#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)

#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)

    + #[
        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)

    + ${15, 25, 35, 45, 51, 53, 55, 57, 59, 65, 75, 85, 95 }$
    + ${50, 52, 54, 56, 58}$
    + $emptyset$
  ]

+ #[
    #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 $

        #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
        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)

    #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
        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 $
      ]

    + #[
        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.

        $ 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)

    + #[
        We compute how many ways there are to not choose 3 and take the complement.

        $ (9 P 4) / (10 P 4) = 40% $
      ]

    + #[
        First we choose 4 distinct numbers from 10 and there is exactly one way
        to list them in increasing order.

        $ vec(10,4) / (10 P 4) = 1 / 24 $
      ]

    + #[
        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 $
        ]
      ]
  ]

+ #[
    #set enum(numbering: "a)", spacing: 2em)

    + #[
        We choose the 9 non-aces from the 48 remaining cards for player 1, then
        we choose the rest accordingly.

        $ vec(48, 9) vec(13, 39) vec(13, 26) vec(13, 13) $
      ]

    + #[
        $4! = 24$
      ]

    + #[
        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)

    + #[
        Same as all the permutations. $10!$
      ]

    + #[
        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 $
      ]
  ]

+ #[
    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 $p_4 + p_5$.

    $
      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
    $
  ]