auto-update(nvim): 2025-01-19 23:49:30

.gitignore

@@ -1,5 +1,6 @@
 result
 main.pdf
+*.pdf
 
 public
 node_modules

documents/by-course/pstat-120a/hw1/main.typ

@@ -1,5 +1,6 @@
 #import "@youwen/zen:0.1.0": *
 #import "@preview/ctheorems:1.1.3": *
+#import "@preview/mitex:0.2.5": *
 
 #show: zen.with(
   title: "Homework 1",
@@ -96,3 +97,176 @@
         $ 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(`\[
+p_4 + p_5 \;=\;\frac{17}{120} \;+\;\frac{15}{120}
+\;=\;\frac{32}{120}
+\;=\;\frac{4}{15}\;\approx\;0.267.
+\]`)
+  ]