Formulas

Permutation

nPr = n! / (n−r)!

Combination

nCr = n! / (n−r)!r!

Binomial Theorem

(x+y)ⁿ = Σ(k=0 to n) (n,k) · xᵏ · yⁿ⁻ᵏ

Multinomial coefficient

n! / (n₁! · n₂! · ... · nᵣ!)

Probability (frequency definition

P(E) = lim(n→∞) n(E)/n

Complement

P(Eᶜ) = 1 − P(E)

Union of 2 events

P(E∪F) = P(E) + P(F) − P(E∩F)

Union of 3 events

P(E∪F∪G) = P(E)+P(F)+P(G) − P(E∩F) − P(E∩G) − P(F∩G) + P(E∩F∩G)

Equally likely outcomes

P(E) = (# outcomes in E) / (# outcomes in S)

Conditional Probability

P(E|F) = P(E∩F) / P(F)- Read as: "Probability of E given F already happened" P(E∩F)- JUst E

Multiplication rule

P(E∩F) = P(F) · P(E|F)

• P(W1)= 6/15 → 6 white out of 15 total

• P(W2∣W1) = 5/14 → given 1 white gone, 5 white left out of 14 total

• P(B3∣W1∩W2) = 9/13 → given 2 whites gone, 9 black out of 13 total

• P(B4∣W1∩W2∩B3) = 8/12 → given 3 balls gone, 8 black out of 12 total

Law of Total Probability

P(E) = P(E|B₁)·P(B₁) + P(E|B₂)·P(B₂) + ... + P(E|Bₖ)·P(Bₖ)

Bayes's Formula (short)

P(E|F) = P(E)·P(F|E) / P(F)

Bayes's Formula (expanded)

P(E|F) = P(E)·P(F|E) / [P(E)·P(F|E) + P(Eᶜ)·P(F|Eᶜ)]

Independent Events

P(E|F) = P(E), equivalently P(E∩F) = P(E)·P(F)

Multinomial Expansion

Given (x+y+z)ⁿ, the coefficient of the term xⁱyʲzᵏ is:

n! / (i! · j! · k!) where i + j + k = n

De Morgan's Law 1 — (E∪F)ᶜ = Eᶜ ∩ Fᶜ

De Morgan's Law 2 — (E∩F)ᶜ = Eᶜ ∪ Fᶜ

— (E∪F)ᶜ = Eᶜ ∩ Fᶜ

— (E∩F)ᶜ = Eᶜ ∪ Fᶜ

Conditional Probability