auto-update(nvim): 2025-01-20 02:19:54
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1 changed files with 80 additions and 123 deletions
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@ -1,6 +1,5 @@
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#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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@ -119,154 +118,112 @@
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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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There are 10 unordered ways to have the 5-card sequence, disregarding
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the suits. To pick suits, we can simply pick 1 of 4 suits for the 5
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cards, then subtract the number of ways that we pick all 5 suits to be
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the same. That is $4^5 - 4$, since there are exactly 4 ways we can
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choose all 5 of our cards to be the same suit.
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$ 13 dot 2 dot 4^4 dot 3 = 19968 $
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$ 10 dot (4^5 - 4) = 10200 $
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]
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+ #[
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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).
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$ vec(13,5) dot 4 - 10 $
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]
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+ #[
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First we choose a rank for our 4 of a kind, then choose any other card.
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$ 13 dot 48 = 624 $
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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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+ #[
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We compute how many ways there are to not choose 3 and take the complement.
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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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$ (9 P 4) / (10 P 4) = 40% $
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]
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2. Probability that the 4 numbers are in strictly increasing order.
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+ #[
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First we choose 4 distinct numbers from 10 and there is exactly one way
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to list them in 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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$ vec(10,4) / (10 P 4) = 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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First we enumerate all of the ways 4 numbers can add up to 13.
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$ 2 dot 4! = 8 / 35 $
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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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+ #[
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We choose the 9 non-aces from the 48 remaining cards for player 1, then
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we choose the rest accordingly.
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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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$ vec(48, 9) vec(13, 39) vec(13, 26) vec(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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+ #[
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$4! = 24$
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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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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.
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$ vec(39,13) vec(26,13) vec(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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+ #[
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Same as all the permutations. $10!$
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]
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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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Treat the five letters $B A C O N$ as a single block to move around. Then our permutations go down to $6!$.
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$ P(E) = (6!) / (10!) = 1 / 5040 $
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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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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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From these observations we note that $p_n = p_0 - n d$. Then
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$
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p_1 = p_0 - d \
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p_0 + p_1 = 0.4 \
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2 p_0 - d = 0.4
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$
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Also, we know the probabilities sum to 1, which gives us an equation
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$
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sum_(n=0)^5 p_n = 6 p_0 - (1 + 2 + 3 + 4 + 5) d = 6 p_0 - 15d = 1 \
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p_0 = (1 + 15d) / 6
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$
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Then we solve the equations simultaneously to obtain
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$
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(1+15d) / 3 - d = 0.4 \
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d = 0.2 / 12 = 1 / 60
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$
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Now we can compute $p_0$,
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$
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p_0 = 5 / 24
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$
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Then we can compute the answer, which is $p_4 + p_5$.
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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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p_4 &= p_0 - 4 / 60 \
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p_5 &= p_4 - 1 / 60 \
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p_4 + p_5 &= 2(5 / 24 - 4 / 60) - 1 / 60 \
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= 16 / 60 &= 4 / 15
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$
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]
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