AMTH 108 Quiz 1 Notes

1⃣ Set Theory (Know symbols instantly)

  • Union: A∪BA \cup BA∪B → A or B

  • Intersection: A∩BA \cap BA∩B or ABABAB → both

  • Complement: AcA^cAc → not A

  • Empty set: ∅\varnothing∅

Facts

  • A∩Ac=∅A \cap A^c = \varnothingA∩Ac=∅

  • A∪Ac=XA \cup A^c = XA∪Ac=X

  • If A⊂BA \subset BA⊂B, then P(A)≤P(B)P(A) \le P(B)P(A)≤P(B)

2⃣ Counting (order matters?)

Multiplication Principle

Total outcomes=n1×n2×⋯\text{Total outcomes} = n_1 \times n_2 \times \cdotsTotal outcomes=n1​×n2​×⋯

Permutations (ORDER MATTERS)

nPk=n!(n−k)!nP_k = \frac{n!}{(n-k)!}nPk​=(n−k)!n!​

Use for:

  • Passwords

  • Officer roles

  • Ordered selections

Combinations (ORDER DOES NOT MATTER)

nCk=(nk)=n!k!(n−k)!nC_k = \binom{n}{k} = \frac{n!}{k!(n-k)!}nCk​=(kn​)=k!(n−k)!n!​

Use for:

  • Poker hands

  • Committees

  • Card selections

🔴 Trap: If order doesn’t matter and you use permutations → wrong

3⃣ Probability Basics

  • 0≤P(E)≤10 \le P(E) \le 10≤P(E)≤1

  • P(X)=1P(X)=1P(X)=1

  • P(∅)=0P(\varnothing)=0P(∅)=0

Equally likely outcomes

P(E)=\frac{\text{# favorable}}{\text{# total}}

4⃣ Complements (VERY IMPORTANT)

P(Ec)=1−P(E)P(E^c)=1-P(E)P(Ec)=1−P(E)

Used when:

  • “At least one”

  • “Contains a digit”

  • Birthday-type problems

5⃣ Two-Event Probability

P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B)=P(A)+P(B)-P(A\cap B)P(A∪B)=P(A)+P(B)−P(A∩B)

6⃣ Three-Event Inclusion–Exclusion

P(A∪B∪C)=  P(A)+P(B)+P(C)−P(AB)−P(AC)−P(BC)+P(ABC)\begin{aligned} P(A\cup B\cup C)=\;&P(A)+P(B)+P(C)\\ &-P(AB)-P(AC)-P(BC)\\ &+P(ABC) \end{aligned}P(A∪B∪C)=​P(A)+P(B)+P(C)−P(AB)−P(AC)−P(BC)+P(ABC)​

7⃣ Venn Diagrams

  • Break into atomic regions

  • Assign probabilities to smallest pieces

  • All regions sum to 1

  • Solve unknowns with equations

8⃣ Conditional Probability (HIGH PRIORITY)

P(E∣A)=P(E∩A)P(A),P(A)≠0P(E\mid A)=\frac{P(E\cap A)}{P(A)}, \quad P(A)\neq0P(E∣A)=P(A)P(E∩A)​,P(A)=0

Meaning: A already happened → shrink sample space

9⃣ Independence (VERY IMPORTANT)

Events A and B are independent if:

P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B)P(A∩B)=P(A)P(B)

or

P(A∣B)=P(A)P(A\mid B)=P(A)P(A∣B)=P(A)

Facts

  • Independent ≠ mutually exclusive

  • If A∩B=∅A\cap B=\varnothingA∩B=∅ and both probs > 0 → NOT independent

  • If A and B independent → A and BcB^cBc also independent