#import "@youwen/zen:0.1.0": * #import "@preview/ctheorems:1.1.3": * #import "@preview/mitex:0.2.5": * #show: zen.with( title: "Homework 1", author: "Youwen Wu", date: "Winter 2025", ) #set enum(spacing: 2em) + #[ #set enum(numbering: "a)", spacing: 2em) + #[ We know that $B$ and $B'$ are disjoint. That is, $B sect B' = emptyset$. Additionally, $ E = (A sect B) subset B \ F = (A sect B') subset B' \ $ Then we note $ forall x in E, x in B, x in.not B' \ forall y in F, y in B, y in.not B' $ So clearly $E$ and $F$ have no common elements, and $ E sect F = emptyset $ ] + #[ $ E union F &= (A sect B) union (A sect B') \ &= (A union A) sect (B union B') \ &= A sect Omega \ &= A $ ] ] + #[ #set enum(numbering: "a)", spacing: 2em) + ${15, 25, 35, 45, 51, 53, 55, 57, 59, 65, 75, 85, 95 }$ + ${50, 52, 56, 58}$ + $emptyset$ ] + #[ #set enum(numbering: "a)", spacing: 2em) + The sample space is every value of the die (1-6) paired with heads and paired with tails. That is, ${1,2,3,4,5,6} times {H,T}$, with cardinality 12. + There are $12^10$ outcomes. + If no participants roll a 5, then we omit any outcome in our sample space where the die outcome is 5, leaving us with 10 outcomes of the die and coin. Now we have $10^10$ outcomes. If at least 1 person rolls a 5, then we note that this is simply the complement of the previous result. So we have $12^10 - 10^10$ outcomes total. ] + #[ #set enum(numbering: "a)", spacing: 2em) + #[ The sample space can be represented as a 6-tuple where the position 1-6 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 $ ] + #[ When the balls are indistinguishable, we can instead represent it as 4-tuples where the position represents the 1st, 2nd, 3rd, or 4th square, and the value represents how many balls landed. Additionally the sum of all the elements must be 6. $ {{x_1, x_2, x_3, x_4} : x_i >= 0, i = 1,...,6 sum_(j=1)^4 x_j = 6} $ ] ] + #[ #set enum(numbering: "a)", spacing: 2em) + #[ We want to determine how many ways to choose 8 people from 27 people, or $vec(27,8) = 2220075$. ] + #[ This is the same as the choosing 4 of the 12 men and 4 of the 15 women, and pairing each group of men with each group of women once. So, $ vec(12,4) times vec(15, 4) = 675675 $ ] + #[ First we determine the amount of ways to choose less than 2 women. $ vec(15, 0) vec(12, 8) + vec(15, 1) times vec(12,7) $ Then the total amount of ways to choose 8 people, from part a, is $vec(27,8)$. Then the chance of forming a committee with less than 2 women is $ (vec(15, 0) vec(12, 8) + vec(15, 1) vec(12,7)) / vec(27,8) $ So our final answer is $ 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(`\[ p_4 + p_5 \;=\;\frac{17}{120} \;+\;\frac{15}{120} \;=\;\frac{32}{120} \;=\;\frac{4}{15}\;\approx\;0.267. \]`) ]