adding flashcards!

2026-06-15 00:31:45 -07:00
parent 713576cc83
commit 26fa6b9723
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+State De Morgan's Law for $n$ sets
+---
+
+For any sets $A_1, A_2, \dots, A_n$:
+
+$$\left(\bigcup_{i=1}^n A_i\right)^c = \bigcap_{i=1}^n A_i^c$$
+
+$$\left(\bigcap_{i=1}^n A_i\right)^c = \bigcup_{i=1}^n A_i^c$$
+
+The complement of a union is the intersection of complements, and vice versa.
+
+===
+
+State the Distributive Law for sets $A$, $B$, and $C$
+---
+
+$$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$
+
+$$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$
+
+===
+
+State the inclusion-exclusion principle for a finite collection of sets $A_1, A_2, A_3, \dots A_n$ where $n=2$
+
+---
+
+$n = 2$ case:
+
+$|A \cup B| = |A| + |B| - |A \cap B|$
+
+---
+
+$n = 3$ case:
+
+$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$
+
+---
+
+For a finite collection of sets $A_1, A_2, A_3, \dots A_n$, we have
+
+$\left| \bigcup_{i=1}^n A_i \right| = \sum_{i=1}^n |A_i| - \sum_{i < j} |A_i \cap A_j| + \sum_{i < j < k} |A_i \cap A_j \cap A_k| - \dots + (-1)^{n-1} |A_1 \cap A_2 \cap \dots \cap A_n|$
+
+===
+
+For the function
+
+$$f: A \to B$$
+
+State the following of $f$:
+
+1. Domain
+2. Co-domain
+3. Range
+
+---
+
+1. $A$
+2. $B$
+3. Set of all possible outputs of $f$ (not necessarily $B$)
+
+===
+
+State definition for each the following:
+
+1. Random experiment
+2. Outcome
+3. Sample space
+4. Event
+
+---
+
+1. A **random experiment** is a process by which we observe something uncertain
+2. An **outcome** is a result of a random experiment
+3. The **sample space** $S$ is the set of all possible outcomes
+4. An **event** is a subset of the sample space
+
+===
+
+State the Axioms of Probability
+
+---
+**Axioms of Probability**
+
+1. For any event $A$, $P(A) \geq 0$
+2. $P(S) = 1$
+3. If $A_1, A_2, A_3, \dots$ are disjoint events, then $P(A_1 \cup A_2 \cup A_3 \cup \dots) = P(A_1) + P(A_2) + P(A_3) + \dots$
+
+===
+
+In a finite sample space $S$, where all outcomes are equally likely, what is the probability of any event $A$?
+
+---
+
+$P(A) = \frac{|A|}{|S|}$
+
+===
+
+What is $P(A^c)$ in terms of $P(A)$?
+---
+
+$$P(A^c) = 1 - P(A)$$
+
+Follows from the axioms: $A$ and $A^c$ are disjoint, and $A \cup A^c = S$, so $P(A) + P(A^c) = P(S) = 1$.
+
+===
+
+Define conditional probability $P(A|B)$
+---
+
+The probability of $A$ given $B$ (when $P(B) > 0$):
+
+$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$
+
+In the equally-likely case:
+
+$$P(A|B) = \frac{|A \cap B|}{|B|}$$
+
+===
+
+State the chain rule of probability for $n$ events
+---
+
+$$P(A_1 \cap A_2 \cap \cdots \cap A_n) = P(A_1)\,P(A_2|A_1)\,P(A_3|A_1,A_2)\cdots P(A_n|A_1,A_2,\dots,A_{n-1})$$
+
+Each factor conditions on all previously listed events.
+
+===
+
+What conditions must hold for three events $A$, $B$, $C$ to be independent?
+---
+
+All four conditions must hold:
+
+1. $P(A \cap B) = P(A)P(B)$
+2. $P(A \cap C) = P(A)P(C)$
+3. $P(B \cap C) = P(B)P(C)$
+4. $P(A \cap B \cap C) = P(A)P(B)P(C)$
+
+Pairwise independence alone is **not** sufficient.
+
+===
+
+What does it mean for $n$ events $A_1, A_2, \dots, A_n$ to be independent?
+---
+
+Every subset of the events must satisfy the product rule. That is, for all $i < j < k < \dots$:
+
+$$P(A_i \cap A_j) = P(A_i)P(A_j)$$
+$$P(A_i \cap A_j \cap A_k) = P(A_i)P(A_j)P(A_k)$$
+$$\vdots$$
+$$P(A_1 \cap A_2 \cap \cdots \cap A_n) = \prod_{i=1}^n P(A_i)$$
+
+===
+
+If $A_1, A_2, \dots, A_n$ are independent, what is $P(A_1 \cup A_2 \cup \cdots \cup A_n)$?
+---
+
+$$P(A_1 \cup A_2 \cup \cdots \cup A_n) = 1 - \prod_{i=1}^n (1 - P(A_i))$$
+
+Derived by taking the complement (none of the events occur) and using independence.
+
+===
+
+What is the difference between disjoint and independent?
+
+---
+
+- Disjoint: $A$ and $B$ cannot occur at the same time. $A \cap B = \empty$
+- Independent: $A$ does not give any information about $B$
+
+===
+
+State the Law of Total Probability using event $B$ and its complement
+---
+
+$$P(A) = P(A|B)\,P(B) + P(A|B^c)\,P(B^c)$$
+
+===
+
+State the general Law of Total Probability for a partition of the sample space
+---
+
+If $B_1, B_2, B_3, \dots$ is a partition of $S$, then for any event $A$:
+
+$$P(A) = \sum_i P(A \cap B_i) = \sum_i P(A|B_i)\,P(B_i)$$
+
+===
+
+State Bayes' Rule for two events $A$ and $B$
+---
+
+For $P(A) \neq 0$:
+
+$$P(B|A) = \frac{P(A|B)\,P(B)}{P(A)}$$
+
+===
+
+State Bayes' Rule when $B_1, B_2, \dots$ form a partition of $S$
+---
+
+For any event $A$ with $P(A) \neq 0$:
+
+$$P(B_j|A) = \frac{P(A|B_j)\,P(B_j)}{\displaystyle\sum_i P(A|B_i)\,P(B_i)}$$
+
+The denominator expands $P(A)$ via the Law of Total Probability.
+
+===
+
+State the meaning of conditionally independent between events $A$ and $B$ given event $C$. Note $P(C) \gt 0$.
+
+---
+
+$$P(A \cap B \mid C) = P(A \mid C)P(B \mid C)$$
+
+===
+
+State the multiplication rule for $P(A \cap B)$
+---
+
+$$P(A \cap B) = P(A|B)\,P(B) = P(B|A)\,P(A)$$
+
+Rearrangement of the definition of conditional probability.
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+For any event $A$, prove $P(A^c) = 1 - P(A)$
+
+---
+
+$$
+1 = P(S)
+= P(A \cup A^c)
+= P(A) + P(A^c)
+$$
+
+===
+
+Prove $P(A \setminus B) = P(A) - P(A \cap B)$
+
+---
+
+TODO