272 lines
9.5 KiB
Text
272 lines
9.5 KiB
Text
#import "@youwen/zen:0.1.0": *
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#import "@preview/ctheorems:1.1.3": *
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#import "@preview/mitex:0.2.5": *
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#show: zen.with(
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title: "Homework 1",
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author: "Youwen Wu",
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date: "Winter 2025",
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)
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#set enum(spacing: 2em)
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+ #[
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#set enum(numbering: "a)", spacing: 2em)
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+ #[
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We know that $B$ and $B'$ are disjoint. That is, $B sect B' =
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emptyset$. Additionally,
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$
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E = (A sect B) subset B \
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F = (A sect B') subset B' \
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$
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Then we note
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$
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forall x in E, x in B, x in.not B' \
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forall y in F, y in B, y in.not B'
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$
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So clearly $E$ and $F$ have no common elements, and
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$
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E sect F = emptyset
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$
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]
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+ #[
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$
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E union F &= (A sect B) union (A sect B') \
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&= (A union A) sect (B union B') \
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&= A sect Omega \
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&= A
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$
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]
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]
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+ #[
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#set enum(numbering: "a)", spacing: 2em)
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+ ${15, 25, 35, 45, 51, 53, 55, 57, 59, 65, 75, 85, 95 }$
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+ ${50, 52, 56, 58}$
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+ $emptyset$
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]
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+ #[
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#set enum(numbering: "a)", spacing: 2em)
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+ 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.
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+ There are $12^10$ outcomes.
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+ 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.
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]
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+ #[
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#set enum(numbering: "a)", spacing: 2em)
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+ #[
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The sample space can be represented as a 6-tuple where the position 1-6
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represents balls numbered 1-6, and the value represents the square it
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was sent to. So it's
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$ {{x_1,x_2,x_3,x_4,x_5,x_6} : x_i in {1,2,3,4}}, i = 1,...6 $
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]
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+ #[
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When the balls are indistinguishable, we can instead represent it as
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4-tuples where the position represents the 1st, 2nd, 3rd, or 4th square,
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and the value represents how many balls landed. Additionally the sum of all
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the elements must be 6.
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$ {{x_1, x_2, x_3, x_4} : x_i >= 0, i = 1,...,6 sum_(j=1)^4 x_j = 6} $
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]
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]
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+ #[
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#set enum(numbering: "a)", spacing: 2em)
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+ #[
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We want to determine how many ways to choose 8 people from 27 people, or $vec(27,8) = 2220075$.
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]
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+ #[
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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,
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$ vec(12,4) times vec(15, 4) = 675675 $
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]
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+ #[
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First we determine the amount of ways to choose less than 2 women.
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$ vec(15, 0) vec(12, 8) + vec(15, 1) times vec(12,7) $
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Then the total amount of ways to choose 8 people, from part a, is $vec(27,8)$.
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Then the chance of forming a committee with less than 2 women is
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$ (vec(15, 0) vec(12, 8) + vec(15, 1) vec(12,7)) / vec(27,8) $
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So our final answer is
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$ 1 - (vec(15, 0) vec(12, 8) + vec(15, 1) vec(12,7)) / vec(27,8) $
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]
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]
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+ #[
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#set enum(numbering: "a)", spacing: 2em)
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+ #[
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First we choose two ranks for our two pairs. Then we choose 2 suits for the
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first pair and 2 suits for the second pair. Then we choose 1 card from the
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remaining 44 cards that aren't of the same rank as the first four.
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$ 13 dot 12 dot vec(4,2) dot vec(4,2) dot 44 $
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]
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+ #[
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First we choose a rank for our three of a kind. Then we choose 3 suits
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for the cards in our three of a kind. Then we choose a rank for our 4th
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card and a rank for our 5th card. Then we choose a suit for our 4th
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card and a suit for our 5th card.
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$ 13 dot vec(4, 3) dot 12 dot 11 dot 4^3 $
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]
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+ #[
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First we choose a rank to start the sequence. Then we choose one of two
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ranks (either above or below). Then the next 3 cards only have one
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possible rank, which is the descending or ascending ranks. Then we need
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to choose a suit for each of our cards, making sure at least one is
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different from the others.
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$ 13 dot 2 dot 4^4 dot 3 = 19968 $
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]
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]
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+ #[
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#set enum(numbering: "a)", spacing: 2em)
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An urn has 10 balls labeled 1 – 10. We draw 4 times _without_ replacement.
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There are #mitex(`\(10P4 = 10\times 9\times 8\times 7 = 5040\)`) equally‐likely ordered draws.
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1. Probability that “3” appears at least once.
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The complement is “3” does _not_ appear at all, i.e.\ all 4 draws come from the other 9 balls.
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#mitex(`
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\[
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P(\text{3 appears}) \;=\; 1 \;-\; \frac{9P4}{10P4}
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\;=\; 1 \;-\;\frac{9\times 8\times 7\times 6}{5040}
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\;=\; 1 \;-\; \frac{3024}{5040}
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\;=\;\frac{2016}{5040}
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\;=\;\frac{2}{5}.
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\]
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`)
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2. Probability that the 4 numbers are in strictly increasing order.
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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
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#mitex(`\[
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P(\text{strictly increasing}) \;=\; \frac{\binom{10}{4}}{10P4}
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\;=\;\frac{210}{5040}
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\;=\;\frac{1}{24}.
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\]`)
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3. Probability that the sum of the 4 draws is 13.
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First find all 4‐element _subsets_ of #mitex(`\(\{1,\dots,10\}\)`) summing to 13:
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#mitex(`\[
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(1,2,3,7),\quad (1,2,4,6),\quad (1,3,4,5).
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\]`)
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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
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#mitex(`\[
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P(\text{sum}=13)\;=\;\frac{72}{5040}\;=\;\frac{1}{70}.
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\]`)
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]
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+ #[
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#set enum(numbering: "a)", spacing: 2em)
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Dealing a 52‐card deck to 4 players (each gets 13).
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The total number of ways is
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#mitex(`\[
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\text{Total deals} \;=\;\frac{52!}{(13!)^4}.
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\]`)
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1. Player 1 gets all four aces.
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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
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#mitex(`\[
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\text{Ways} \;=\; \binom{48}{9}\;\times\;\binom{39}{13}\,\binom{26}{13}\,\binom{13}{13}.
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\]`)
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2. Each player’s entire 13‐card hand is “all one suit.”
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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:
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#mitex(`\[
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\text{Ways} \;=\;4!\;=\;24.
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\]`)
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3. Players 1 and 2 together get all the hearts.
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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:
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- 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).
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- The remaining 13 non‐hearts plus the 13 hearts go to Players 1+2, and we then choose which 13 go to Player 1.
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In binomial‐coefficient form:
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#mitex(`\[
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\text{Ways}
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\;=\;
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\binom{39}{13}\,\binom{26}{13}\,\binom{26}{13}.
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\]`)
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]
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+ #[
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#set enum(numbering: "a)", spacing: 2em)
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Forming 10‐letter “words” from the letters \(\{B,A,C,O,N,R,U,L,E,S\}\).
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1. Number of 10‐letter arrangements.
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All 10 letters are distinct, so there are
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#mitex(`\[
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10!\;=\;3{,}628{,}800
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\]`)
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possible orderings.
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2. Probability that the block “BACON” appears consecutively in that order.
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Treat the five letters *B A C O N* as a single block plus the other 5
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letters #mitex(`\(\{R,U,L,E,S\}\)`). That gives #mitex(`\(6\)`) total “items” to permute, so
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#mitex(`\(6!\)`) orderings. There is only 1 way to arrange the block “BACON”
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internally (since we want that exact order). Hence the favorable count is
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#mitex(`\(6!=720\)`). Therefore
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#mitex(`\[
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P(\text{“BACON” together})
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\;=\;\frac{6!}{10!}
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\;=\;\frac{720}{3{,}628{,}800}
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\;=\;\frac{1}{5040}.
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\]`)]
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+ #[
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#set enum(numbering: "a)", spacing: 2em)
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A succinct way to see the solution is to note that the six probabilities
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#mitex(`\[
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p_0,\,p_1,\,p_2,\,p_3,\,p_4,\,p_5
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\]`)
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form an arithmetic (nonincreasing) sequence, so one can write
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#mitex(`\[
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p_n \;=\;p_0 - n\,d\quad\text{for }n=0,1,2,3,4,5,
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\]`)
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where #mitex(`\(d \ge 0.\)`) The conditions then translate into the two equations
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1. The probabilities sum to 1:
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#mitex(`\[
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p_0 + p_1 + p_2 + p_3 + p_4 + p_5 \;=\;6\,p_0 - (0+1+2+3+4+5)\,d
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\;=\;6\,p_0 \;-\;15\,d\;=\;1.
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\]`)
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2. Exactly 40% of policyholders file fewer than two claims:
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#mitex(`\[
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p_0 + p_1
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\;=\;(p_0) + (p_0 - d)
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\;=\;2\,p_0 - d
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\;=\;0.40.
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\]`)
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Solving these simultaneously gives
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#mitex(`\[
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p_0 \;=\;\frac{5}{24},
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\quad
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d \;=\;\frac{1}{60}.
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\]`)
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Hence one can compute
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#mitex(`\[
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p_4 \;=\;p_0 - 4d \;=\;\tfrac{5}{24} - \tfrac{4}{60} \;=\;\tfrac{17}{120},
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\quad
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p_5 \;=\;p_0 - 5d \;=\;\tfrac{5}{24} - \tfrac{5}{60} \;=\;\tfrac{15}{120}.
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\]`)
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The probability that a policyholder files more than three claims (i.e.\ 4 or 5) is
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#mitex(`\[
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p_4 + p_5 \;=\;\frac{17}{120} \;+\;\frac{15}{120}
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\;=\;\frac{32}{120}
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\;=\;\frac{4}{15}\;\approx\;0.267.
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\]`)
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]
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