and Passing
In a class, 60% of students are athletes. Among athletes, 70% pass an exam. Among non-athletes, 50% pass.
- $P(\text{pass} \mid \text{athlete}) = 0.70$
- $P(\text{athlete and passes}) = P(\text{athlete}) \cdot P(\text{pass} \mid \text{athlete}) = 0.60 \times 0.70 = \mathbf{0.42}$
42% of all students are athletes who pass the exam.
Example 7.6 — Drawing Cards With Replacement
Two cards are drawn with replacement from a standard 52-card deck.
$P(\text{both aces}) = P(\text{ace on 1st}) \times P(\text{ace on 2nd}) = \dfrac{4}{52} \times \dfrac{4}{52} = \dfrac{16}{2704} = \dfrac{1}{169}$
Because the first card is replaced before drawing the second, the draws are independent — the outcome of the first draw does not affect the second.
A fair coin is flipped twice. Are the two flips independent? Find $P(HH)$.
Show Answer
AP Exam Tip: Independence and mutual exclusivity are frequently confused. If $P(A) > 0$ and $P(B) > 0$, mutually exclusive events cannot be independent: if $A$ and $B$ are disjoint, then knowing $A$ occurred means $B$ definitely did not — so $P(B \mid A) = 0 \neq P(B)$. The two concepts are opposites, not synonyms.
7.4 Tree Diagrams and Two-Way Tables
For multi-step probability problems, tree diagrams organize sequential outcomes systematically. Two-way tables display joint frequencies for two categorical variables and make conditional probability calculations straightforward.
Example 7.7 — Drawing Without Replacement (Tree Diagram)
A bag has 5 red and 3 blue marbles (8 total). Two marbles are drawn without replacement. Find $P(\text{both red})$.
Tree diagram approach:
- 1st draw Red (probability 5/8) → 2nd draw Red (probability 4/7): $P(RR) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$
- 1st draw Red (5/8) → 2nd draw Blue (3/7): $P(RB) = \dfrac{5}{8} \times \dfrac{3}{7} = \dfrac{15}{56}$
- 1st draw Blue (3/8) → 2nd draw Red (5/7): $P(BR) = \dfrac{3}{8} \times \dfrac{5}{7} = \dfrac{15}{56}$
- 1st draw Blue (3/8) → 2nd draw Blue (2/7): $P(BB) = \dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{6}{56} = \dfrac{3}{28}$
Check: $\dfrac{20+15+15+6}{56} = \dfrac{56}{56} = 1$ ✓
$P(\text{both red}) = \dfrac{5}{14} \approx 0.357$
Example 7.8 — Two-Way Table: Gender and Sport Preference
A survey of 200 students records gender and whether they prefer sports or non-sports activities:
| Plays Sport | No Sport | Total | |
|---|---|---|---|
| Male | 60 | 40 | 100 |
| Female | 50 | 50 | 100 |
| Total | 110 | 90 | 200 |
- $P(\text{sport} \mid \text{female}) = \dfrac{50}{100} = 0.50$
- $P(\text{sport} \mid \text{male}) = \dfrac{60}{100} = 0.60$
- $P(\text{sport}) = \dfrac{110}{200} = 0.55$ (marginal probability)
Are gender and sport preference independent?
Test: $P(\text{sport} \mid \text{female}) = 0.50 \neq P(\text{sport}) = 0.55$. Since the conditional probability differs from the marginal, gender and sport preference are not independent.
Using the two-way table in Example 7.8, find $P(\text{male and plays sport})$.
Show Answer
Tree diagram for drawing 2 marbles from a bag of 5 red and 3 blue without replacement. Branch probabilities shown at each level.
Figure 7.3 — Tree Diagram: Drawing Without Replacement
Practice Problems
A spinner has 8 equal sections numbered 1–8. Find: $P(\text{prime})$, $P(\text{even or} > 5)$, $P(\text{multiple of 3})$.
Show Solution
Even: $\{2,4,6,8\}$; $> 5$: $\{6,7,8\}$; overlap: $\{6,8\}$. $P(\text{even or} > 5) = 4/8 + 3/8 - 2/8 = 5/8$.
Multiples of 3: $\{3,6\}$, so $P(\text{mult of 3}) = 2/8 = 1/4$.
$P(A) = 0.55$, $P(B) = 0.40$, $P(A \cup B) = 0.70$. Find $P(A \cap B)$. Are $A$ and $B$ mutually exclusive?
Show Solution
Since $P(A \cap B) = 0.25 \neq 0$, the events are not mutually exclusive.
In a class, 40% play video games ($V$), 30% play sports ($S$), 15% do both. Find $P(V \cup S)$ and $P(\text{neither } V \text{ nor } S)$.
Show Solution
$P(\text{neither}) = 1 - P(V \cup S) = 1 - 0.55 = \mathbf{0.45}$.
A standard 52-card deck: find $P(\text{heart} \mid \text{face card})$. (Face cards are J, Q, K — 12 total; 3 are hearts.)
Show Solution
$P(\text{heart} \mid \text{face card}) = \dfrac{3}{12} = \dfrac{1}{4} = 0.25$.
$P(A) = 0.6$, $P(B) = 0.4$, $P(A \cap B) = 0.24$. Are $A$ and $B$ independent? Show your work.
Show Solution
Since $P(A \cap B) = 0.24 = P(A) \cdot P(B)$, the events $A$ and $B$ are independent.
Of 300 adults, 180 drink coffee and 120 do not. Among coffee drinkers, 90 prefer mornings; among non-coffee drinkers, 30 prefer mornings. Are coffee drinking and morning preference independent?
Show Solution
$P(\text{morning} \mid \text{coffee}) = 90/180 = 0.50$.
Since $0.50 \neq 0.40$, the events are not independent — coffee drinkers are more likely to prefer mornings.
Three cards are drawn without replacement from a standard 52-card deck. Find $P(\text{all three are aces})$.
Show Solution
AP FRQ: A medical test is 95% accurate for those WITH disease and 90% accurate for those WITHOUT. In the population, 2% have the disease. A person tests positive. Using 10,000 people, find $P(\text{disease} \mid \text{positive})$.
Show Solution
With disease (200): test positive = $0.95 \times 200 = 190$; negative = 10.
Without disease (9,800): test positive (false) = $0.10 \times 9800 = 980$; negative = 8,820.
Total positive = $190 + 980 = 1,170$.
$P(\text{disease} \mid \text{positive}) = \dfrac{190}{1170} \approx \mathbf{0.162}$.
Only about 16% of those who test positive actually have the disease — this is the base rate effect.
📋 Chapter Summary
Core Probability Rules
$0 \leq P(A) \leq 1$ for any event $A$. $P(S) = 1$ where $S$ is the sample space. $P(A^c) = 1 - P(A)$.
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$. If mutually exclusive: $P(A \cup B) = P(A) + P(B)$.
$P(A \cap B) = P(A) \cdot P(B|A)$. If independent: $P(A \cap B) = P(A) \cdot P(B)$.
$P(B|A) = \dfrac{P(A \cap B)}{P(A)}$ — probability of $B$ given that $A$ has occurred.
Key Concepts
Events $A$ and $B$ are independent if $P(B|A) = P(B)$, equivalently $P(A \cap B) = P(A)P(B)$.
Events $A$ and $B$ cannot both occur: $P(A \cap B) = 0$. Note: mutually exclusive events are NOT independent (unless one has probability 0).
Organize joint probabilities. Marginal probabilities are row/column totals. Joint probabilities are cell entries.
Multiply probabilities along branches to find joint probabilities. Add branch probabilities for mutually exclusive outcomes.