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  1. \documentclass[a4paper,12pt]{article}
  2. \usepackage[english,vietnamese]{babel}
  3. \usepackage{amsmath}
  4. \usepackage{amssymb}
  5. \usepackage{enumerate}
  6. \usepackage{mathtools}
  7. \usepackage{pgfplots}
  8. \usepackage{venndiagram}
  9.  
  10. \newcommand{\E}{\mathbf E}
  11. \newcommand{\C}{\mathrm C}
  12. \newcommand{\var}{\mathrm{var}}
  13. \newcommand{\exercise}[1]{\noindent\textbf{#1.}}
  14. \renewcommand*{\thefootnote}{\fnsymbol{footnote}}
  15.  
  16. \title{Probability Homework}
  17. \author{Nguyễn Gia Phong}
  18. \date{Fall 2019}
  19.  
  20. \begin{document}
  21. \maketitle
  22. \section{Basic Probability 1}
  23.  
  24. \exercise 1  Problems regarding de Morgan's law
  25. \begin{enumerate}[(a)]
  26.   \item Consider rolling a six-sided die, where
  27.     \begin{align*}
  28.       &\begin{cases}
  29.         A = \{2, 4, 6\}\iff A^\C = \{1, 3, 5\}\\
  30.         B = \{4, 5, 6\}\iff B^\C = \{1, 2, 3\}
  31.       \end{cases}\\
  32.       \Longrightarrow &\begin{cases}
  33.         (A \cup B)^\C = \{2, 4, 5, 6\}^\C = \{1, 3\} = A^\C \cap B^\C\\
  34.         (A \cap B)^\C = \{4, 6\}^\C = \{1, 2, 3, 5\} = A^\C \cup B^\C\\
  35.       \end{cases}
  36.     \end{align*}
  37.   \item By de Morgan's law,
  38.     \begin{align*}
  39.       P\left(A^\C \cap B^\C\right) &= P\left((A \cup B)^\C\right)\\
  40.       &= 1 - P\left(A \cup \left(A^\C \cap B\right)\right)\\
  41.       &= 1 - P(A) - P\left(A^C \cap B\right)
  42.       \tag{since $A \cap \left(A^\C \cap B\right) = \varnothing$}\\
  43.       &= 1 -P(A) -P\left((A\cap B)\cup\left(A^\C\cap B\right)\right) +P(A\cap B)
  44.       \tag{since $(A\cap B)\cap\left(A^\C\cap B\right) = \varnothing$}\\
  45.       &= 1 - P(A) - P(B) + P(A\cap B)
  46.     \end{align*}\label{1.b}
  47.   \item Consider events A and B such that $P(A) = 1/2$, $P(A\cup B) = 3/4$,
  48.     $P\left(B^\C\right) = 5/8$.
  49.  
  50.     \begin{align*}
  51.       & P\left(A^\C\cap B\right) = P(A\cup B) - P(A)
  52.       = \frac 3 4 - \frac 1 2 = \frac 1 4\\
  53.       & P\left(A^\C\cap B^\C\right) = P\left((A\cup B)^\C\right)
  54.       = P(\Omega) - P(A\cup B) = 1 - \frac 3 4 = \frac 1 4\\
  55.       & P\left(A\cap B^\C\right) = P\left(B^\C\right)-P\left((A\cup B)^\C\right)
  56.       = \frac 5 8 - \frac 1 4 = \frac 3 8\\
  57.       & P(A\cap B) = P(A) - P\left(A\cap B^\C\right)
  58.       = \frac 1 2 - \frac 3 8 = \frac 1 8\\
  59.       & P\left(A^\C\cup B^\C\right) = P\left((A\cap B)^\C\right)
  60.       = P(\Omega) - P(A\cap B) = 1 - \frac 1 8 = \frac 7 8
  61.     \end{align*}
  62. \end{enumerate}
  63.  
  64. \exercise 2  A four-sided die is rolled repeatedly,
  65. until the first time (if ever) that an even number is obtained.
  66. What is the sample space for this experiment?
  67.  
  68. Let the outcome be a n-dimensional vector, whose elements are values
  69. of each roll in chronological order.  The sample space would then be
  70. \[\Omega = \{v \in \{1, 3\}^m\times\{2, 4\} \mid m \in \mathbb N\}\]
  71.  
  72. \exercise 3  A ball is drawn at random from a box containing 6 red balls,
  73. 4 white balls, and 5 blue balls.
  74.  
  75. Let $\Omega$ be the sample space then $n(\Omega) = 6 + 4 + 5 = 15$.
  76. Let the R, W and B be the event where a red, white and blue ball is drawn
  77. respectively, each of these events are mutually exclusive.  Suppose each ball
  78. is equally likely to be drawn, we get
  79. \[\begin{cases}
  80.   n(R) = 6\\
  81.   n(W) = 4\\
  82.   n(B) = 5
  83. \end{cases}
  84. \Longrightarrow\begin{dcases}
  85.   P(R) = \frac{n(R)}{n(\Omega)} = \frac{2}{5}\\
  86.   P(W) = \frac{n(W)}{n(\Omega)} = \frac{4}{15}\\
  87.   P(B) = \frac{n(B)}{n(\Omega)} = \frac{1}{3}
  88. \end{dcases}\]
  89.  
  90. \begin{enumerate}[(a)]
  91.   \item For a ball that is not red to be drawn, the probability is
  92.     \[P\left(R^\C\right) = P(\Omega) - P(R) = 1 - \frac 2 5 = \frac 3 5\]
  93.   \item For a ball that is either red or white to be drawn, the probability is
  94.     \[P(R\cup W) = P(R) + P(W) = \frac{2}{5} + \frac{4}{15} = \frac 2 3\]
  95. \end{enumerate}
  96.  
  97. \exercise 4  Given $P(C_a) = 0.8$, $P(C_b) = 0.6$ and $P(C_a\cap C_b) = 0.5$.
  98.  
  99. We can easily prove that $P(C_a\cup C_b) = P(C_a) + P(C_b) - P(C_a\cap C_b)$
  100. (similar to what we did in exercise 1.b).  Thus the probability that the student
  101. will get at least one offer from these two companies is $0.8 + 0.6 - 0.5 = 0.9$.
  102.  
  103. \exercise 5  Let G and C be the events that the selected student is a genius and
  104. is a chocolate lover, respectively, then $P(G) = 0.6$, $P(C) = 0.7$ and
  105. $P(G\cap C) = 0.4$.  The probability that a randomly selected student is
  106. neither a genius nor a chocolate lover is
  107. \[P\left((G\cup C)^\C\right) = 1 - P(G) - P(C) + P(G\cap C)
  108. = 1 - 0.6 - 0.7 + 0.4 = 0.1\]
  109.  
  110. \exercise 6  First, consider Rick's choice of entrance.  We denote
  111. the outcome that he chooses each gate as $R_A$, $R_B$, $R_C$ and $R_D$,
  112. then $P\{R_A\} = 1/3$ and $P\{R_B\} = P\{R_C\} = P\{R_D\} = 2/9$.
  113. The sample space is $\Omega_R = \{R_A, R_B, R_C, R_D\}$.
  114.  
  115. Similarly, denote Brenda's and Ali's choices as $B_Y$ and $A_X$ respectively,
  116. where $X$, $Y$ (and later $Z$) are one of the four entrances
  117. $\omega = \{A, B, C, D\}$, we get
  118. \[\begin{dcases}
  119.   P\{B_A\} = P\{B_B\} = P\{B_C\} = P\{B_D\} = \frac 1 4\\
  120.   P\{A_A\} = P\{A_B\} = \frac{2}{35}\\
  121.   P\{A_C\} = \frac 2 7\\
  122.   P\{A_D\} = \frac 3 5
  123. \end{dcases}\]
  124. The sample spaces of these two models are $\Omega_B = \{B_A, B_B, B_C, B_D\}$
  125. and $\Omega_A = \{A_A, A_B, A_C, A_D\}$.
  126.  
  127. Now consider the probability model of the choices of the three friends.
  128. The sample space is $\Omega = \Omega_R\times\Omega_B\times\Omega_A$.
  129. Since the three friends chooses their entrance independently,
  130. for all $\mathbf v = \langle R_Z, B_Y, A_X\rangle$ in $\Omega$,
  131. \[P\{\mathbf v\} = P\{R_Z\} \cdot P\{B_Y\} \cdot P\{A_X\}\]
  132.  
  133. \begin{enumerate}[(a)]
  134.   \item The event that at least two friends choose entrance B is
  135.     \[a = (\Omega_R \times \{B_B\} \times \{A_B\})
  136.     \cup (\{R_B\} \times \Omega_B \times \{A_B\})
  137.     \cup (\{R_B\} \times \{B_B\} \times \Omega_A)\]
  138.  
  139.     Notice that
  140.     \begin{align*}
  141.       &(\Omega_R \times \{B_B\} \times \{A_B\})
  142.       \cap (\{R_B\} \times \Omega_B \times \{A_B\})\\
  143.       =\,&(\Omega_R \times \{B_B\} \times \{A_B\})
  144.       \cap (\{R_B\} \times \Omega_B \times \{A_B\})
  145.       \cap (\{R_B\} \times \{B_B\} \times \Omega_A)\\
  146.       =\,&\{R_B, B_B, A_B\}
  147.     \end{align*}
  148.  
  149.     Therefore the probability of this event is
  150.     \begin{align*}
  151.       P(a) &= P(\Omega_A \times \{B_B\} \times \{A_B\})\\
  152.       &+ P(\{R_B\} \times \Omega_B \times \{A_B\})\\
  153.       &- P\{R_B, B_B, A_B\}\\
  154.       &+ P(\{R_B\} \times \{B_B\} \times \Omega_A)\\
  155.       &- P\{R_B, B_B, A_B\}\\
  156.       &= P\{B_B\} \cdot P\{A_B\}\\
  157.       &+ P\{R_B\} \cdot P\{A_B\}\\
  158.       &+ P\{R_B\} \cdot P\{B_B\}\\
  159.       &- 2 \cdot P\{R_B\} \cdot P\{B_B\} \cdot P\{A_B\}\\
  160.       &= \frac{1}{4} \cdot \frac{2}{35}
  161.        + \frac{2}{9} \cdot \frac{2}{35}
  162.        + \frac{2}{9} \cdot \frac{1}{4}
  163.        - \frac{2}{9} \cdot \frac{1}{4} \cdot \frac{2}{35}
  164.        = \frac{8}{105}
  165.     \end{align*}
  166.   \item The only four cases where all friends choose the same entrance are
  167.     $\{\langle R_X, B_X, A_X\rangle \mid X = \omega\}$.
  168.     Hence the probability of this event is
  169.     \begin{align*}
  170.       P(b) &= \sum_{X \in \omega} P\{R_X\}\cdot P\{B_X\}\cdot P\{A_X\}\\
  171.       &= \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{2}{35}
  172.        + \frac{2}{9} \cdot \frac{1}{4} \cdot \frac{2}{35}
  173.        + \frac{2}{9} \cdot \frac{1}{4} \cdot \frac{2}{7}
  174.        + \frac{2}{9} \cdot \frac{1}{4} \cdot \frac{3}{5}
  175.        = \frac{2}{35}
  176.     \end{align*}
  177. \end{enumerate}
  178.  
  179. \exercise 7  We roll two fair six-sided dice.
  180. \begin{enumerate}[(a)]
  181.   \item The event that doubles are rolled has six outcomes, thus its probability
  182.     is 6/36 = 1/6.
  183.   \item Among the six outcomes where the result is four or less
  184.     (\{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)\}),
  185.     there are two that are doubles, hence the probability would then be 1/3.
  186.   \item Let $\omega = \{1, 2, 3, 4, 5, 6\}$,
  187.     the sample space is $\Omega = \omega^2$.  For one die roll is a six,
  188.     the event is $C = (\{6\}\times\omega) \cup (\omega\times\{6\})$.
  189.     Since $(\{6\}\times\omega) \cap (\omega\times\{6\}) = \{(6, 6)\}$,
  190.     \begin{align*}
  191.       P(C) &= P(\{6\}\times\omega) + P(\omega\times\{6\}) - P\{(6, 6)\}\\
  192.       &= \frac{n(\{6\}\times\omega) + n(\omega\times\{6\}) - n\{(6, 6)\}}
  193.               {n(\Omega)}\\
  194.       &= \frac{6 + 6 - 1}{36} = \frac{11}{36}
  195.     \end{align*}
  196. \end{enumerate}
  197.  
  198. \exercise 8  A baby rolls two six-sided dice.  Assumed that the dice are fair.
  199. Let $\omega = \{1, 2, 3, 4, 5, 6\}$, the sample space is $\Omega = \omega^2$.
  200. \begin{enumerate}[(a)]
  201.   \item There are six outcomes where the result of seven:
  202.     \[A = \{(m, 7 - m) \mid m \in \omega\}\]
  203.     hence this event's probability is
  204.     \[P(A) = \frac{n(A)}{n(\Omega)} = \frac{6}{36} = \frac{1}{6}\]
  205.   \item There are two outcomes where the result of eleven:
  206.     $B = \{(5, 6), (6, 5)\}$, thus $P(B) = 1/18$.
  207.     As $A$ and $B$ are disjoint, the probability of not getting
  208.     a sum of seven or eleven is
  209.     \begin{align*}
  210.       P\left((A\cup B)^\C\right) &= P(\Omega) - P(A\cup B)\\
  211.       &= 1 - (P(A) + P(B))\\
  212.       &= 1 - \frac{1}{6} - \frac{1}{18} = \frac{7}{9}
  213.     \end{align*}
  214. \end{enumerate}
  215.  
  216. \exercise 9  Given $n(\Omega) = 25$, $n(C) = 9$, $n(D) = 8$
  217. and $n\left((C\cup D)^\C\right) = 10$.
  218.  
  219. By the Venn diagram, $a = C\cap D^\C$, $b = C\cap D$ and $c = C^\C\cap D$.
  220. Let $E = C\cup D$ and $d = E^\C$, $n(d) = 10$ and $n(E) = n(\Omega)-n(d) = 15$.
  221. Assume that the boy fairly randomly selected,
  222. \begin{align*}
  223.   P(E) &= \frac{n(E)}{n(\Omega)}
  224.   = \frac{15}{25}
  225.   = \frac{3}{5}\\
  226.   P(a) &= P\left((D\cap d)^\C\right)
  227.   = 1 - P(D\cap d)
  228.   = 1 - P(D) - P(d)\\
  229.   &= 1 - \frac{n(D) + n(d)}{n(\Omega)}
  230.   = 1 - \frac{8 + 10}{25}
  231.   = \frac{7}{25}\\
  232.   P(c) &= P\left((C\cap d)^\C\right)
  233.   = 1 - P(C\cap d)
  234.   = 1 - P(C) - P(d)\\
  235.   &= 1 - \frac{n(C) + n(d)}{n(\Omega)}
  236.   = 1 - \frac{9 + 10}{25}
  237.   = \frac{6}{25}\\
  238.   P(b) &= \frac{n(b)}{n(\Omega)}
  239.   = \frac{n(E) - n(a) - n(c)}{n(\Omega)}
  240.   = P(E) - P(a) - P(c)\\
  241.   &= \frac{3}{5} - \frac{7}{25} - \frac{6}{25}
  242.   = \frac{2}{25}
  243. \end{align*}
  244.  
  245. \exercise{10}
  246. \begin{enumerate}[(a)]
  247.   \item Venn diagram:
  248.     \begin{venndiagram3sets}[
  249.       labelOnlyB={5},
  250.       labelABC={2},
  251.       labelNotABC={0},
  252.       overlap=1cm]
  253.       \setpostvennhook{
  254.         \draw[-stealth] (labelA) -- ++(135:2.5cm) node[left]{15};
  255.         \draw[-stealth] (labelB) -- ++(45:2.5cm) node[right]{8};
  256.         \draw[-stealth] (labelC) -- ++(-90:2cm) node[below]{12};
  257.         \draw[-stealth] (0,0) -- ++(-135:1.5cm) node[below]{27};
  258.         \draw[-stealth] (1.4,1.9) -- ++(180:2.5cm) node[left]{4};
  259.         \draw[-stealth] (2.4,3.6) -- ++(-157:3.8cm);
  260.         \draw[-stealth] (4,3) -- ++(-70:4cm) node[below]{21};
  261.         \draw[-stealth] (1,2.5) -- ++(-40:5.3cm);
  262.         \draw[-stealth] (3.3,1.5) -- ++(-50:3cm);
  263.       }
  264.     \end{venndiagram3sets}
  265.   \item The number of tourists who had not visited Burundi:
  266.     \[n\left(B^\C\right) = n(\Omega) - n(B) = 27 - 8 = 19\]
  267.   \item The number of tourists who had not visited Cameroon
  268.     unless they had visited all three countries:
  269.     \[n\left((A\cap B)\cup C^\C\right)
  270.     = n(\Omega) - n(C) + n(A\cap B\cap C)
  271.     = 27 - 12 + 2 = 17\]
  272.   \item For the randomly selected tourist to have visited at least
  273.     two countries, that person must not visited only one country.
  274.     Thus the event can be denoted as
  275.     \[d = \Omega\setminus((A\setminus B\setminus C)\cup
  276.     (B\setminus C\setminus A)\cup(C\setminus A\setminus B))\]
  277.     Since the selection is random, the event's probability can be calculated as
  278.     \[P(d) = 1 - \frac{n((A\setminus B\setminus C)\cup
  279.     (B\setminus C\setminus A)\cup(C\setminus A\setminus B))}{n(\Omega)}
  280.     = 1 - \frac{21}{27} = \frac{2}{9}\]
  281. \end{enumerate}
  282. \pagebreak
  283.  
  284. \section{Basic Probability 2}
  285. \exercise 1  Let $A$ be the event that the chosen transistor is defective,
  286. $B$ be the event that the chosen one is partially defective
  287. and $C$ be the event that the chosen one is acceptable.
  288. $A$, $B$ and $C$ are disjoint and $A\cup B\cup C = \Omega$, thus
  289. \[n(\Omega) = n(A) + n(B) + n(C) = 5 + 10 + 25 = 40\]
  290.  
  291. The probability that the chosen transistor does not immediately fail is
  292. $P\left(A^\C\right) = 1 - P(A) = 1 - n(A)/n(\Omega) = 1 - 5/40 = 7/8$.
  293.  
  294. Given this condition, the probability the chosen transistor is acceptable is
  295. \[P\left(C|A^\C\right) = \frac{P\left(C\cap A^\C\right)}{P\left(A^C\right)}
  296. = \frac{P(C)}{7/8}
  297. = \frac{8n(C)}{7n(\Omega)} = \frac{8\cdot 25}{7\cdot 40} = \frac 5 7\]
  298.  
  299. \exercise 2  Denote the outcomes of tossing a coin as $H$ (head) and $T$ (tail).
  300. \begin{enumerate}[(a)]
  301.   \item Consider tossing a coin $n$ times, the sample space is
  302.     $\Omega = \{H, T\}^n$.  Let $A$ be the event of getting at least a head,
  303.     $A^\C$ would then be getting all tails ($\{T\}^n$).  Suppose the chance of getting
  304.     head and tail are equal,
  305.     \[P\left(A^\C\right) = \frac{n\left(A^\C\right)}{n(\Omega)} = \frac{1}{2^n}
  306.     \Longrightarrow P(A) = 1 - P\left(A^\C\right) = \frac{2^n - 1}{2^n}\]
  307.   \item For $n = 4$, $P(A) = \dfrac{2^4 - 1}{2^4} = \dfrac{15}{16}$.
  308.   \item Consider rolling a six-sided die $n$ times, the sample space is
  309.     \[\Omega = \{1, 2, 3, 4, 5, 6\}^n\]
  310.     Let $B$ be the event of getting a six, $B^\C = \{1, 2, 3, 4, 5\}^n$.
  311.     Therefore the probability of $B$ is
  312.     \[P(B) = 1 - P\left(B^\C\right)
  313.     = 1 - \frac{n\left(B^\C\right)}{n(\Omega)} = 1 - \frac{5^n}{6^n}\]
  314.     For $n = 4$, $P(B) = 671/1296$.
  315.   \item For $P(B) = 0.99$,
  316.     \[1 - \left(\frac{5}{6}\right)^n = 0.99
  317.     \iff \left(\frac{5}{6}\right)^n = 0.01
  318.     \iff n = \log_{5/6}0.01 \approx 25\]
  319. \end{enumerate}
  320.  
  321. \exercise 3  Let $B$ be the event that the woman rides the bicycle to work,
  322. $B^\C$ would be that she ride the scooter.  Let $L$ be that she is late,
  323. \begin{align*}
  324.   P(B) &= 0.7\\
  325.   P\left(B^\C\right) &= 0.3\\
  326.   P(L|B) &= 0.03\\
  327.   P\left(L|B^\C\right) &= 0.02
  328. \end{align*}
  329. \begin{enumerate}[(a)]
  330.   \item By Total Probability Theorem, the probability the woman
  331.     is late for work is
  332.     \[P(L) = P(B)\cdot P(L|B) + P\left(B^\C\right)\cdot P\left(L|B^\C\right)
  333.     = 0.7\cdot 0.03 + 0.3\cdot 0.02 = 0.027\]
  334.   \item The probability she is not late for work is
  335.     \[P\left(L^\C\right) = 1 - P(L) = 1 - 0.027 = 0.973\]
  336.     Since the woman is expected to be on time roughly 223 days a year,
  337.     she goes to work $223 / P\left(L^\C\right) \approx 229$ days a year.
  338. \end{enumerate}
  339.  
  340. \exercise 4  Consider flipping the coin twice, the sample space is
  341. \[\Omega = \{(H, H), (H, T), (T, H), (T, T)\}\]
  342. where $H$ stands for head and $T$ stands for tail.
  343.  
  344. Denote getting a head from the first flip as $H_1$ and getting a head
  345. from the second one as $H_2$.  Assume that $P(H_1) = P(H_2) = 0.6$.
  346. It is obvious that these two events are independent, or in other words
  347. \[P(H_1\cap H_2) = P(H_1)\cdot P(H_2)\]
  348.  
  349. Similarly,
  350. \begin{align*}
  351.   P\{(H, T)\} &= P\left(H_1\cap H_2^\C\right)
  352.   = P\left(H_1\right)\cdot\left(1 - P\left(H_2\right)\right) = 0.24\\
  353.   P\{(T, H)\} &= P\left(H_1^\C\cap H_2\right)
  354.   = \left(1 - P\left(H_1\right)\right)\cdot P\left(H_2\right) = 0.24
  355. \end{align*}
  356.  
  357. Therefore if Minh and Nam flip the coin twice for both head and tail
  358. and choose K-pop when they get a head first and US music otherwise,
  359. the genre would be chosen equally even.
  360.  
  361. \exercise 5  Place three maths, two history and four biology book on a shelf.
  362. \begin{enumerate}[(a)]
  363.   \item There would be $(3 + 2 + 4)! = 362880$ ways to do it
  364.     without any further restriction.
  365.   \item If each subject needs to stay together,
  366.     there are $3! 2! 4! 3! = 1728$ ways.
  367.   \item If only biology books must stay together,
  368.     we can do it in $4!(3 + 2 + 1)!$ or 17280 ways.
  369. \end{enumerate}
  370.  
  371. \exercise 6  Seat six people around a table.
  372. \begin{enumerate}[(a)]
  373.   \item If they can sit anywhere, there are $6!/6 = 120$ arrangements.
  374.   \item If two particular people must sit next to each other,
  375.     there are $2\cdot 5!/5$ or 48 arrangements; thus if those two cannot
  376.     sit side-by-side, the figure is $120 - 48 = 72$.
  377. \end{enumerate}
  378.  
  379. \exercise 7  Let $\omega$ be the set of cards in a standard 52-card deck.
  380. Shuffle the deck an draw seven cards,
  381. the sample space of this probability model is $\Omega = \binom{\omega}{7}$,
  382. $n(\Omega) = \binom{52}{7}$.
  383. \begin{enumerate}[(a)]
  384.   \item Let $A$ be the event that exactly three of the drawn ones are aces,
  385.     \[n(A) = \binom{4}{3}\binom{48}{4}
  386.     \Longrightarrow P(A) = \frac{n(A)}{n(\Omega)} = \frac{9}{1547}\]
  387.   \item Let $K$ be the event that exactly two of the drawn ones are kings,
  388.     \[n(K) = \binom{4}{2}\binom{48}{5}
  389.     \Longrightarrow P(K) = \frac{n(K)}{n(\Omega)} = \frac{594}{7735}\]
  390.   \item The probability that exactly three aces and two kings are drawn is
  391.     \[n(A\cap K) = \binom{4}{3}\binom{4}{2}\binom{44}{2}
  392.     \Longrightarrow P(A\cap K) = \frac{n(A\cap K)}{n(\Omega)}
  393.     = \frac{1419}{8361535}\]
  394.  
  395.     Thus probability that either exactly three aces or two kings are drawn is
  396.     \[P(A\cup K) = P(A) + P(K) - P(A\cap K) = \frac{137868}{1672307}\]
  397. \end{enumerate}
  398.  
  399. \exercise 8  Let $M$ be the event that a red marble is picked
  400. and $C$ be the event of getting head from tossing the coin, we have
  401. \begin{align*}
  402.   P(M|C) &= 0.6\\
  403.   P\left(M|C^\C\right) &= 0.2\\
  404.   P(C) = P\left(C^\C\right) &= 0.5
  405. \end{align*}
  406.  
  407. \begin{enumerate}[(a)]
  408.   \item By Total Probability Theorem, the probability a red marble is picked is
  409.     \[P(M) = P(C)\cdot P(M|C) + P\left(C^\C\right)\cdot P\left(M|C^\C\right)
  410.     = 0.4\]
  411.   \item The probability that a blue marble is picked is
  412.     \[P\left(M^\C\right) = 1 - P(M) = 0.6\]
  413.   \item The probability of getting a head if the red marble is picked is
  414.     \[P(C|M) = \frac{P(C\cap M)}{P(M)} = \frac{P(C)\cdot P(M|C)}{P(M)}
  415.     = \frac{0.5\cdot 0.6}{0.4} = 0.75\]
  416. \end{enumerate}
  417.  
  418. \exercise 9  Consider $n$ random people and their birthdays, assuming
  419. that all 366 birthdays are equally likely\footnote{If you are wondering
  420. how one could be equally likely to be born on the leap day, then well,
  421. the distribution of birthdays on other days is in fact not uniform either.
  422. \textit{Don't complicate it, don't drive yourself insane!}}.
  423. The size of the sample space is $n(\Omega_n) = 366^n$.
  424.  
  425. Let $A_n$ be the event that no two of these $n$ people to celebrate
  426. their birthday on the same day, $n(A_n) = \prod_{i=0}^{n-1}(366-i)$.
  427. Thus the probability of this is
  428. \[P(A_n) = \frac{n(A_n)}{n(\Omega_n)} = \prod_{i=1}^{n-1}\frac{366 - i}{366}\]
  429.  
  430. Since $P(A_{23}) < 0.5 < P(A_{22})$, $n$ needs to be at least 23
  431. for the probability to be less than 0.5.
  432.  
  433. \exercise{10} The reasoning is not correct because:
  434. \begin{itemize}
  435.   \item If he is not to be released, the answer from the guard will be
  436.     both of other prisoners, and everyones' fate will be known.
  437.   \item Otherwise, in case the guard only gives one name,
  438.     our protagonist will sure be released.
  439. \end{itemize}
  440. \pagebreak
  441.  
  442. \section{Discrete Random Variable 1}
  443. \subsection{Discrete Random Variable and PMF}
  444. \exercise 1  Consider a fair coin.
  445. \begin{enumerate}[(a)]
  446.   \item Toss it twice and let $X$ be the number of heads,
  447.     $X$ would be a binomial random variable
  448.     \begin{align*}
  449.       X\colon \Omega &\to \{0, 1, 2\}\\
  450.       \omega &\mapsto x
  451.     \end{align*}
  452.     whose probability mass function is
  453.     \[p_X(x) = \binom{2}{x}\cdot 0.5^x\cdot 0.5^{2-x}
  454.     = \frac{1}{2x!(2-x)!}\]
  455.  
  456.     Therefore PMF of $X$ for each case is
  457.     \begin{align*}
  458.       p_X(0) = p_X(2) &= \frac{1}{2\cdot 0!2!} = \frac{1}{4}\\
  459.       p_X(1) &= \frac{1}{2\cdot 1!1!} = \frac{1}{2}
  460.     \end{align*}
  461.   \item Toss it thrice and let $Y$ be the number of heads,
  462.     $Y$ would be a binomial random variable
  463.     \begin{align*}
  464.       Y\colon \Omega &\to \{0, 1, 2, 3\}\\
  465.       \omega &\mapsto y
  466.     \end{align*}
  467.     whose probability mass function is
  468.     \[p_Y(y) = \binom{3}{y}\cdot 0.5^y\cdot 0.5^{3-y}
  469.     = \frac{3}{4y!(3-y)!}\]
  470.  
  471.     Therefore PMF of $Y$ for each case is
  472.     \begin{align*}
  473.       p_Y(0) = p_Y(3) &= \frac{3}{4\cdot 0!3!} = \frac{1}{8}\\
  474.       p_Y(1) = p_Y(2) &= \frac{3}{4\cdot 1!2!} = \frac{3}{8}
  475.     \end{align*}
  476. \end{enumerate}
  477.  
  478. \exercise 2  Toss a pair of fair siz-sided dice
  479. and let $X$ be the sum of the points
  480. \begin{align*}
  481.   X\colon \Omega &\to [2, 12]\cap\mathbb Z\\
  482.   \omega &\mapsto x
  483. \end{align*}
  484. with $\Omega = S^2 = \{1, 2, 3, 4, 5, 6\}^2 \Longrightarrow n(\Omega) = 36$.
  485.  
  486. \begin{enumerate}[(a)]
  487.   \item $X$ is a random variable whose PMF is
  488.     \begin{align*}
  489.       p_X(2) &= P\{(1,1)\} = \frac{1}{36}\\
  490.       p_X(3) &= P\{(1,2), (2,1)\} = \frac{1}{18}\\
  491.       p_X(4) &= P\{(1,3), (2,2), (3,1)\} = \frac{1}{12}\\
  492.       p_X(5) &= P\{(1,4), (2,3), (3,2), (4,1)\} = \frac{1}{9}\\
  493.       p_X(6) &= P\{(1,5), (2,4), (3,3), (4,2), (5,1)\} = \frac{5}{36}\\
  494.       p_X(7) &= P\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\} = \frac{1}{6}\\
  495.       p_X(8) &= P\{(2,6), (3,5), (4,4), (5,3), (6,2)\} = \frac{5}{36}\\
  496.       p_X(9) &= P\{(3,6), (4,5), (5,4), (6,3)\} = \frac{1}{9}\\
  497.       p_X(10) &= P\{(4,6), (5,5), (6,4)\} = \frac{1}{12}\\
  498.       p_X(11) &= P\{(5,6), (6,5)\} = \frac{1}{18}\\
  499.       p_X(12) &= P\{(6,6)\} = \frac{1}{36}
  500.     \end{align*}
  501.   \item The graph of $p_X(x)$:
  502.  
  503.     \begin{tikzpicture}
  504.       \begin{axis}[xlabel={$x$}, ylabel={$p_X(x)$}]
  505.         \addplot[ycomb, samples at={2,3,...,12}]{1/6-abs(x-7)/36};
  506.       \end{axis}
  507.     \end{tikzpicture}
  508. \end{enumerate}
  509. \pagebreak
  510.  
  511. \exercise 3  Denote the event of winning, tying and losing the first game as
  512. $A_2$, $A_1$ and $A_0$ respectively, we get $P(A_2) = P(A_1) = 0.2$
  513. and $P(A_0) = 0.6$.  Similarly, let $B_2$, $B_1$ and $B_0$ in that order
  514. be the event MIT soccer team winning, tying and losing the second game,
  515. we get $P(B_2) = P(B_1) = 0.35$ and $P(B_0) = 0.3$.
  516.  
  517. Let $A$ and $B$ be the random variable satisfying
  518. \begin{align*}
  519.   A &= \begin{cases}
  520.     2\text{ if }A_2\\
  521.     1\text{ if }A_1\\
  522.     0\text{ if }A_0
  523.   \end{cases}
  524.   \Longrightarrow\begin{cases}
  525.     p_A(2) = p_A(1) = 0.2\\
  526.     p_A(0) = 0.6
  527.   \end{cases}\\
  528.   B &= \begin{cases}
  529.     2\text{ if }B_2\\
  530.     1\text{ if }B_1\\
  531.     0\text{ if }B_0
  532.   \end{cases}
  533.   \Longrightarrow\begin{cases}
  534.     p_B(2) = p_B(1) = 0.35\\
  535.     p_B(0) = 0.3
  536.   \end{cases}
  537. \end{align*}
  538. then the number of points the team earns over the weekend is $X = A + B$.
  539.  
  540. Since the outcome of the two games are independent,
  541. \begin{align*}
  542.   p_X(0) &= p_A(0)\cdot p_B(0) = 0.18\\
  543.   p_X(1) &= p_A(0)\cdot p_B(1) + p_A(1)\cdot p_B(0) = 0.27\\
  544.   p_X(2) &= p_A(0)\cdot p_B(2) + p_A(1)\cdot p_B(1) + p_A(2)\cdot p_B(0)
  545.           = 0.34\\
  546.   p_X(3) &= p_A(1)\cdot p_B(2) + p_A(2)\cdot p_B(1) = 0.14\\
  547.   p_X(4) &= p_A(2)\cdot p_B(2) = 0.07
  548. \end{align*}
  549.  
  550. \subsection{Expectation of Random Variables}
  551. \exercise 4  Given a random variable
  552. \[X = \begin{cases}
  553.   -2&\text{ with probability of 1/3}\\
  554.   3&\text{ with probability of 1/2}\\
  555.   1&\text{ with probability of 1/6}
  556. \end{cases}\]
  557. \begin{align*}
  558.   \E[X] &= \sum_{x\in\{-2, 1, 3\}}xp_X(x) = 1\\
  559.   \E[2X+5] &= \sum_{x\in\{-2, 1, 3\}}(2x + 5)p_X(x) = 7\\
  560.   \E\left[X^2\right] &= \sum_{x\in\{-2, 1, 3\}}x^2 p_X(x) = 6
  561. \end{align*}
  562.  
  563. \exercise 5  Consider the genders of the three children, and assume
  564. that both genders\footnote{You SJWs really need to calm down.
  565. This is just a mathematical problem.} are equally likely.
  566.  
  567. Let $X$ be the number of girls, $X$ is a binomial random variable
  568. whose probability mass function is
  569. \begin{multline*}
  570.   p_X(x) = \binom{3}{x}\cdot 0.5^x\cdot 0.5^{3-x} = \frac{3}{4x!(3-x)!}\\
  571.   \Longrightarrow \E[X] = \sum_{x=0}^3\frac{3x}{4x!(3-x)!} = \frac 3 2
  572. \end{multline*}
  573.  
  574. \exercise 6  Consider rolling a fair six-sided die, the sample space is
  575. $\Omega = \{1, 2, 3, 4, 5, 6\}$.  Let $X$ be a random variable given by
  576. \begin{multline*}
  577.   X(\omega) = \begin{cases}
  578.     -1&\text{ if }\omega \in \{1, 2, 3\}\\
  579.     2&\text{  if }\omega \in \{4, 5\}\\
  580.     8&\text{  if }\omega = 6
  581.   \end{cases}
  582.   \Longrightarrow\begin{dcases}
  583.     p_X(-1) = \frac{3}{6} = \frac{1}{2}\\
  584.     p_X(2) = \frac{2}{6} = \frac{1}{3}\\
  585.     p_X(8) = \frac{1}{6}
  586.   \end{dcases}\\
  587.   \Longrightarrow \E[X] = \frac{-1}{2} + \frac{2}{3} + \frac{4}{3} = \frac{3}{2}
  588. \end{multline*}
  589.  
  590. Practically, this means that at the end of the day,
  591. it is very unlikely that the house will win.
  592.  
  593. \exercise 7  Let $X$ be the prize in dollars on a randomly chosen
  594. lottery ticket, its PMF is
  595. \begin{align*}
  596.   p_X(100) &= \frac{5}{10\,000} = \frac{1}{2\,000}\\
  597.   p_X(25) &= \frac{20}{10\,000} = \frac{1}{5\,000}\\
  598.   p_X(5) &= \frac{200}{10\,000} = \frac{1}{500}\\
  599.   p_X(0) &= \frac{10\,000-200-20-5}{10\,000} = \frac{391}{400}
  600. \end{align*}
  601.  
  602. Thus the expected value for a ticket's value in dollars is
  603. \[\E[X] = \sum_x x\cdot p_X(x) = \frac{13}{200}\]
  604. or 6.5 cents.
  605.  
  606. \exercise 8  Let $X$ be the prize in dollars on a randomly chosen
  607. raffle ticket, its PMF is
  608. \begin{align*}
  609.   & p_X(1998) = p_X(999) = \frac{1}{5000}\\
  610.   & p_X(498) = \frac{2}{5000} = \frac{1}{2500}\\
  611.   & p_X(98) = \frac{5}{5000} = \frac{1}{1000}\\
  612.   & p_X(-2) = \frac{5000-5-2-1-1}{5000} = \frac{4991}{5000}
  613. \end{align*}
  614.  
  615. Thus the expected value in dollars to get when buying a ticket is
  616. \[\E[X] = \sum_x x\cdot p_X(x) = \frac{-11}{10}\]
  617. or to lose \$1.1.
  618.  
  619. \subsection{Variance and Standard Deviation}
  620. \exercise{9} Given the outcome $X$ from rolling a fair six-sided die.
  621. \begin{align*}
  622.   \E[X] &= \frac{1+2+3+4+5+6}{6} = \frac{7}{2}\\
  623.   \Longrightarrow \var(X) &= \E\left[(X - \E[X])^2\right]
  624.   = \sum_x \frac{(x - \frac{7}{2})^2}{6} = \frac{35}{24}\\
  625.   \Longrightarrow \sigma_X &= \sqrt{\var(X)} = \sqrt\frac{35}{24}
  626. \end{align*}
  627.  
  628. \exercise{10} Based on the result of exercise 2,
  629. \[E[X] = 7,\qquad\var(X) = \frac{35}{6},\qquad\sigma_X = \sqrt\frac{35}{6}\]
  630.  
  631. \exercise{11} Given the integral random variable $X$ with PMF
  632. \[p_X(x) = \begin{dcases}
  633.   \frac{1}{9} &\text{ if } x \in [-4, 4]\\
  634.   0 &\text{ otherwise}
  635. \end{dcases}\]
  636.  
  637. Let $S = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$,
  638. \begin{multline*}
  639.   \E[X] = \sum_{x\in\mathbb Z}x\cdot p_X(x)
  640.   = \sum_{x\in S}\frac{x}{9} + \sum_{x\in\mathbb Z\setminus S}x\cdot 0 = 0\\
  641.   \Longrightarrow \var(X) = \E\left[X^2\right]
  642.   = \sum_{x\in S}\frac{x^2}{9} = \frac{20}{3}
  643. \end{multline*}
  644.  
  645. \exercise{12} Given the integral random variable $X$ with PMF
  646. \[p_X(x) = \begin{dcases}
  647.   \frac{x^2}{a} &\text{ if } x \in [-3, 3]\\
  648.   0 &\text{ otherwise}
  649. \end{dcases}\]
  650. \begin{enumerate}[(a)]
  651.   \item Let $S = \{-3, -2, -1, 0, 1, 2, 3\}$. Since
  652.     \begin{multline*}
  653.       \sum_{x\in\mathbb Z}p_X(x) = 1
  654.       \iff \sum_{x\in S}\frac{x^2}{a} + \sum_{x\in\mathbb Z\setminus S}0 = 1
  655.       \iff a = \sum_{x\in S}x^2 = 28\\
  656.       \Longrightarrow \E[X] = \sum_{x\in S}\frac{x^3}{28} = 0
  657.     \end{multline*}
  658.   \item Let $Z = (X - \E[X])^2 = X^2$,
  659.     the range of $Z$ is $\{z^2\mid z\in\mathbb Z\}$.
  660.     For all $z > 9$, it is trivial that $p_Z(z) = 0$.  Otherwise,
  661.     \begin{align*}
  662.       p_Z(0) &= P(Z = 0) = P(X=0) = p_X(0) = \frac{0^2}{28} = 0\\
  663.       p_Z(1) &= P(X = \pm 1) = p_X(-1) + p_X(1)
  664.       = \frac{(-1)^2}{28} + \frac{1^2}{28} = \frac{1}{14}\\
  665.       p_Z(4) &= P(X = \pm 2) = p_X(-2) + p_X(2)
  666.       = \frac{(-2)^2}{28} + \frac{2^2}{28} = \frac{2}{7}\\
  667.       p_Z(9) &= P(X = \pm 3) = p_X(-3) + p_X(3)
  668.       = \frac{(-3)^2}{28} + \frac{3^2}{28} = \frac{9}{14}
  669.     \end{align*}
  670.   \item The variance of $X$ is
  671.     \[\var(X) = \E\left[(X - \E[X])^2\right]
  672.     = \E[Z] = 1\cdot\frac{1}{14} + 4\cdot\frac{2}{7} + 9\cdot\frac{9}{14} = 7\]
  673. \end{enumerate}
  674. \pagebreak
  675.  
  676. \section{Discrete Random Variable 2}
  677. \subsection{Conditional PMF and Expectation}
  678. \exercise 1  Compute conditional PMF:
  679. \begin{enumerate}[(a)]
  680.   \item Let $X$ be the roll if a fair six-sided die and $A$ be the event that
  681.     the roll is an number greater or equal to 4, we have $A = \{X \geq 4\}$
  682.     and $P(A) = 0.5$, thus
  683.     \[p_{X|A}(x) = \frac{P(\{X = x\}\cap\{X \geq 4\})}{P(A)}\]
  684.  
  685.     For $x \in \{1, 2, 3\}$, $\{X = x\}\cap\{X \geq 4\} = \varnothing$
  686.     so $p_{X|A}(x) = 0/0.5 = 0$.
  687.  
  688.     For $x \in \{4, 5, 6\}$, $\{X = x\}\cap\{X \geq 4\} = \{x\}$,
  689.     \[p_{X|A}(x) = \frac{1/6}{0.5} = \frac{1}{3}\]
  690.   \item Let $X$ represent number of heads from the three-time toss
  691.     of a fair coin and $B = \{X \geq 2\}$,
  692.     $P(B) = \binom{3}{2}0.5^3 + \binom{3}{3}0.5^3 = 0.5$.
  693.     \begin{multline*}
  694.       p_{X|B}(x) = \frac{P(\{X = x\}\cap B)}{P(B)}
  695.       = \frac{P(\{X = x\}\cap\{X \geq 2\})}{0.5}\\
  696.       \Longrightarrow\begin{dcases}
  697.         p_{X|B}(0) = p_{X|B}(1) = 0\\
  698.         p_{X|B}(2) = \frac{\binom{3}{2}0.5^3}{0.5} = \frac{3}{4}\\
  699.         p_{X|B}(3) = \frac{\binom{3}{3}0.5^3}{0.5} = \frac{1}{4}
  700.       \end{dcases}
  701.     \end{multline*}
  702.   \item Let $X$ be the roll of a pair of fair dice and $C = \{X = 7\}$.
  703.     As shown in the previous section, $P(C) = 1/6$ and thus
  704.     \[p_{X|C}(x) = \begin{cases}
  705.       1\text{ if }x = 7\\
  706.       0\text{ if }x \neq 7
  707.     \end{cases}\]
  708. \end{enumerate}
  709.  
  710. \exercise 2  Consider the destination of the message and denote the event
  711. it arrives at Liberty City, Chicago and San Fierro as $B$, $C$ and $F$
  712. respectively, we have $B\cup C\cup F = \Omega$.  The expected transit time is
  713. \begin{align*}
  714.   \E[X] &= P(B)\E[X|B] + P(C)\E[X|C] + P(F)\E[X|F]\\
  715.   &= 0.5\cdot 0.05 + 0.3\cdot 0.1 + 0.2\cdot 0.3 = 0.115
  716. \end{align*}
  717.  
  718. \exercise 3  Let $V$ and $T$ be respectively the speed (in mph) and time
  719. (in hours) Alyssa get to class.  Denote the event she walk to class as $W$,
  720. $P(W) = 0.6$,
  721. \[\begin{cases}
  722.   \E[V|W] = 5\\
  723.   \E\left[V|W^\C\right] = 30
  724. \end{cases}
  725. \Longrightarrow\begin{dcases}
  726.   \E[T|W] = \frac{2}{5}\\
  727. \E\left[T|W^\C\right] = \frac{1}{15}
  728. \end{dcases}\]
  729. \begin{enumerate}[(a)]
  730.   \item The expected value of Alyssa's speed is
  731.     \[\E[V] = P(W)\E[V|W] + P\left(W^\C\right)\E\left[V|W^\C\right]
  732.     = 0.6\cdot 5 + (1 - 0.6)30 = 15\]
  733.   \item The expected value of the time Alyssa she takes to get to class is
  734.     \[\E[T] = P(W)\E[T|W] + P\left(W^\C\right)\E\left[T|W^\C\right]
  735.     = 0.6\cdot\frac{2}{5} + (1 - 0.6)\frac{1}{15} = \frac{4}{15}\]
  736. \end{enumerate}
  737.  
  738. \exercise 4  With $X$ being the number of tries until the program works
  739. correctly and $p$ being the probability each try succeed,
  740. we have $\mathrm{range}(X) = \mathbb N^*$ and $p_X(x) = (1 - p)^{x - 1}p$.
  741. The mean of $X$ is
  742. \begin{align*}
  743.   \E[X] &= \sum_{x\in\mathbb N^*}x\cdot p_X(x)\\
  744.   &= \sum_{x\in\mathbb N^*}x(1 - p)^{x - 1}p\\
  745.   &= -p\sum_{x\in\mathbb N^*}\frac{\mathrm d (1 - p)^x}{\mathrm d p}\\
  746.   &= -p\frac{\mathrm d}{\mathrm d p}\left(\sum_{x\in\mathbb N}(1 - p)^{x} - 1\right)\\
  747.   &= -p\frac{\mathrm d}{\mathrm d p}\left(\frac{1}{p} - 1\right)\\
  748.   &= \frac{1}{p}
  749. \end{align*}
  750.  
  751. Similarly,
  752. \begin{align*}
  753.   \E\left[X^2\right] &= p\sum_{x\in\mathbb N^*}x^2(1 - p)^{x - 1}\\
  754.   &= -p\frac{\mathrm d}{\mathrm d p}
  755.   \left(\frac{1 - p}{p}\sum_{x\in\mathbb N^*}x(1 - p)^{x - 1}p\right)\\
  756.   &= -p\frac{\mathrm d}{\mathrm d p}\left(\frac{1}{p^2} - \frac{1}{p}\right)\\
  757.   &= \frac{2}{p^2} - \frac{1}{p}
  758. \end{align*}
  759.  
  760. Therefore the variance of $X$ is
  761. \[\var(X) = \E\left[X^2\right] - (\E[X])^2 = \frac{1}{p^2} - \frac{1}{p}\]
  762.  
  763. \subsection{Joint PMF and independent variables}
  764. \exercise 5  Consider two independent coin tosses,
  765. each with a 3/4 probability of a head,
  766. and let $X$ be the number of heads obtained,
  767. $X$ is a binomial random variable.
  768. \[\E[X] = 0p_X(0) + 1p_X(1) + 2p_X(2)
  769. = \binom{2}{1}\frac{3}{4}\left(1 - \frac{3}{4}\right)
  770. + 2\binom{2}{2}\left(\frac{3}{4}\right)^2
  771. = \frac{3}{2}\]
  772.  
  773. \exercise 6  Let $X$ be the number of red traffic lights Alyssa encounters,
  774. $X$ is a binomial random variable whose PMF is
  775. \[p_X(x) = \binom{4}{x}0.5^4\]
  776.  
  777. The mean of $X$ is
  778. \[\E[X] = \frac{1}{16}\sum_{x=0}^4 x\binom{4}{x} = 2\]
  779.  
  780. The variance of $X$ is
  781. \[\var(X) = \E\left[X^2\right] - (\E[X])^2
  782. = \frac{1}{16}\sum_{x=0}^4 x^2\binom{4}{x} - 4 = 1\]
  783.  
  784. \exercise 7  Let $X_i$ be 1 if the $i$th person gets his or her own hat and 0
  785. otherwise, then for all positive integer $i \leq n$
  786. \[\E[X_i] = p_{X_i}(1) = \frac{(n - 1)!}{n!} = \frac{1}{n}\]
  787. since if we fix one hat to its owner, there are $(n - 1)!$ arrangements
  788. for the rest.  Due to the linearity property of expectation,
  789. \[\E[X] = \E\left[\sum_{i=1}^n X_i\right] = \sum_{i=1}^n\E[X_i] = 1\]
  790.  
  791. \exercise 8  Consider four independent rolls of a six-sided die.
  792. Let $X$ and $Y$ be the number of ones and twos obtained respectively,
  793. both are binomial random variables:
  794. \[p_X(k) = p_Y(k)
  795. = \binom{4}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{4 - k}
  796. = \binom{4}{k}\frac{5^{4 - k}}{1296}\]
  797.  
  798. Given $Y = y$, $X$ is the number of ones in the remaining $4 - y$ rolls,
  799. each of which can take the values other than two equally likely:
  800. \[p_{X|Y}(x|y)
  801. = \binom{4 - y}{x}\left(\frac{1}{5}\right)^x\left(\frac{4}{5}\right)^{4 - y - x}
  802. = \binom{4 - y}{k}\frac{4^{4 - y - x}}{625}\]
  803.  
  804. Thus the joint PMF of $X$ and $Y$ is
  805. \[p_{X,Y}(x, y) = p_Y(y)p_{X|Y}(x|y)
  806. = \binom{4}{x}\binom{4 - y}{k}\frac{5^{4 - x}4^{4 - y - x}}{810\,000}\]
  807.  
  808. \exercise 9  Given the joint PMF of two discrete random variables $X$ and $Y$
  809. \[p_{X,Y}(x, y) = \begin{cases}
  810.   c(2x + y) &\text{where }(x, y)\in\{0, 1, 2\}\times\{0, 1, 2, 3\}\\
  811.   0 &\text{otherwise}
  812. \end{cases}\]
  813. \begin{enumerate}[(a)]
  814.   \item Consider all cases:
  815.     \begin{align*}
  816.       \sum_x\sum_y p_{X,Y}(x, y) = 1
  817.       &\iff \sum_{x=0}^2\sum_{y=0}^3 c(2x + y) = 1\\
  818.       &\iff \sum_{x=0}^2 c(8x + 6) = 1\\
  819.       &\iff c(24 + 18) = 1\\
  820.       &\iff c = \frac{1}{42}
  821.     \end{align*}
  822.   \item $P(X = 2, Y = 1) = (2\cdot 2 + 1)/42 = 5/42$.
  823.   \item Similarly, $P(X \geq 1, Y \leq 2) = 4/7$.
  824.   \item The marginal PMF of $X$:
  825.     \[p_X(x) = \sum_y p_{X,Y}(x, y)
  826.     = \sum_{y=0}^3\frac{2x + y}{42} = \frac{4x + 3}{21}\]
  827.   \item The marginal PMF of $Y$:
  828.     \[p_Y(y) = \sum_x p_{X,Y}(x, y)
  829.     = \sum_{x=0}^2\frac{2x + y}{42} = \frac{2 + y}{14}\]
  830.   \item Since $p_X(2)p_Y(1) \neq p_{X,Y}(2, 1)$,
  831.     the two variables are dependent.
  832.   \item Given $X = 2$,
  833.     \[p_{Y|X}(y|2) = \frac{p_{X,Y}(2, y)}{p_X(2)} = \frac{4 + y}{22}
  834.     \Longrightarrow p_{Y|X}(1|2) = \frac{5}{22}\]
  835.   \item Given $Y = 2$,
  836.     \[p_{X|Y}(x|2) = \frac{p_{X,Y}(x, 2)}{p_Y(2)} = \frac{x + 1}{6}
  837.     \Longrightarrow p_{X|Y}(3|2) = \frac{2}{3}\]
  838. \end{enumerate}
  839.  
  840. \exercise{10} Given the joint PMF of two discrete random variables $X$ and $Y$
  841. \[p_{X,Y}(x, y) = \begin{cases}
  842.   cxy &\text{where }(x, y)\in\{1, 2, 3\}\times\{1, 2, 3\}\\
  843.   0 &\text{otherwise}
  844. \end{cases}\]
  845. \begin{enumerate}[(a)]
  846.   \item Consider all cases:
  847.     \begin{align*}
  848.       \sum_x\sum_y p_{X,Y}(x, y) = 1
  849.       &\iff \sum_{x=1}^3\sum_{y=1}^3 cxy = 1\\
  850.       &\iff 36c = 1\\
  851.       &\iff c = \frac{1}{36}
  852.     \end{align*}
  853.   \item $P(X = 2, Y = 3) = 1/6$.
  854.   \item Similarly, $P(1\leq X\leq 2, Y\leq 2) = 1/4$.
  855.   \item By the result of (e), $P(X\geq 2) = 5/6$,
  856.     $P(Y < 2) = P(Y = 1) = P(X = 1) = 1/6$ and $P(Y = 3) = 1/2$.
  857.   \item The marginal PMF of $X$:
  858.     \[p_X(x) = \sum_y p_{X,Y}(x, y)
  859.     = \sum_{y=1}^3\frac{xy}{36} = \frac{x}{6}\]
  860.     The marginal PMF of $Y$:
  861.     \[p_Y(y) = \sum_x p_{X,Y}(x, y)
  862.     = \sum_{x=1}^3\frac{xy}{36} = \frac{y}{6}\]
  863.   \item Since $p_{X,Y}(x, y) = p_X(x)p_Y(y)$, $X$ and $Y$ are independent.
  864. \end{enumerate}
  865. \end{document}
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