STAT400 Practice Problems - 0213,0223,0233.pdf - STAT400...

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Unformatted text preview: STAT400 Haoran Li 2020 Fall Contents 1 Practice problems 1.1 Practice 1 - 9/3/2020 . 1.2 Practice 2 - 9/10/2020 1.3 Practice 3 - 9/17/2020 1.4 Practice 4 - 9/24/2020 1.5 Practice 5 - 10/1/2020 1.6 Practice 6 - 10/9/2020 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2 4 7 9 11 13 1 Practice problems 1.1 Practice 1 - 9/3/2020 Exercise 1.1 ([1] Section 2.1, Exercise 1). Four universities (1, 2, 3, and 4) are participating in a holiday basketball tournament. In the first round, 1 will play 2 and 3 will play 4. Then the two winners will play for the championship, and the two losers will also play. One possible outcome can be denoted by 1324 (1 beats 2 and 3 beats 4) a. List all outcomes in b. Let S A denote the event that 1 wins the tournament. List outcomes in A B denote the event that 2 gets into the championship game. List outcomes in B d. What are the outcomes in A [ B and in A \ B ? What are the outcomes in A0 ? c. Let Solution. The first two digits represent the two winners of the first round, the last two digits represent the two losers of the first round, the order of the first two digits determine the first and second place from the second round, the order of the last two digits determine the third and fourth place from the second round. Note that what the first two digits are automatically determine what the last two digits are and vice versa a. The first two digits must consist one from S ( f ; g and the other from f ; g 1 2 3 4 ) ; ; ; ; ; ; ; ; 3124; 3142; 4123; 4132; 3214; 3241; 4213; 4231 1324 1342 1423 1432 2314 2341 2413 2431 = b. The first digit must be 1 A = f1324; 1342; 1423; 1432g c. The first two digits must contain 2 B = f2314; 2341; 3214; 3241; 2413; 2431; 4213; 4231g d. Note that if 1 wons the first place is equivalent of saying 2 doesn’t make into the championship game, hence A; B are disjoint, therefore A \ B = ? A [ B = f1324; 1342; 1423; 1432; 2314; 2341; 3214; 3241; 2413; 2431; 4213; 4231g and A0 = f2314; 2341; 2413; 2431; 3124; 3142; 4123; 4132; 3214; 3241; 4213; 4231g Exercise 1.2 ([1] Section 2.1, Exercise 3). Three components are connected to form a system as shown in the accompanying diagram. Because the components in the 2-3 subsystem are connected in parallel, that subsystem will function if at least one of the two individual components functions. For the entire system to function, component 1 must function and so must the 2-3 subsystem. The experiment consists of determining the condition of each component [S (success) for a functioning component and F (failure) for a nonfunctioning component] a. Which outcomes are contained in the event A that exactly two out of the three components function? 2 b. Which outcomes are contained in the event B that at least two of the components function? c. Which outcomes are contained in the event d. List outcomes in C that the system functions? C 0 , A [ C , A \ C , B [ C , and B \ C Solution. Write T1 T2 T3 for the outcome, where Ti 2 fS; F g stands for whether component functioning successfully or not, thus the sample space is i is S fSSS; SSF; SF S; F SS; SF F; F SF; F F S; F F F g = a. b. A = fSSF; SF S; F SS g B = fSSF; SF S; F SS; SSS g c. Since component 1 and subsystem 2-3 are connected in series, the whole system functions if both component 1 and subsystem function, and since component 2,3 are connected in parallel, subsystem 2-3 functions if at least one of components 2,3 works C = fSSF; SF S; SSS g d. C 0 = fF SS; SF F; F F S; F SF; F F F g A [ C = fSSSF; SF S; F SS; SSS g A \ C = fSSF; SF S g B [ C = fSSF; SF S; F SS; SSS g B \ C = fSSF; SF S; SSS g 3 1.2 Practice 2 - 9/10/2020 Exercise 1.3. Suppose that vehicles taking a particular highway exit can either turn right (denoted as R), turn left (denoted as L), or go straight (denoted as S ). Consider observing three successive cars who takes this exit (a) Let A denote the event that only one car turns right. List all the outcomes in the event A. (An event is represented as a union of outcomes in sample space occurrence of which makes this event occur) (b) List all outcomes for the event that all three cars take different directions. Call this event as B (c) Describe the elements in the sample space and and compute the size of the sample space (d) List the outcomes in A \ B, A \ B0 Solution. Represent the outcomes as three letter sequences, where the first letter indicates the first car’s choice, the second indicates the second car’s choice and the third letter indicates the third car’s choice. For example, RLS means the first car turns right, the second car turns left and the third car goes straight (a) A = fRLS; RSL; RSS; RLL; SRL; LRS; SRS; LRL; SLR; LSR; SSR; LLRg (b) B = fRSL; RLS; SRL; SLR; LRS; LSRg (c) Suppose the outcome is covered with cards, and you try to reveal what they are one by one R? ? ? ? L? S ? ? ? ? 4 RR ? RRR; RRL; RRS RL ? RLR; RLL; RLS RS ? RSR; RSL; RSS LR ? LRR; LRL; LRS LL ? LLR; LLL; LLS LS ? LSR; LSL; LSS SR ? SRR; SRL; SRS SL ? SLR; SLL; SLS SS SSR; SSL; SSS ?   different outcomes A \ B B as B is contained in A. A \ B 0 A n B fRLL; LRL; LLR; RSS; SRS; SSRg You can see there are in total (d) 3 3 3 = 27 = = = Exercise 1.4. Let P denote the probability function which is defined as a function that satisfies Axiom1, Axiom 2, Axiom 3 in Section 2.2 of Devore’s book (these are the axioms we talked about in class). Let A, B , C denote three events defined on a sample space, justify that the following two expressions are correct. Note: Using Venn diagrams to display the sets is fine, please make sure to provide enough explanations and refer to the right axioms (a) P (A [ B ) = P (A \ B 0 ) + P (B ) (b) P (A) = P (A ( B [ C )) + P (A \ C ) + P (A \ B ) P (A \ B \ C ) Solution. (a) A [ B = (A n B ) [ B , since A n B and B are disjoint, thus by Axiom 3 P (A [ B ) = P (A n B ) + P (B ) = P (A \ B 0 ) + P (B ) (b) P (A) = P (1 [ 2 [ 3 [ 4) = P (1) + P (2) + P (3) + P (4) where (1.1) P (1) = P (A n (B [ C )) P (2) = P (A \ B ) P (A \ B \ C ) P (3) = P (A \ C ) P (A \ B \ C ) P (4) = P (A \ B \ C ) Then equation (1.1) gives the result A B 1 2 4 3 C Figure 1.1: The Venn diagram Exercise 1.5. A purse contains five 10$, four 5$, and eight 1$ bills. The bills are selected one by one -without replacement- in random order, Let A denote the event that at least two bills are selected to obtain the first 10$ bill. Describe this event as the complement of another event which is much a smaller set and easier to comprehend. Next week you will compute the probability of the event A 5 Solution. The complement event would be the event that a drawal 10$ bill is selected at the first Exercise 1.6 ([1] Section 2.2, Exercise 26). A certain system can experience three different types of defects. Let event that the system has a defect of type i. Suppose that A (i = 1; 2; 3) denote the i P (A1 ) = :12 P (A2 ) = :07 P (A3 ) = :05 P (A1 [ A2 ) = :13 P (A1 [ A3 ) = :14 P (A2 [ A3 ) = :10 P (A1 \ A2 \ A3 ) = :01 a. What is the probability that the system does not have a type 1 defect? b. What is the probability that the system has both type 1 and type 2 defects? c. What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect? d. What is the probability that the system has at most two of these defects? Solution. a. b. P (A01 ) = 1 P (A1 ) = 1 :12 = :88 P (A1 \ A2 ) = P (A1 ) + P (A2 ) P (A1 [ A2 ) = :12 + :07 :13 = :06 c. According to a. P (A1 \ A2 \ A03 ) = P (A1 \ A2 A3 ) = P (A1 \ A2 ) P (A1 \ A2 \ A3 ) = :06 :01 = :05 d. Denote the event that the system has at most two of these defects as E , then E 0 will be the event that all three types of defects occurs at the same time, i.e. A1 \ A2 \ A3 , hence P (E ) = 1 P (E 0 ) = 1 P (A1 \ A2 \ A3 ) = 1 :01 = :99 6 1.3 Practice 3 - 9/17/2020 Exercise 1.7. A purse contains five 10$, four 5$, and eight 1$ bills. If the bills are selected one by one in random order, what is the probability that at least two bills must be selected to obtain the first 10$ bill. Hint: Describe this event as the complement of another event whose probability is relatively easier to compute Solution. Let A be the event that at least two bills to get the first 10$ bill, then A0 would be the event of getting the 10$ bill at the first drawal. Let the sample space S be all the possible outcomes from the first drawal, then S f 1; = 10 ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; 102 103 104 105 51 52 53 54 11 12 13 14 15 16 17 18 g A0 = f101 ; 102 ; 103 ; 104 ; 105 g The subscripts indicates different dollar bills with same value. Hence 1 P (A0 ) = P (A0 ) = 5 17 and P (A) = 12 17 Exercise 1.8. Solve the following counting problems (a) Given a group consisting of five couples, we would like to choose two couples and an additional person for a study for which people will be asked about their opinion on the same matter. In how many ways this selection can be done (b) Given 260 seats in our classroom, 250 students can be seated in these seats in P250;260 different ways. Explain why this is true. Note that: P250;260 denotes permutations of size 250 that can be formed from 260 items (c) Using the counting techniques you have mastered in our class, show that total number of subsets of a set of size n is 2n Solution.  (a) We have 52 ways to choose two couples out of five couples, and then there are still three couples(6 persons) remaining, so we have 61 = 6 ways to choose an additional person.   5 6 Thus in total there are 2 1 = 10  6 = 60 different choices (b) Recall P250;260 means the number of ways of ordering a subset of size 250 from the whole set of size 260. Let’s write 1;    ; 260 on 260 cards, each representing a seat, giving 250 of these cards to the 250 students constitute as a seating plan, which is the same as ordering a subset of size 250 from f1;    ; 260g, thus justifying the answer Or you can reason like this: For student 1 there are 260 seats to choose from, after making a selection for student 1, there are 259 seats to choose from for student 2, and after that there are 258 choices for seating the third student, etc. there are only 11 choices for the last student, so in total there are 260  259  258    11 = 260! 10! = P250 260 ; different ways to assign seats for 250 students (c) Assume the set is S = f1;    ; ng, every subset can be represented as a sequence of n digits of 0’s and 1’s, put 1 on the k-th digit if element k is in the subset, 0 if not, then is a well defined one-to-one correspondence between the subsets of S and these sequences, so we just need to figure out how many such sequences are there. For the first digit there are 2 choices(put either 0 or 1), there are 2 choices for the second digit, etc. The total number of such sequences is 2 |  {z    } 2 n 2 = 2 of them 7 n Exercise 1.9. The composer Beethoven wrote 9 symphonies, 5 piano concertos and 32 piano sonatas (a) A music director would like to choose one symphony and 2 piano sonatas from Beethoven, in how many ways this selection can be done. Justify the steps (b) If every year at the birthday of Beethoven a concert featuring Beethoven’s music is organized and three pieces of music is going to be presented, how many different programs are possible. order in which music pieces are played is not significant and the same piece of music can not be played more than once in a program (c) When a program consisting of three (distinct) pieces of Beethoven music is formed, compute the probability that the program consists of two piano sonatas and one symphony. Justify to yourselves that any three pieces of music are equally likely to come together Solution.  (a) The number of ways of choosing the symphony is there are  32 2 choices of piano sonatas, so in total  9 1  9 1 32 2 = 9  , for each choice of symphony = 9   32 31 2 = 9   16 31 = 4464 (b) 3 pieces from the total 46 pieces of music which is   46 3 = 46     3 45 2 44 1 = 46   15 22 = 15180 (c) The probability of a 3-piece program consists of exactly one symphony and 2 sonatas is P = 9 1  32 2 46 3  = 8 4464 15180  : 0 294 1.4 Practice 4 - 9/24/2020 Exercise 1.10. Using the multiplication rule P (A \ B ) = P (AjB )  P (B ), given events A; B; C (a) First, show that P (A \ B \ C ) = P (AjB \ C )  P (B \ C ) (b) Then, prove that P (A \ B \ C ) = P (AjB \ C )  P (B jC )  P (C ) (c) Derive a similar rule for the probability of intersection of k events, i.e. P (A1 j    )      P (A P (A1 \ A2   \ A k) = k) Solution. (a) P (A \ B \ C ) = P (A \ (B \ C )) = P (AjB \ C )  P (B \ C ) (b) Since P (B \ C ) = P (B jC )  P (C ), P (A \ B \ C ) = P (AjB \ C )  P (B jC )  P (C ) (c) P (A1 \ A2 \ A3 \    \ A ) = P (A1 \ (A2 \ A3 \    \ A )) = P (A1 jA2 \ A3 \    \ A )  P (A2 \ A3 \    \ A ) = P (A1 jA2 \ A3 \    \ A )  P (A2 jA3 \    \ A )  P (A3 \    \ A k k k k k .. = . = P (A1 jA2 \ A3 \    \ A k) k k)  P A2 jA3 \    \ A      P A ( k) ( k) Exercise 1.11. Suppose an individual is going to be randomly selected from the population of all adult males in the US. Let A be the event that the selected individual is taller than 6ft in height, and B the event that the selected individual is a professional Basketball player. Assume additionally that all basketball players are taller than 6 feet (a) Compare P (AjB ) to you think so (b) Also compare P (B jA) in magnitude, which one do you think is larger? explain why P (B ) to P (B jA) in magnitude? (c) Assume that 35% of adult men living in the US are taller than 6 feet, and 0.001% of these men are professional basketball players, represent these (two) numbers as probabilities (or conditional probabilities) about the events A and B (d) If we would like to compute what is missing? P (B ), can we derive it from the given information? If not, Solution. (a) By the assumption, we know that B  A, thus A \ B = B P (AjB ) = P (A \ B ) P (B ) P (B jA) = P (A \ B ) P (A) = P (B ) P (B ) = 1 = P (B ) P (A)  1 Since not every person taller than 6 feet is a basketball player, it may well be P (B ) ) P (B ) < 1, hence in general P (AjB ) > P (B jA) P (A) 9 P (A) > (b) In general there could be people no taller than 6 feet, thus P (A) < 1 ) P (B jA) = P (B ) (c) P (A) = 0:35, P (B jA) = 0:00001 (d) P (B jA) = P (B ) P (A) )P B ( ) = P (A)  P (B jA) = 0:35  0:00001 = 35  10 P (B ) > P (A) 7 Exercise 1.12. At a certain gas station, 25% of customers use regular gas, 45% use plus gas, and 30% use premium gas. Among those who use regular gas, only 40% fill their tanks. Of those who use plus, 60% fill their tanks, whereas of those using premium, 90% fill their tanks (a) What is the probability that next customer will request premium gas and not fill his tank? While solving this problem, think about how the above percentages translate into probabilities of some events (b) What is the probability that next customer fills his tank? (c) Given that the next customer fills his tank, what is the probability that premium gas is requested? Solution. Let A; B; C deonte the events that the next customer use regular, plus and premium gas respectively, F denote the event that the next customer fill the tank. The given informations translates to P (A) = 0:25, P (B ) = 0:45, P (C ) = 0:3, P (F jA) = 0:4, P (F jB ) = 0:6, P (F jC ) = : 0 9 (a) P (C \ F 0 ) = P (F 0 jC )P (C ) = (1 P (F jC ))P (C ) = (1  : : P F jA P A P F jB P B P F jC P C : 0 9) 0 3 = 0 03 P (F ) = P (F \ A) + P (F \ B ) + P (F \ C ) = ( ) ( ) + ( ) ( ) + ( 0:64. Note that this is the law of total probability P (F \ C ) P (F jC )P (C ) 0:27 (c) P (C jF ) = = = . Note that this is Baye’s theorem P (F ) P (F ) 0:64 (b) 10 ) ( ) = 1.5 Practice 5 - 10/1/2020 Exercise 1.13. Show that for any two independent events independent Solution. P (B )[1 A0 ; B are independent since P (A0 \ B ) = P (B ) P (A)] = P (B )P (A0 ) A; B ; the events A and B are also P (A \ B ) = P (B ) P (A)P (B ) = Exercise 1.14. Let A and B be two disjoint events defined on a sample space, and assume that P (A) = 1=3, P (B ) = 1=3. Are A and B independent events or not? You should provide a proof to your answer Solution. Since 0 = P (?) = P (A \ B ) 6= P (A)P (B ) = 1 3  13 = 1 9 , A; B are not independent Exercise 1.15. An urn contains four slips of paper, each having the same dimensions, which are associated with some prices. The paper with number one on it wins Price 1; paper with number 2 wins Price 2; paper with number 3 wins Price 3; paper with number four on it wins all prices 1,2,3. In an experiment only one slip of paper is going to be drawn from the urn. Let A1 denote the event of winning Price 1. and let A2 denote the event of winning Price 2, and let A3 denote the event of winning price 3 associated to this experiment (a) Show that A1 and A2 are independent events, similarly show that A1 and A3 are independent events; A2 and A3 are independent events. This is called pairwise independence (b) However show that P (A1 \ A2 \ A3 ) 6= P (A1 )  P (A2 )  P (A3 ), therefore the three events are not mutually independent (c) Think of an analogous problem where in the urn there are FIVE cards of equal dimensions, where the first four cards are associated to single prices Price 1, Price 2, Price 3, Price 4 respectively and the fifth card wins all four prices. Let A1 denote the event of winning the first price, and A2 be the event of winning Price 2, are A1 and A2 independent or not? Solution. P (A1 ) = P (A2 ) = P (A3 ) = 12 , P (A1 \ A2 ) = P (A1 \ A3 ) = P (A2 \ A3 ) = P (A1 \ A2 \ A3 ) = (a) 1 4 P (A1 \ A2 ) = P (A1 )P (A2 ) = A1 ; A2 ; A3 are pairwise independent 1 = 4 1 2  12 (b) 1 = 4 P (A1 \ A2 \ A3 ) 6= P (A1 )P (A2 )P (A3 ) = (c) 1 = 5 P (A1 \ A2 ) = P (A1 )P (A2 ) = 2 5  25 = 1 4 = 1 2 , similarly for  12  12 = A1 ; A3 and A2 ; A3 , hence 1 8 4 25 Exercise 1.16. Answer if each X defined as below is a random variable or not. If random variable, try to create a random variable X 0 based on X X is not a (a) X is defined as the genders of next two consecutive births in a particular hospital (b) X is defined as the outcome of a coin toss experiment in which a coin is tossed three times (c) X is defined as 0 when a randomly chosen adult man living in the US has a college (d) X denotes the trials until a success occurs in a gambling machine. degree,and is defined as 5 if he does not have a college degree Solution. (a) X is not a random variable. Let X 0 be the number of boys of the next two consecutive births in a particular hospital 11 (b) X is not a random variable. Let X 0 be the number of heads of the three tosses (c) X is a random variable (d) X is not a random variable. Let X 0 be the number failures until a success occurs in a gambling machine Exercise 1.17. Let X be the number of boy babies in next 5 consecutive births a midwife attends. Assume that this midwife believes that probability of a boy baby birth is 0.45 while probability of a girl baby birth is 0.55. Compute all of the following probabilities: P (X = 0), P (X = 1), P (X = 2), P (X = 3), P (X = 4), P (X = 5), P (X = 6) according to her belief. Note that outcomes of the births are independent from each other. (You can leave the results as products of numbers) Solution. Let Ai = B or G, i = 1;    ; 5 to denote the i-th baby’s gender, then Ai ’s are mutually independent. Since these events are independent, P (BGBGB ) = P (B )P (G)P (B )P (G)P (B ) = P (B )3 P (G)2 , and P (BBBGG) = P (B )3 P (G)2 , so the order of the gender doesn’t matter, what really matters is the number of each gender 1. P (X = 6) = 0 since it’s just impossible to have six kids! 2. P (X = 5) = P (BBBBB ) = P (B...
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