Chapter 1 — Foundations
What is probability?
Probability measures how likely an event is to happen, from 0 (impossible) to 1 (certain).
Notation: We use P for probability. So P(rolling a 4) means "the probability of rolling a 4."
Picture: one die — 6 equally likely outcomes. One is “4.”
[1] [2] [3] [4] [5] [6]
Example: P(rolling a 4) = 1 favorable ÷ 6 total = 1/6.
Quick check
What is the probability of rolling a 4 on one fair die? (P = probability)
Chapter 1 — Foundations
Sample space and outcomes
The sample space is the set of all possible outcomes. Each outcome should be equally likely when we use the formula.
Pictorial:
- One die: 6 outcomes → [1] [2] [3] [4] [5] [6]
- Two dice: First die 6 × Second die 6 = 36 outcomes. Picture a 6×6 grid: (1,1), (1,2), …, (6,6).
- Coin: [Heads] [Tails] → 2 outcomes.
Two dice — how many ways to get sum 7? (Table: first die × second die.)
| Sum | Pairs |
|---|---|
| 7 | (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) |
So P(sum is 7) = 6 favorable ÷ 36 total = 1/6.
Quick check
Two dice. How many outcomes in the sample space? (sample space = all possible outcomes)
Chapter 2 — Complementary probability
“At least one” and the complement
What does “at least one” mean? Example: “At least one heads in 3 flips” means we want 1 head, or 2 heads, or 3 heads — we only don’t want zero heads (all tails). So “at least one heads” is the opposite of “no heads” = “all tails.”
Why use 1 − P(none)? Every outcome is either “all tails” or “at least one heads” — nothing in between. So P(at least one heads) + P(all tails) = 1. So P(at least one heads) = 1 − P(all tails). We only need to find P(all tails), then subtract from 1.
Venn diagram: P(A) + P(not A) = 1, so P(not A) = 1 − P(A). Here A is an event; P(A) = probability that A happens.
3 flips: the only way to get zero heads is TTT. So P(all tails) = 1/8, and P(at least one heads) = 1 − 1/8 = 7/8.
| Outcome | All tails (TTT) | At least one heads (HTT, THT, TTH, HHT, HTH, THH, HHH) |
|---|---|---|
| Probability | 1/8 | 7/8 |
Another example: “At least one 6 in 2 dice” = opposite of “no 6.” P(no 6 on either die) = (5/6)(5/6) = 25/36. So P(at least one 6) = 1 − 25/36 = 11/36.
Chapter 3 — Independent events
Multiplying probabilities
Two events are independent if one does not affect the other. Then multiply: P(A and B) = P(A) × P(B). (A and B are events; P = probability.)
Picture: coin (2 outcomes) × die (6 outcomes) = 12 equally likely pairs.
Examples: Coin + die: P(heads and 5) = (1/2)(1/6) = 1/12. Three coins: P(heads, heads, heads) = (1/2)³ = 1/8.
If events are not independent (e.g. no replacement), see next slide.
Chapter 3 — Independent events
“And” vs “or”
- And (both happen): multiply probabilities when independent: P(A and B) = P(A) × P(B).
- Or (at least one happens): add when events cannot happen together (mutually exclusive): P(A or B) = P(A) + P(B).
Example (no overlap): One die. P(rolling 2 or 5) = P(2) + P(5) = 1/6 + 1/6 = 2/6 = 1/3. A roll can’t be both 2 and 5, so we just add.
When A and B don’t overlap, just add the probabilities.
When A and B can both happen — why subtract P(A and B)? If we do P(A) + P(B), every outcome that is in both A and B gets counted in P(A) and again in P(B). We counted those outcomes twice. So we subtract P(A and B) once to fix the double counting.
The overlap is counted in both circles — so we subtract it once to avoid double counting.
Die example: which outcomes are in Even? in >4? (Overlap = 6.)
| Outcome | 2 | 4 | 5 | 6 |
|---|---|---|---|---|
| Even? | ✓ | ✓ | — | ✓ |
| >4? | — | — | ✓ | ✓ |
Only 6 is in both (overlap). So “even or >4” = {2, 4, 5, 6} = 4 outcomes → P = 4/6. Formula: 3/6 + 2/6 − 1/6 = 4/6.
Chapter 4 — Dependent events
Without replacement (dependent)
What does “dependent” mean? The second draw depends on what happened on the first. You don’t put the first card back — so the deck actually changes. After you draw one ace, there are only 3 aces left and 51 cards total. The chance for the second card is different than it was before. That’s “dependent.”
With replacement (independent): You put the first card back and shuffle. The deck is the same again — 4 aces, 52 cards. The second draw doesn’t depend on the first. That’s “independent.”
Without replacement: first draw changes what’s left.
So P(both aces, no replacement) = (4/52)(3/51) = 1/221. With replacement: (4/52)(4/52) = 1/169.
Chapter 5 — Counting for probability
When order doesn’t matter
Example 1 — Picking a team: 3 people: Sam, Jordan, and Alex. Pick 2 for your team.
Example 2 — Marbles: Bag: 5 red + 4 blue (9 total). Grab 2. P(both red)?
• Pairs of red marbles (both red): A-B, A-C, A-D, A-E, B-C, B-D, B-E, C-D, C-E, D-E. That’s 10 red-only pairs.
• Pairs from all 9 marbles (any two): There are 36 different pairs in total. Logic: If we pick “first” then “second”: 9 choices, then 8 → 9×8 = 72. But the pair (marble A, marble B) is the same as (marble B, marble A). So we counted every pair twice. So 72÷2 = 36.
Probability both red: P(both red) = (number of red-only pairs) ÷ (number of any pairs) = 10 ÷ 36 = 5/18.
Adding the formula (scaffolding): The number of ways to choose r things from n things (when order doesn’t matter) has a name and a formula. We write it as C(n, r) or say “n choose r.”
So in our examples: 3 choose 2 = 3, 5 choose 2 = 10, 9 choose 2 = 36. For small n you can list; for larger numbers use this formula (or the pattern like 9×8÷2 for “9 choose 2”).
Chapter 5 — Counting for probability
When order matters
When order matters (e.g. 1st, 2nd, 3rd place), we count arrangements: first choice × second × third …
Example: 5 runners, gold/silver/bronze. Number of ways = 5 × 4 × 3 = 60. P(one specific order, e.g. runner A 1st, runner B 2nd, runner C 3rd) = 1/60.
Adding the formula (scaffolding): When order does matter, we use P(n, r) — here P stands for permutations (arrangements), not probability: number of ways to arrange r items from n: first place n choices, second n−1, third n−2, … so
Same as above: P(5, 3) = 5×4×3 = 60. Start with the simple multiplication; use P(n, r) when you need the general rule.
Chapter 6 — Expected value
Expected value (average outcome)
Expected value is the long-run average — what you'd get if you repeated the same experiment many, many times. To find it: multiply each outcome by its probability, then add them all up.
Fair die: each outcome 1–6 has probability 1/6.
| Outcome | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| P | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 |
| Outcome×P | 1/6 | 2/6 | 3/6 | 4/6 | 5/6 | 6/6 |
E = (1+2+3+4+5+6)/6 = 3.5. Game: win $10 (P=1/5), lose $2 (P=4/5) → E = 10(1/5)+(−2)(4/5) = 2/5 ($0.40 per game).
Chapter 7 — Common setups
Dice and coins (quick reference)
One die: 6 outcomes. Two dice: 36. Coins: n flips → 2ⁿ outcomes.
Two dice — number of ways to get each sum
| Sum | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Ways | 1 | 2 | 3 | 4 | 5 | 6 | 5 | 4 | 3 | 2 | 1 |
(Sum 2 = (1,1); sum 12 = (6,6); sum 7 has 6 ways.)
So P(sum is 7) = 6/36 = 1/6. P(doubles) = 6/36. Coins: P(all heads in n flips) = 1/2ⁿ.
Chapter 7 — Common setups
Cards (standard 52-card deck)
| Event | Count | P |
|---|---|---|
| Heart | 13 | 13/52 |
| Ace | 4 | 4/52 |
| Face card | 12 | 12/52 |
| Face and heart | 3 | 3/52 |
P(face or heart) = 12/52 + 13/52 − 3/52 = 22/52. P(both aces, no replacement) = (4/52)(3/51) = 1/221.
Chapter 8 — Tables and conditional
Two-way tables
When data is in a table (e.g. Pass/Fail by section or group), each cell is a count; row and column totals go at the edges; grand total is the bottom-right.
Example: 50 students — Pass/Fail by section
| Pass | Fail | Total | |
|---|---|---|---|
| Section A | 12 | 8 | 20 |
| Section B | 18 | 12 | 30 |
| Total | 30 | 20 | 50 |
Real-life example: “Given that a student is in Section A, what’s the chance they passed?” Focus only on Section A students. There are 20 Section A students; 12 of them passed. So P(Pass | Section A) = 12/20 — of the people in Section A, what fraction passed?
Another: “Given that a student passed, what’s the chance they’re in Section A?” Now look only at the 30 who passed. Of those, 12 are in Section A. So P(Section A | Pass) = 12/30.
In general: P(B | A) = P(A and B) ÷ P(A) — “of the A’s, what fraction are also B?”
Chapter 9 — AMC 8 style
Problem-solving strategy
Quick checklist
| Step | Do this |
|---|---|
| 1 | Identify sample space (equally likely?) |
| 2 | “At least one” → 1 − P(none) |
| 3 | Small n? List outcomes |
| 4 | No calculator — simplify fractions |
Probability + counting ≈ 40–45% of AMC 8. Practice short, clear setups.
Chapter 9 — AMC 8 style
Sample AMC-style question (idea)
“Bag: 3 red, 2 blue. Draw 2 without replacement. P(different colors)?”
| Method | Calculation |
|---|---|
| 1: R then B + B then R | (3/5)(2/4) + (2/5)(3/4) = 6/20+6/20 = 3/5 |
| 2: 1 − both same | 1 − (3/5)(2/4) − (2/5)(1/4) = 12/20 = 3/5 |
Chapter 10 — 9th grade placement (SUSD)
What placement tests often ask
| Topic | Example |
|---|---|
| Basic P | Dice, coins, marbles: favorable ÷ total |
| Complement | P(not A) = 1 − P(A). “At least one” = 1 − P(none), e.g. 1 − P(all tails) |
| Independent | Multiply: P(A and B) = P(A)×P(B) |
| Tables / conditional | Two-way table; P(B | A) = given A |
| Word problems | Name: experiment, sample space, event |
Practice timed, no-calculator to match test conditions.
Chapter 10 — 9th grade placement (SUSD)
Quick checklist before the test
| ✓ | Idea |
|---|---|
| 1 | P = favorable / total (equally likely) |
| 2 | P(not A) = 1 − P(A) |
| 3 | Independent: P(A and B) = P(A)·P(B) |
| 4 | Dependent (no replacement): deck changes; multiply step 1 × step 2 given step 1 |
| 5 | “At least one” → 1 − P(none) |
| 6 | Expected value = Σ(outcome × P) |
Underline “with/without replacement,” “at least one,” “given” when you read the problem.
Practice
Mini drill (concepts only)
1. Two dice. P(sum > 9)? (Favorable: 10, 11, 12 → 3+2+1 = 6. So 6/36 = 1/6.)
2. Three coins. P(at least one tails)? (1 − 1/8 = 7/8.)
3. Deck: one card. P(king or heart)? (4+13−1)/52 = 16/52 = 4/13.)
4. 4 red, 3 blue; pick 2 without replacement. P(both red)? (4/7)(3/6) = 12/42 = 2/7.)
Try similar problems from AMC 8 past years and your placement prep materials.
Summary
- Basic: P = favorable / total; outcomes in the sample space should be equally likely.
- Complement: P(not A) = 1 − P(A). “At least one” means “not none” → use 1 − P(none).
- Independent: P(A and B) = P(A)·P(B). Dependent: no replacement — what’s left changes; multiply first step × second step (given first).
- Counting: “n choose r” = ways to pick r from n when order doesn’t matter. When order matters, multiply: e.g. 5×4×3 for 3 places.
- Expected value: Sum of (outcome × probability).
- AMC 8 & placement: Clear setup, small numbers, no calculator; practice under time.
End of chapter test
Answer each question, then click Submit answers.
P(not A) = ?
Two spinners each have 6 equal sectors (1–6). P(sum = 7) = ?
A target: P(hit) = 2/3 per shot. Two shots, independent. P(at least one hit) = ?
A box has 3 gold and 9 silver tokens. Two drawn without replacement. P(both gold) = ?
In a set of 20 badges: 7 are stars, 5 are stripes, 2 are both. One badge picked at random. P(star or stripe) = ?
Spinner A has 4 sectors; Spinner B has 5. Both fair. P(A lands 3 and B lands 4) = ?
Game: $0 (P=0.5), $5 (P=0.3), $10 (P=0.2). Expected value = ?
Jar: 4 red, 3 blue jelly beans. Two drawn without replacement. P(both red) = ?
Multiple choice: 5 options, 1 correct. Random guess on 4 questions. P(at least one correct) = ?
Bowl: 4 orange, 2 green fruits. Pick 2 at random (order doesn't matter). P(both orange) = ?
Spinner 1–6, equally likely. Spin twice with replacement. P(both even) = ?
Two-way table: 45 Pass, 35 Fail. Total 80. P(randomly chosen student passed) = ?
Two fair spinners (1–6). P(at least one lands on 6) = ?
Events A and B are independent. P(A) = 2/5, P(B) = 3/8. P(A and B) = ?
5 runners. Gold, silver, bronze (order matters). P(one specific finish order) = ?