Topic A: Compound Probability
Lessons 1–3 · AND/OR probability with and without replacement.
01–03 — Lessons 1–3
AND, OR & Complement Rules
- P(A and B) = P(A)·P(B) (if A and B are independent).
- P(A or B) = P(A) + P(B) − P(A and B) (Addition Rule).
- P(A or B) = P(A) + P(B) (if mutually exclusive — can't happen together).
- P(not A) = 1 − P(A) (Complement Rule).
- With replacement: events are independent. Without replacement: dependent.
P(ace) = 4/52 = 1/13. P(ace or king) = 4/52 + 4/52 = 8/52 (mutually exclusive).
Practice — Lessons 1–3
Q1 · Multiple Choice
P(A) = 0.5 and P(B) = 0.3. If A and B are mutually exclusive, P(A or B) = ?
Q2 · Multiple Choice
You flip a fair coin twice. P(both heads) = ?
Topic B: Conditional Probability
Lessons 4–7 · P(A|B) and the multiplication rule.
04–07 — Lessons 4–7
Conditional Probability & Independence
- P(A|B) = P(A∩B) / P(B) — probability of A given B has occurred.
- Independent events: P(A|B) = P(A). Knowing B gives no info about A.
- Multiplication rule: P(A∩B) = P(A)·P(B|A).
- Use two-way tables to calculate conditional probabilities.
- Test for independence: check if P(A|B) = P(A).
60 students: 30 like math, 20 like both math and science, 40 like science.
P(science|math) = P(math∩science)/P(math) = (20/60)/(30/60) = 20/30 = 2/3.
Practice — Lessons 4–7
Q3 · Multiple Choice
P(A) = 0.4, P(A∩B) = 0.12, P(B) = 0.3. Are A and B independent?
Q4 · Short Answer
P(B) = 0.5 and P(A∩B) = 0.2. Find P(A|B).
Topic C: Permutations & Combinations
Lessons 8–10 · Counting principles, nPr, and nCr.
08–10 — Lessons 8–10
Counting, Permutations & Combinations
- Fundamental Counting Principle: multiply the number of choices for each step.
- Permutation nPr: ordered arrangements of r items from n. nPr = n!/(n−r)!
- Combination nCr: unordered selections of r items from n. nCr = n!/[r!(n−r)!]
- Use permutations when ORDER matters (1st, 2nd, 3rd place).
- Use combinations when order does NOT matter (choosing a committee).
- n! = n·(n−1)·(n−2)···2·1
Choose 3 from 8 people for a committee (no order): ₈C₃ = 8!/(3!5!) = 56.
Rank 3 from 8 for 1st/2nd/3rd: ₈P₃ = 8!/5! = 336.
Practice — Lessons 8–10
Q5 · Multiple Choice
You need to choose a president and vice-president from 10 students. Use:
Q6 · Short Answer
How many ways can you choose 2 from 6 students for a pair project? (Use ₆C₂)