diff --git a/documents/by-course/pstat-120a/hw1/main.typ b/documents/by-course/pstat-120a/hw1/main.typ index 730f70a..d19eb44 100644 --- a/documents/by-course/pstat-120a/hw1/main.typ +++ b/documents/by-course/pstat-120a/hw1/main.typ @@ -1,6 +1,5 @@ #import "@youwen/zen:0.1.0": * #import "@preview/ctheorems:1.1.3": * -#import "@preview/mitex:0.2.5": * #show: zen.with( title: "Homework 1", @@ -119,154 +118,112 @@ ] + #[ - 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. + There are 10 unordered ways to have the 5-card sequence, disregarding + the suits. To pick suits, we can simply pick 1 of 4 suits for the 5 + cards, then subtract the number of ways that we pick all 5 suits to be + the same. That is $4^5 - 4$, since there are exactly 4 ways we can + choose all 5 of our cards to be the same suit. - $ 13 dot 2 dot 4^4 dot 3 = 19968 $ + $ 10 dot (4^5 - 4) = 10200 $ + ] + + + #[ + We can simply choose 5 cards from the 13 per rank, multiply by the 4 suits, and then substract the amount of ways we can get a straight (which is 10). + + $ vec(13,5) dot 4 - 10 $ + ] + + + #[ + First we choose a rank for our 4 of a kind, then choose any other card. + + $ 13 dot 48 = 624 $ ] ] + #[ #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. + + #[ + We compute how many ways there are to not choose 3 and take the complement. - The complement is “3” does _not_ appear at all, i.e.\ all 4 draws come from the other 9 balls. - #mitex(` - \[ - P(\text{3 appears}) \;=\; 1 \;-\; \frac{9P4}{10P4} - \;=\; 1 \;-\;\frac{9\times 8\times 7\times 6}{5040} - \;=\; 1 \;-\; \frac{3024}{5040} - \;=\;\frac{2016}{5040} - \;=\;\frac{2}{5}. - \] - `) + $ (9 P 4) / (10 P 4) = 40% $ + ] - 2. Probability that the 4 numbers are in strictly increasing order. + + #[ + First we choose 4 distinct numbers from 10 and there is exactly one way + to list them in 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(`\[ - P(\text{strictly increasing}) \;=\; \frac{\binom{10}{4}}{10P4} - \;=\;\frac{210}{5040} - \;=\;\frac{1}{24}. - \]`) + $ vec(10,4) / (10 P 4) = 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(`\[ - P(\text{sum}=13)\;=\;\frac{72}{5040}\;=\;\frac{1}{70}. - \]`) + + #[ + First we enumerate all of the ways 4 numbers can add up to 13. + $ 2 dot 4! = 8 / 35 $ + ] ] + #[ #set enum(numbering: "a)", spacing: 2em) - Dealing a 52‐card deck to 4 players (each gets 13). - The total number of ways is - #mitex(`\[ -\text{Total deals} \;=\;\frac{52!}{(13!)^4}. -\]`) + + #[ + We choose the 9 non-aces from the 48 remaining cards for player 1, then + we choose the rest accordingly. - 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(`\[ - \text{Ways} \;=\; \binom{48}{9}\;\times\;\binom{39}{13}\,\binom{26}{13}\,\binom{13}{13}. - \]`) + $ vec(48, 9) vec(13, 39) vec(13, 26) vec(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(`\[ - \text{Ways} \;=\;4!\;=\;24. - \]`) + + #[ + $4! = 24$ + ] - 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(`\[ - \text{Ways} - \;=\; - \binom{39}{13}\,\binom{26}{13}\,\binom{26}{13}. - \]`) + + #[ + Select 13 cards from the 39 non-hearts for player 4, select 13 from the 26 non-hearts for player 3, then select 13 hearts from the 26 cards left distributed amongst player 1 and 2. + + $ vec(39,13) vec(26,13) vec(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. + + #[ + Same as all the permutations. $10!$ + ] - 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(`\[ - P(\text{“BACON” together}) - \;=\;\frac{6!}{10!} - \;=\;\frac{720}{3{,}628{,}800} - \;=\;\frac{1}{5040}. - \]`)] + + #[ + Treat the five letters $B A C O N$ as a single block to move around. Then our permutations go down to $6!$. + + $ P(E) = (6!) / (10!) = 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 + From these observations we note that $p_n = p_0 - n d$. Then + $ + p_1 = p_0 - d \ + p_0 + p_1 = 0.4 \ + 2 p_0 - d = 0.4 + $ + Also, we know the probabilities sum to 1, which gives us an equation + $ + sum_(n=0)^5 p_n = 6 p_0 - (1 + 2 + 3 + 4 + 5) d = 6 p_0 - 15d = 1 \ + p_0 = (1 + 15d) / 6 + $ + Then we solve the equations simultaneously to obtain + $ + (1+15d) / 3 - d = 0.4 \ + d = 0.2 / 12 = 1 / 60 + $ + Now we can compute $p_0$, + $ + p_0 = 5 / 24 + $ + Then we can compute the answer, which is $p_4 + p_5$. - 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(`\[ -p_4 + p_5 \;=\;\frac{17}{120} \;+\;\frac{15}{120} -\;=\;\frac{32}{120} -\;=\;\frac{4}{15}\;\approx\;0.267. -\]`) + $ + p_4 &= p_0 - 4 / 60 \ + p_5 &= p_4 - 1 / 60 \ + p_4 + p_5 &= 2(5 / 24 - 4 / 60) - 1 / 60 \ + = 16 / 60 &= 4 / 15 + $ ]