#import "@youwen/zen:0.1.0": *
#import "@preview/ctheorems:1.1.3": *

#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) $
      ]
  ]