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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={$p_X(x)$}]
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$Xfrom 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 variableX$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$p_Z(z) = 0. Otherwise, 661. \begin{align*} 662. p_Z(0) &= P(Z = 0) = P(X=0) = p_X(0) = \frac{0^2}{28} = 0\\ 663. p_Z(1) &= P(X = \pm 1) = p_X(-1) + p_X(1) 664. = \frac{(-1)^2}{28} + \frac{1^2}{28} = \frac{1}{14}\\ 665. p_Z(4) &= P(X = \pm 2) = p_X(-2) + p_X(2) 666. = \frac{(-2)^2}{28} + \frac{2^2}{28} = \frac{2}{7}\\ 667. p_Z(9) &= P(X = \pm 3) = p_X(-3) + p_X(3) 668. = \frac{(-3)^2}{28} + \frac{3^2}{28} = \frac{9}{14} 669. \end{align*} 670. \item The variance ofX$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 LetV$and$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$Xis 742. \begin{align*} 743. \E[X] &= \sum_{x\in\mathbb N^*}x\cdot p_X(x)\\ 744. &= \sum_{x\in\mathbb N^*}x(1 - p)^{x - 1}p\\ 745. &= -p\sum_{x\in\mathbb N^*}\frac{\mathrm d (1 - p)^x}{\mathrm d p}\\ 746. &= -p\frac{\mathrm d}{\mathrm d p}\left(\sum_{x\in\mathbb N}(1 - p)^{x} - 1\right)\\ 747. &= -p\frac{\mathrm d}{\mathrm d p}\left(\frac{1}{p} - 1\right)\\ 748. &= \frac{1}{p} 749. \end{align*} 750. 751. Similarly, 752. \begin{align*} 753. \E\left[X^2\right] &= p\sum_{x\in\mathbb N^*}x^2(1 - p)^{x - 1}\\ 754. &= -p\frac{\mathrm d}{\mathrm d p} 755. \left(\frac{1 - p}{p}\sum_{x\in\mathbb N^*}x(1 - p)^{x - 1}p\right)\\ 756. &= -p\frac{\mathrm d}{\mathrm d p}\left(\frac{1}{p^2} - \frac{1}{p}\right)\\ 757. &= \frac{2}{p^2} - \frac{1}{p} 758. \end{align*} 759. 760. Therefore the variance ofX$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$Y809. $p_{X,Y}(x, y) = \begin{cases} 810. c(2x + y) &\text{where }(x, y)\in\{0, 1, 2\}\times\{0, 1, 2, 3\}\\ 811. 0 &\text{otherwise} 812. \end{cases}$ 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. \itemP(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$Y841. $p_{X,Y}(x, y) = \begin{cases} 842. cxy &\text{where }(x, y)\in\{1, 2, 3\}\times\{1, 2, 3\}\\ 843. 0 &\text{otherwise} 844. \end{cases}$ 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. \itemP(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}
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