From 7a073a861b2e92c9efb8d03b53b91a3796602d2b Mon Sep 17 00:00:00 2001 From: Youwen Wu Date: Mon, 27 Jan 2025 14:07:50 -0800 Subject: [PATCH] auto-update(nvim): 2025-01-27 14:07:50 --- .../by-course/math-8/course-notes/main.typ | 129 ++++++++++++++++++ documents/by-course/pstat-120a/hw1/main.typ | 30 +++- documents/by-course/pstat-120a/hw2/main.typ | 109 +++++++++++++++ .../by-course/pstat-120a/hw2/package.nix | 37 +++++ 4 files changed, 304 insertions(+), 1 deletion(-) create mode 100644 documents/by-course/pstat-120a/hw2/main.typ create mode 100644 documents/by-course/pstat-120a/hw2/package.nix diff --git a/documents/by-course/math-8/course-notes/main.typ b/documents/by-course/math-8/course-notes/main.typ index 316c6ac..a6e2894 100644 --- a/documents/by-course/math-8/course-notes/main.typ +++ b/documents/by-course/math-8/course-notes/main.typ @@ -467,6 +467,135 @@ $ exists x in U, P(x) $ rational number $z$. ] += Lecture #datetime(day: 27, month: 1, year: 2025).display() + +== Basic properties of sets + +#definition[ + A set is a collection of elements. +] + +#definition[ + The cardinality of a set is the amount of elements in the set. +] + +#example[ + Prove $A subset.eq B$ iff. $A sect B = A$. + + #proof[ + Assume $A subset.eq B$. Let $x in A sect B$. Then $x in A$ and $x in B$. So + $x in A$ and $A sect B subset.eq A$. Now let $x in A$. Since $A subset.eq + B$, then $x in B$. Thus $x in A$ and $x in B$, so $x in A sect B$. Then $A + subset.eq A sect B$. + + Now we show the other direction. Assume $A sect B = A$. Then $x in A sect + B$ implies $x in A$ and $x in B$. In particular $forall x in A, x in B$. + Thus $a subset.eq B$. + ] +] + +#theorem[ + Let $U$ be the universe, and let $A$ and $B$ be subsets of $U$. Then + + $(A^c)^c = A$ + + $A union A^c = U$ + + $A sect A^c = emptyset$ + + $A - B = A sect B^c$ + + $A subset.eq B <=> B^c subset.eq A^c$ + + $A sect B = emptyset <=> A subset.eq B^c$ + + $(A union B)^c = A^c sect B^c$ + + $(A sect B)^c = A^c union B^c$ +] + +#example[part of @basic-sets][ + Show that $(A sect B)^c = A^c union B^c$. + + $ + &x in (A sect B)^c \ + &<=> x in.not A sect B \ + &<=> x in A^c, x in B^c \ + &<=> x in A^c union B^c + $ +] + +#definition[ + The Cartesian product of sets $A$ and $B$: + + $ A times B = {(a,b) : a in A "and" b in B} $ +] + +#fact[ + If $A$ has cardinality $m$, $B$ has cardinality $n$, then $A times B$ has + cardinality $n dot m$. +] + +#example[ + Prove that $A times emptyset = emptyset$. + + #proof[ + Suppose that $A times emptyset != emptyset$. Then $exists (a,b), a in A, b in +emptyset$. But $b in emptyset$ is a contradiction by its definition. + ] +] + +== Index families of sets + +#definition[A set of sets is called a family or collection of sets.] + +#definition[ + Let $Delta$ be a nonempty index set such that $forall alpha in Delta, exists + A_alpha$. The family of sets $cal(A) = {A_alpha : alpha in Delta}$ is an + index family of sets. +] + +#example[ + Let $A in {1,3}$ and consider $cal(P)(A)$. +] + +#example[ + Define $A_n = (n,n+2)$, the open interval from $n$ to $n+2$, for each $n in + NN$. + + $cal(A) = {(n, N + 2) : n in NN} = {A_n : n in NN}$ + + Let $Delta = NN$ and then $alpha in Delta <=> n in NN$. +] + +#definition[ + The union over $cal(A)$ is the set + $ union.big_(A in cal(A)) A = {x : x in A, exists A in cal(A)} $ +] + +#definition[ + The intersection over $cal(A)$ is the set + $ sect.big_(A in cal(A)) A = {x : x in A, forall A in cal(A)} $ +] + +#example[ + Let $cal(A) = {{1}, {1,2},{2,3}}$. Then + $ + &union.big_(A in cal(A)) A = {1,2,3} \ + §.big_(A in cal(A)) A = emptyset + $ +] + +#example[ + Let $A_n = [-1/n,1/n]$, the closed interval from $-1/n$ to $1/n$ for each $N in NN$. Consider the family of sets $cal(A) = {A_n : n in NN}$. Then + + $ + &union.big_(A in cal(A)) A = union.big_(n=1)^infinity A_n = A_1 union A_2 union A_3 union dots.c = {x : x in [-1,1]} \ + §.big_(A in cal(A)) A = sect.big_(n=1)^infinity A_n = 0 + $ +] + +#example[ + Let $A_n = [0,n)$ for each $n in NN$ and let $cal(A) = {A_n : n in NN}$. Then + + $ + &union.big_(A in cal(A)) A = [0, infinity] \ + §.big_(A in cal(A)) = [0,1) + $ +] + = Solutions to selected exercises and problems Solutions to selected problems and exercises. diff --git a/documents/by-course/pstat-120a/hw1/main.typ b/documents/by-course/pstat-120a/hw1/main.typ index d19eb44..12dab87 100644 --- a/documents/by-course/pstat-120a/hw1/main.typ +++ b/documents/by-course/pstat-120a/hw1/main.typ @@ -9,6 +9,19 @@ #set enum(spacing: 2em) +#let correction = content => { + set text(fill: red) + box(stroke: 1pt, inset: 5pt, content) +} + +#correction[ + There were 7 points off, so: + + Initial score: 47/54 + + Corrected score: 52/52 +] + + #[ #set enum(numbering: "a)", spacing: 2em) @@ -44,7 +57,7 @@ #set enum(numbering: "a)", spacing: 2em) + ${15, 25, 35, 45, 51, 53, 55, 57, 59, 65, 75, 85, 95 }$ - + ${50, 52, 56, 58}$ + + ${50, 52, 54, 56, 58}$ + $emptyset$ ] @@ -64,6 +77,10 @@ represents balls numbered 1-6, and the value represents the square it was sent to. So it's $ {{x_1,x_2,x_3,x_4,x_5,x_6} : x_i in {1,2,3,4}}, i = 1,...6 $ + + #correction[ + -1. We should probably write this more explicitly as ${1,2,3,4}^6$. + ] ] + #[ When the balls are indistinguishable, we can instead represent it as @@ -100,6 +117,11 @@ + #[ #set enum(numbering: "a)", spacing: 2em) + #correction[ + -4. + These are all correct, but need to be divided by $vec(52,5)$ for the final probability. Oops... + ] + + #[ First we choose two ranks for our two pairs. Then we choose 2 suits for the first pair and 2 suits for the second pair. Then we choose 1 card from the @@ -159,6 +181,12 @@ + #[ First we enumerate all of the ways 4 numbers can add up to 13. $ 2 dot 4! = 8 / 35 $ + #correction[ + -2. Correct way: directly find the how many outcomes sum to 13 + $ {{1,2,3,7},{1,2,4,6},{1,3,4,5}} $ + So the answer is simply these 3 outcomes divided by total ways to choose 4 numbers from 10: + $ 3 / vec(10,4) approx 0.0143 $ + ] ] ] diff --git a/documents/by-course/pstat-120a/hw2/main.typ b/documents/by-course/pstat-120a/hw2/main.typ new file mode 100644 index 0000000..6bb1ccf --- /dev/null +++ b/documents/by-course/pstat-120a/hw2/main.typ @@ -0,0 +1,109 @@ +#import "@youwen/zen:0.1.0": * +#import "@preview/ctheorems:1.1.3": * +#import "@preview/mitex:0.2.5": * + +#show: zen.with( + title: "Homework 2", + author: "Youwen Wu", + date: "Winter 2025", +) + +#set enum(spacing: 2em) + +#let correction = content => { + set text(fill: red) + box(stroke: 1pt, inset: 5pt, content) +} + +#mitex(` +\textbf{Problem 1} + +Let $P(A)$ denote the probability that a customer watches exactly one category of programs. From the problem: +\begin{itemize} + \item 70\% watch more than one category: $P(A^c) = 0.7 \Rightarrow P(A) = 0.3$. + \item 20\% watch sports: $P(S) = 0.2$. + \item Of those watching more than one category, 15\% watch sports: $P(S | A^c) = 0.15$. +\end{itemize} + +We need $P(A \cap S^c)$, the probability a customer watches exactly one category and it is not sports: +\[ +P(A \cap S^c) = P(A) - P(A \cap S). +\] + +Since $P(A \cap S) = 0$ (sports watchers are counted under $P(A^c)$): +\[ +P(A \cap S^c) = P(A) = 0.3. +\] +\textbf{Solution:} $P(A \cap S^c) = 0.3$. + +\textbf{Problem 2} + +We need $P(6|3,4)$, the probability the 6-sided die was chosen given rolls 3 and 4. + +Using Bayes' Theorem: +\[ +P(6|3,4) = \frac{P(3,4|6) P(6)}{P(3,4)}. +\] +Assuming equal probabilities of choosing any die ($P(4) = P(6) = P(12) = \frac{1}{3}$), and independent rolls: +\[ +P(3,4|6) = P(3|6) P(4|6) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}. +\] +\[ +P(3,4) = \frac{1}{3}\left(\frac{1}{16} + \frac{1}{36} + \frac{1}{144}\right). +\] +Simplify and compute: +\[ +P(6|3,4) = \frac{\frac{1}{36} \cdot \frac{1}{3}}{\frac{1}{3}\left(\frac{1}{16} + \frac{1}{36} + \frac{1}{144}\right)}. +\] +Numerical calculation gives $P(6|3,4) \approx 0.51$. + +\textbf{Problem 3} + +Probability a marble is blue after the second draw: +\[ +P(\text{Blue on second draw}) = \frac{b}{n} \cdot \frac{b+k}{n+k} + \frac{g}{n} \cdot \frac{g+k}{n+k}. +\] +Simplify and substitute as needed. + +\textbf{Problem 4} + +(a) Probability of drawing two candies with the same flavor is: +\[ +P(\text{same flavor}) = \sum_{pockets} P(\text{flavor from pocket})^2 \cdot P(\text{pocket})^2. +\] + +(b) Bayesian calculations apply. Let heads/tails represent sequences, use Bayes' theorem. + +(c) Set probabilities equal: +\[ +\frac{2+x}{2+7+x} = \frac{5}{5+2}. +\] +Solve for $x$. + +\textbf{Problem 5} + +For independence: check $P(A \cap B) = P(A)P(B)$. +Repeat for other pairs. + +\textbf{Problem 6} + +Partition definition and law of total probability: +\[ +P(A|B) = \sum P(A|B_i)P(B_i|B). +\] +Proof by substitution. + +\textbf{Problem 7} + +(a)-(c) Condition on defendant guilt and independence. +Use $P(\text{Guilty}) = 0.7$ and $P(\text{Innocent}) = 0.3$. + +\textbf{Problem 8} + +(a) Use total probability: +\[ +P(A) = \sum P(A|word_i)P(word_i). +\] + +(b) Enumerate possible word lengths. +`) diff --git a/documents/by-course/pstat-120a/hw2/package.nix b/documents/by-course/pstat-120a/hw2/package.nix new file mode 100644 index 0000000..7879e7e --- /dev/null +++ b/documents/by-course/pstat-120a/hw2/package.nix @@ -0,0 +1,37 @@ +{ + pkgs, + typstPackagesCache, + typixLib, + cleanTypstSource, + flakeSelf, + ... +}: +let + src = cleanTypstSource ./.; + commonArgs = { + typstSource = "main.typ"; + + fontPaths = [ + # Add paths to fonts here + # "${pkgs.roboto}/share/fonts/truetype" + ]; + + virtualPaths = [ + # Add paths that must be locally accessible to typst here + # { + # dest = "icons"; + # src = "${inputs.font-awesome}/svgs/regular"; + # } + ]; + + XDG_CACHE_HOME = typstPackagesCache; + SOURCE_DATE_EPOCH = builtins.toString flakeSelf.lastModified; + }; + +in +typixLib.buildTypstProject ( + commonArgs + // { + inherit src; + } +)