From 27717552932f7ebc116d57b6ee11b0f8281085a4 Mon Sep 17 00:00:00 2001 From: Youwen Wu Date: Sun, 19 Jan 2025 23:49:30 -0800 Subject: [PATCH] auto-update(nvim): 2025-01-19 23:49:30 --- .gitignore | 1 + documents/by-course/pstat-120a/hw1/main.typ | 174 ++++++++++++++++++++ 2 files changed, 175 insertions(+) diff --git a/.gitignore b/.gitignore index 701aa37..2113ba0 100644 --- a/.gitignore +++ b/.gitignore @@ -1,5 +1,6 @@ result main.pdf +*.pdf public node_modules diff --git a/documents/by-course/pstat-120a/hw1/main.typ b/documents/by-course/pstat-120a/hw1/main.typ index 57c6e6c..730f70a 100644 --- a/documents/by-course/pstat-120a/hw1/main.typ +++ b/documents/by-course/pstat-120a/hw1/main.typ @@ -1,5 +1,6 @@ #import "@youwen/zen:0.1.0": * #import "@preview/ctheorems:1.1.3": * +#import "@preview/mitex:0.2.5": * #show: zen.with( title: "Homework 1", @@ -96,3 +97,176 @@ $ 1 - (vec(15, 0) vec(12, 8) + vec(15, 1) vec(12,7)) / vec(27,8) $ ] ] + ++ #[ + #set enum(numbering: "a)", spacing: 2em) + + + #[ + 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 + remaining 44 cards that aren't of the same rank as the first four. + + $ 13 dot 12 dot vec(4,2) dot vec(4,2) dot 44 $ + ] + + + #[ + First we choose a rank for our three of a kind. Then we choose 3 suits + for the cards in our three of a kind. Then we choose a rank for our 4th + card and a rank for our 5th card. Then we choose a suit for our 4th + card and a suit for our 5th card. + + $ 13 dot vec(4, 3) dot 12 dot 11 dot 4^3 $ + ] + + + #[ + First we choose a rank to start the sequence. Then we choose one of two + ranks (either above or below). Then the next 3 cards only have one + possible rank, which is the descending or ascending ranks. Then we need + to choose a suit for each of our cards, making sure at least one is + different from the others. + + $ 13 dot 2 dot 4^4 dot 3 = 19968 $ + ] + ] + ++ #[ + #set enum(numbering: "a)", spacing: 2em) + An urn has 10 balls labeled 1 – 10. We draw 4 times _without_ replacement. + There are #mitex(`\(10P4 = 10\times 9\times 8\times 7 = 5040\)`) equally‐likely ordered draws. + + 1. Probability that “3” appears at least once. + + The complement is “3” does _not_ appear at all, i.e.\ all 4 draws come from the other 9 balls. + #mitex(` + \[ + P(\text{3 appears}) \;=\; 1 \;-\; \frac{9P4}{10P4} + \;=\; 1 \;-\;\frac{9\times 8\times 7\times 6}{5040} + \;=\; 1 \;-\; \frac{3024}{5040} + \;=\;\frac{2016}{5040} + \;=\;\frac{2}{5}. + \] + `) + + 2. Probability that the 4 numbers are in strictly increasing order. + + To be strictly increasing, one simply chooses which 4 distinct numbers (out of 10) and then there is exactly _one_ way to list them in increasing order. Hence the favorable cases are #mitex(`\(\binom{10}{4}\)`). So + #mitex(`\[ + P(\text{strictly increasing}) \;=\; \frac{\binom{10}{4}}{10P4} + \;=\;\frac{210}{5040} + \;=\;\frac{1}{24}. + \]`) + + 3. Probability that the sum of the 4 draws is 13. + + First find all 4‐element _subsets_ of #mitex(`\(\{1,\dots,10\}\)`) summing to 13: + #mitex(`\[ + (1,2,3,7),\quad (1,2,4,6),\quad (1,3,4,5). + \]`) + There are exactly 3 such sets. Each set of 4 distinct numbers can appear in \(4!\) different orders among the draws. Thus the number of favorable ordered draws is \(3\times 4!=72.\) Therefore + #mitex(`\[ + P(\text{sum}=13)\;=\;\frac{72}{5040}\;=\;\frac{1}{70}. + \]`) + ] + ++ #[ + #set enum(numbering: "a)", spacing: 2em) + Dealing a 52‐card deck to 4 players (each gets 13). + + The total number of ways is + #mitex(`\[ +\text{Total deals} \;=\;\frac{52!}{(13!)^4}. +\]`) + + 1. Player 1 gets all four aces. + We must choose the remaining 9 cards in Player 1’s hand from the 48 non‐aces, and then distribute the remaining 39 cards among Players 2, 3, 4. Hence + #mitex(`\[ + \text{Ways} \;=\; \binom{48}{9}\;\times\;\binom{39}{13}\,\binom{26}{13}\,\binom{13}{13}. + \]`) + + 2. Each player’s entire 13‐card hand is “all one suit.” + Since each suit has exactly 13 cards, this can only happen if one suit goes entirely to Player 1, another suit to Player 2, etc. There are 4 suits and 4 players, so the number of ways is simply the number of ways to _assign_ each suit to a distinct player: + #mitex(`\[ + \text{Ways} \;=\;4!\;=\;24. + \]`) + + 3. Players 1 and 2 together get all the hearts. + There are 13 hearts and 39 other cards. Players 3 and 4 must then share the 39 non‐hearts only, while the 13 hearts + 13 of the non‐hearts go to Players 1 and 2. One convenient count is: + - Choose which 26 of the 39 non‐hearts go to Players 3+4, then choose 13 of those for Player 3 (and 13 for Player 4). + - The remaining 13 non‐hearts plus the 13 hearts go to Players 1+2, and we then choose which 13 go to Player 1. + In binomial‐coefficient form: + #mitex(`\[ + \text{Ways} + \;=\; + \binom{39}{13}\,\binom{26}{13}\,\binom{26}{13}. + \]`) + ] + ++ #[ + #set enum(numbering: "a)", spacing: 2em) + Forming 10‐letter “words” from the letters \(\{B,A,C,O,N,R,U,L,E,S\}\). + + 1. Number of 10‐letter arrangements. + All 10 letters are distinct, so there are + #mitex(`\[ + 10!\;=\;3{,}628{,}800 + \]`) + possible orderings. + + 2. Probability that the block “BACON” appears consecutively in that order. + Treat the five letters *B A C O N* as a single block plus the other 5 + letters #mitex(`\(\{R,U,L,E,S\}\)`). That gives #mitex(`\(6\)`) total “items” to permute, so + #mitex(`\(6!\)`) orderings. There is only 1 way to arrange the block “BACON” + internally (since we want that exact order). Hence the favorable count is + #mitex(`\(6!=720\)`). Therefore + #mitex(`\[ + P(\text{“BACON” together}) + \;=\;\frac{6!}{10!} + \;=\;\frac{720}{3{,}628{,}800} + \;=\;\frac{1}{5040}. + \]`)] + ++ #[ + #set enum(numbering: "a)", spacing: 2em) + A succinct way to see the solution is to note that the six probabilities + #mitex(`\[ +p_0,\,p_1,\,p_2,\,p_3,\,p_4,\,p_5 +\]`) + form an arithmetic (nonincreasing) sequence, so one can write + #mitex(`\[ +p_n \;=\;p_0 - n\,d\quad\text{for }n=0,1,2,3,4,5, +\]`) + where #mitex(`\(d \ge 0.\)`) The conditions then translate into the two equations + + 1. The probabilities sum to 1: + #mitex(`\[ + p_0 + p_1 + p_2 + p_3 + p_4 + p_5 \;=\;6\,p_0 - (0+1+2+3+4+5)\,d + \;=\;6\,p_0 \;-\;15\,d\;=\;1. + \]`) + + 2. Exactly 40% of policyholders file fewer than two claims: + #mitex(`\[ + p_0 + p_1 + \;=\;(p_0) + (p_0 - d) + \;=\;2\,p_0 - d + \;=\;0.40. + \]`) + + Solving these simultaneously gives + #mitex(`\[ +p_0 \;=\;\frac{5}{24}, +\quad +d \;=\;\frac{1}{60}. +\]`) + Hence one can compute + #mitex(`\[ +p_4 \;=\;p_0 - 4d \;=\;\tfrac{5}{24} - \tfrac{4}{60} \;=\;\tfrac{17}{120}, +\quad +p_5 \;=\;p_0 - 5d \;=\;\tfrac{5}{24} - \tfrac{5}{60} \;=\;\tfrac{15}{120}. +\]`) + The probability that a policyholder files more than three claims (i.e.\ 4 or 5) is + #mitex(`\[ +p_4 + p_5 \;=\;\frac{17}{120} \;+\;\frac{15}{120} +\;=\;\frac{32}{120} +\;=\;\frac{4}{15}\;\approx\;0.267. +\]`) + ]