Fluent Essays

Introduction to Probability Summer 2021 Homework Assignments Fan Xue Contents Instructions 1 Homework Assignment 1 2 Homework Assignment 2 6 Homework Assignment 3 0 Homework Assignment 4 0 Homework Assignment 5 5 Instructions Replace the highlighted portions of the text with your specific information. Remove the highlighting as you do so. 1 Homework Assignment 1 If there are LATEXerrors, please take care of them. Set 1, problem 1 How many plate numbers can be created in California format DLLLDDD, where D stands for digit and L stands for letter? What if no repetitions are allowed in the digits? Solution: In the plate numbers, there are 4 digits and 3 letters. As we know that no repetitions are allowed in digits. the answer should be 10∗9∗8∗7∗26∗26∗26 = 88583040 So, there are 88583040 plate numbers can be created. You only answered the second question. 2.5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Set 1, problem 2 A zip code consists of 5 digits. If the first cannot be the digit 0, how many zip codes can be generated? Solution: There are ten possible digits(0 to 9). If they can be repeated, and the first cannot be the digit 0. Then, the answer should be 9 ∗ 10 ∗ 10 ∗ 10 ∗ 10 = 90000 There are 90000 zip codes can be generated. 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Set 1, problem 6 If you need to split a class of 32 students in 8 groups of 4, in how many ways can you do it? Solution: 32! 4! 28! 4! 24! 4! 20! 4! 16! 4! 12! 4! 8! 4! 4! 4! 8! = 5.92 ∗ 1019 there are 5.92 ∗ 1019 ways. Please explain how you get these numbers. You may mean the correct answer but the number you given in the numerator is incorrect. 2/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Set 1, problem 7 If a student is required to solve 8 out of 10 assigned problems, how many choices does she have? What if the first 4 problems are mandatory? Solution: a. 10! 8!2! = 45 ( 10 8 ) b. 6! 4!2! = 15 ( 6 4 ) She has 45 choices to solve 8 out of 10 assigned problems. If the first 4 problems are mandatory, she has 15 choices. Please explain how you get these numbers. 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Set 2, problem 1 An urn contains 3 balls: 1 red, 1 green, and 1 blue. Consider an experiment that consists of taking 1 ball from the urn, replacing it in the box, and drawing a second ball. Describe the sample space of this experiment and for the experiment obtained not replacing the first ball in the urn. Solution: The event consists of drawing of total of 2 balls from the box since the ball is replaced after taking, we take any of the 3 balls during the second attempt and the sample space will be:{ (r,r)(r,b)(r,g)(b,r)(b,b)(b,g)(g,r)(g,b)(g,g) } the first color indicated the ball taken during the first attempt, and the second color indicated the ball taken after replacing the ball in the box. Now, if the ball is not replaced in the box after taking it, only two balls will be left if we take 1 ball, hence only 2 possibilities will be there in the second attempt, so, the sample space will be: {(red,blue)(red,green)(blue,red)(blue,green)(green,red)(green,blue)} 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Set 2, problem 4 A hospital administrator codes incoming patients suffering heart disease according to whether they have insurance (code 1 if they do and 0 if they do not) and based on their condition, which can be rated as good G, fair F, or serious S. Consider an experiment that consists of producing codes for such a patient. a. Give the sample space of this experiment. b. Let S be the event that the patient is in serious condition and list all outcomes in S c. Let U be the event that the patient is uninsured and all outcomes in U d. List the outcomes in Sc ∪U . Solution: a. Sample space consists all possible outcomes of this experiment. That is, all possible combinations of insurance coding and condition coding S = {(1,f), (1,g), (1,s), (o,f), (0,g), (0,s)}= {0, 1}×{f,g,s} b. Given that a person is in serious condition. There are only two possible outcomes in S: insured or not insured S = {(1,s), (0,s)} c. Given that a person does not have insurance, there are only three possible outcomes: serious, good or fair U = {(0,g), (0,f), (0,s)} d. Bc∪A is the event that the patient has insurance(does not have insurance) or is in serious condition S = Bc = {(1,g), (1,f), (1,s)} Bc ∪A = (1,g), (1,f), (1,s), (0,s) what are all these new sets A,B? And what is S now? 4/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Set 2, problem 5 Suppose that two event E, Fare such that E ∩F = ∅ and satisfy P(E)=0.3 and P(F)=0.5. What are the probabilities that a. either E or F occurs? b. E occurs but F does not? c. both E and F occur? What happens if you don’t assume that E ∩F = ∅? Solution: a. P(either E or F) = P(E) + P(F) = 0.3 + 0.5 = 0.8 use mathematical notation and why? b. P(E occurs but F does not) = P(E ∩Fc) = P(E) - P(E ∩F) = 0.3 − 0 = 0.3 c. P(both E and F) = P(E ∩F) = 0 d. if we don’t assume that E ∩F = ∅, we cannot find the probabilities. 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Set 2, problem 6 At a summer camp three activities are offered: tennis, swimming, and archery. The courses are open to the 120 summer camp participants. There are 31, 24, and 17 in these activities, respectively. There are 4 participants in all activities, 11 in tennis and swimming, 9 in tennis and archery, 7 in swimming and archery. a. If a participant is chosen at random, what is the probability that he or she is taking part exactly one activity? b. If a participant is chosen at random, what is the probability that he or she is taking part in no activity? c. If two participants are chosen at random, what is the probability that they are both taking part in exactly one activity? Solution: Total = 120 Tennis = 31 Swimming = 24 Archery = 17 Tennis and Swimming = 11 Tennis and Archery = 9 Swimming and Archery = 7 All three activities = 4 Only Tennis = 31 - (9 - 4 + 4 + 7) = 15 Only Swimming = 24 - (7-4 + 4 + 7) = 10 Only Archery = 17 - (9 - 4 + 4 + 7 - 4) = 5 Those not taking any part = 120 - (15 + 7 + 10 + 5 + 4 + 3 + 5) = 120 - 49 = 71 a. P(Exactly 1 activity) = (15 + 10 + 5) / 120 = 30 / 120 = 0.25 b. P(No activity) = 71 / 120 = 0.592 c. P(two will be in exactly 1 activity) = 0.25 * 0.25 = 0.0625 Explanations are needed. Use set notation: T for tennis, S for swimming, and A for archery. Then we have for instance that only tennis is given by the set = T ∩Sc ∩Ac. If you provide your work, I will 4 give you full credit. 0/5 5 Homework Assignment 2 Set 3, problem 1 If 8 identical computers are to be divided among 4 lab rooms, how many divisions are possible? How many if each room must receive at least 1 computer? Solution: We need to divide 9 identical computers to 4 rooms. This problem type of dividing identical can be visualized as dividing n identical 0′s into r groupings. For example 00101000100 can represent dividing 2 computers to first, 1 computer to second 3 to third and 2 to fourth lab. 8 zeros and division represented by 1′s. n be total and r be groups. Then the problem becomes selecting r − 1 1′s to divide from total of n + r− 1. So (n + r− 1)Cr−1 is the number of ways we can divide n identical things into r groups. Here n = 8, 4 = 4 11C3 = 11 ∗ 10 ∗ 9/(3 ∗ 2) = 165 possible ways When each room must receive 1 computer the formula changes to (n − 1)Cr−1. This is because in the other formula we assign one to each and then assign the remaining n−r + r−1 = n−1. 7C3 = 7 ∗ 6 ∗ 5/(3 ∗ 2) = 35 ways. You need to take a closer look at how to use LATEXsince your formulæ do not display correctly. 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Set 3, problem 2 Suppose that 10 fish are caught in a pond that contains 5 distinct types of fish. a. How many different outcomes are possible, where an outcome consists in the number of fish caught of each of the 5 types? b. How many outcomes are possible if 3 of the 10 fish caught are trout? c. How many when at least 2 of the 10 are trout? Solution:a. So the number of possible combinations can be thought of as dividing 10 identical items into 5 groups since all combinations are possible n = 10,r = 5. 14C4 = 14 ∗ 13 ∗ 12 ∗ 11/(1 ∗ 2 ∗ 3 ∗ 4) = 1001 b. 3 of 10 fish are trought that means remaining 7 fish to be assigned to 4 groups of fish. n = 7,r = 4 10C3 = 10 ∗ 9 ∗ 8/(3 ∗ 2) = 120possibilities c. It says at least 2 trouts but it can be more so we assign 2 to trouts and then divide the remaining 8 into 5 types. n = 8,r = 5. 12C4 = 12 ∗ 11 ∗ 10 ∗ 9/(2 ∗ 3 ∗ 4) = 495possibilities Check your LATEX. Your formulæ do not show correctly! 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Set 3, problem 8 Two cards are randomly selected from an ordinary playing deck (52 cards). What is the probability that one of the cards is an ace and the other one is either a ten, a jack, a queen, or a king? This combinations is called a black jack. Solution:we have to find the probability that they form a back jack. overall outcomes = 52C2. From two cards, first card should be an ace, and the total number of ways = 4 ways. The second card should be a either ”ten” or ”jack” or ”queen” or ”king”. So, the total number of ways = 4∗4 = 16 ways. The overall favourable outcomes = 1stcard∗ 2ndcard = 4 ∗ 16 = 64 ways. To find the probability to have a blackjack. Probabilitytohaveablackjack = overallfavourableoutcomes overalloutcomes = 64 52C2 = 64 26 ∗ 51 = 64 1326 = 0.04826 Thus, probability to have a blackjack = 0.04826. Check your LATEX. 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Set 3, problem 7 A retail establishment accepts either the American Express or the VISA credit card. A total of 24% of its customers carry an American Express card, 61% carry a VISA card, and 11% carry both cards. What percentage of its customers carry a credit card that the establishment will accept? Solution: Let A be the event the customer has an American Express card, and let V be the event the customer has a VISA card. Then P(A) = 0.24,P(V ) = 0.61, and P(AV ) = 0.11. The question asks for P(A∪V ). Using the Inclusion-Exclusion formula P(A∪V ) = P(A) + P(V ) −P(AV ) = 0.74 . 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Set 4, problem 1 If two fair dice are rolled, what is the conditional probability that the first one lands on 6 given that the sum of the dice is i for i = 1, . . . , 12? If you use copy and paste, do it from the source code of the HTML page. Solution: Let A be the event that the first die is a six and let Bi be the event that the sum of the two die is i. There are 6 possible outcomes for A. A = {(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}. Use the command \{ if you want to generate a {. 7 All of these have a sum of at least 7, so P(ABi) = 0 for all i ≤ 6. For every i ≥ 7. Thus P(ABi) = 136 for all i ≥ 7. Also for all i ≥ 7 there are 13 − i possible outcomes in Bi so P(Bi) = 13−i36 for all i ≥ 7. for i = 2; P(A|Bi) = 0 for i = 3; P(A|Bi) = 0 for i = 4; P(A|Bi) = 0 for i = 5; P(A|Bi) = 0 for i = 6; P(A|Bi) = 0 for i = 7; P(A|Bi) = 1 6 for i = 8; P(A|Bi) = 1 5 for i = 9; P(A|Bi) = 1 4 for i = 10; P(A|Bi) = 1 3 for i = 11; P(A|Bi) = 1 2 for i = 12; P(A|Bi) = 1 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Set 4, problem 4 In a certain community, 36% of the families own a dog and 22% of the families that own a dog also own a cat. In addition, 30% of the families own a cat. What is a. the probability that a randomly selected family owns both a dog and a cat? b. the conditional probability that a randomly selected family owns a dog given that it owns a cat? Solution: Let A be the event that family owns a dog, and let B be the event that family owns a cat P(A) = 0.36,P(B|A) = 0.22,P(B) = 0, 3 a. P(A∩B) = P(A) ∗P(B|A) = 0.36 ∗ 0.22 = 0.0792 b. P(A|B) = P(A∩B) P(B) = 0.0792 0.3 = 0.264 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Set 4, problem 2 Consider an urn containing 12 balls, of which 8 are white. A sample of size 4 is to be drawn with replacement (without replacement). What is the conditional probability (in each case) that the first and third balls drawn will be white given that the sample drawn contains exactly 3 white balls? Solution: The total number of balls in the urn is 12. Out of 12 balls 8 balls are white and 4 balls are black. Let A be the event of the first and third balls are white, and let B be the event that the sample drawn contain exactly 3 white balls. The conditional probability is P(A|B) = P(AB) P(B) . A∩B : The first and third balls are white and the sample contains exactly 3 white balls. The possibil- ities are (WBWW) and (WWWB). B : The sample contains exactly 3 white balls. The possibilities are (BWWW)(WBWW)(WWBW)(WWWB) P(A|B) = P(AB) P(B) = (( 8 12 )( 4 12 )( 8 12 )( 8 12 ) + ( 8 12 )( 8 12 )( 8 12 )( 4 12 )) 4(( 8 12 )( 4 12 )( 8 12 )( 8 12 )) = (( 2048 20736 ) + ( 2048 20736 )) 4( 2048 20736 ) = ( 4096 20736 ) (4 2048 20736 ) = 4096 8192 = 0.5 Therefore, the conditional probability that the first and third balls drawn will be white given that the sample drawn contains exactly 3 white balls in the case of with replacement is 0.5 The problem is also asking the same question when balls are drawn with no replacement. 2.5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Set 4, problem N4 Urn A contains 2 white and 4 red balls, whereas urn B contains 1 white and 1 red ball. A ball is ran- domly chosen from urn A and put into urn B, and a ball is then randomly selected from urn B. What is a. the probability that the ball selected from urn B is white? b. the conditional probability that the transferred ball was white given that a white ball is selected from urn B? Solution: Let W be white balls, and R be red balls. A ball is randomly chosen from urnA and put into urnB, and a ball is then randomly selected from urnB a. case1 : If ball chosen from urn A is white. Probability of chosing a white ball from urn A is P = 2 6 urn B already contains 1W and 1R. After a white ball from urn A put into urn B, then the number of white balls in urn B will be 2. Now, the probability of selecting a white ball from urn B is P(whiteballfromurnB) = 2 3 . Hence, the probability that the ball selected from urn B is white and the ball chosen from urn A is white is P(W1,W2) = 2 6 ∗ 2 3 = 2 9 case2 : 9 if ball chosen from urn A is red. Probability of choosing a red ball from urn is P = 4 6 . Then, we place the red ball in urn B. Now, urn B contains 1W and 2R, now the probability of selecting a white ball from urn B is P(whiteballfromurnB) = 1 3 . Hence, the probability that a ball selected from urn B is white and the ball chosen from urn A is red is P(R1,W2) = 4 6 ∗ 1 3 = 2 9 . We add these two mutually exclusive cases to get the required probability is P(ballselectedfromurnBwhite) = 2 9 + 2 9 = 4 9 . b. The conditional probability that the transferred ball was white. Given that a white ball is selected from urn B is P( transferredballiswhite whiteballisselectedfromurnB ) = P(transferredballiswhite∩whiteballisselectedfromurnB) P(whiteballisselectedfromurnB) = 4 18 4 9 = 4 18 ∗ 9 4 = 1 2 = 0.5 You can formulate your solutions in much simpler terms using Bayes’ formula. Also check your usage of LATEX. 5/5 10 Homework Assignment 3 Set 5, problem 1 A worker has asked her supervisor for a letter of recommendation for a new job. She estimates that there is an 80% chance that she will get the job if she receives a strong recommendation, a 40% chance if she receives a moderately good recommendation, and a 10% chance if she receives a weak recommendation. She further estimates that the probabilities that the recommendation will be strong, moderate, and weak are 0.7, 0.2, and 0.1, respectively. a. How certain is she that she will receive the new job offer? b. Given that she does receive the offer, how likely should she feel that she received a strong recom- mendation? A moderate recommendation? A weak recommendation? c. Given that she does not receive the job offer, how likely should she feel that she received a strong recommendation? A moderate recommendation? A weak recommendation? Solution: Let O be get-offer, NO be non-offer, ST be strong recommendation, G be good recommenda- tion, and W be week recommendation. From given problem data, we can get P(O ∩ST) = P(O|ST)P(ST) = (0.8)(0.7) = 0.56 P(O ∩G) = P(O|G)P(G) = (0.4)(0.2) = 0.08 P(O ∩W) = P(O|W)P(W) = (0.1)(0.1) = 0.01 a. P(O) = P(O ∩ST) + P(O ∩G) + P(O ∩W) = 0.56 + 0.08 + 0.01 = 0.65 b. P(ST|O) = P(ST ∩O) P(O) = 0.56 0.65 = 0.86 P(G|O) = P(G∩O) P(O) = 0.08 0.65 = 0.12 P(W |O) = P(W ∩O) P(O) = 0.01 0.65 = 0.02 c. from the data, we can get P(NO ∩ST) = 0.7 −P(O ∩ST) = 0.7 − 0.56 = 0.14 P(NO ∩G) = 0.2 −P(O ∩G) = 0.2 − 0.08 = 0.12 P(NO ∩W) = 0.1 −P(O ∩W) = 0.1 − 0.01 = 0.09 P(NO) = P(NO ∩ST) + P(NO ∩G) + P(NO ∩W) = 0, 14 + 0.12 + 0.09 = 0.35 P(ST|NO) = P(ST ∩NO) P(NO) = 0.14 0.35 = 0.4 P(G|NO) = P(G∩NO) P(NO) = 0.12 0.35 = 0.34 P(W |NO) = P(W ∩NO) P(NO) = 0.09 0.35 = 0.26 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 Set 5, problem 2 Barbara and Dianne go target shooting. Suppose that each of Barbara’s shots hits a wooden duck target with probability P1, while each shot of Dianne’s hits it with probability P2. Suppose that they shoot simultaneously at the same target. If the wooden duck is knocked over (indicating that it was hit), what is the probability that a. both shots hit the duck? -b. Barbara’s shot hit the duck? Solution: Let B be the Barbara hits wooden duck, D be Diana hits wooden duck, and H be the duck is hit P(B) = P1,P(D) = P2 a. P(Bothhitduck|duckishit) = P(BD|H) = P(BDH) H = P(H|BD)P(BD) H P(H|BD) = 1,P(BD) = P1P2 P(H) = P(B ∪D) = P(B) + P(D) −P(BD) = P1 + P2 −P1P2 Probability = (1)(P1)(P2) (P1 + P2) − (P1)(P2) b. P(B|H) = P(B ∩H) P(H) = P(H|B)P(B) P(H) = 1(P1) (P1 + P2) − (P1P2) P(B|H) = P1 (P1 + P2) − (P1P2) Both shoot at the duck at the same time and their hits are independent 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Set 5, problem 3 In successive rolls of a pair of fair dice, what is the probability of getting 2 sums of seven before 6 even numbers? Solution: Let A, B and C be the events that get a seven, and even number, or neither on a given roll, respectively. Thus P(A) = 6 36 = 1 6 ,P(B) = 18 36 = 1 2 ,P(C) = 1 −P(A) −P(B) = 1 3 We have P(2 sevens before 6 evens) = 5∑ i=0 P(i evens before 2 sevens) = 5∑ i=0 i∑ j=0 P(ievensbefore2sevens|jevensbeforefirstseven)P(jevensbeforefirstseven) 1 = 5∑ i=0 i∑ j=0 P(i− jevensbeforefirstseven)P(jevensbeforefirstseven) For each non-negative integer j let Ej be the event that you roll j evens before the first seven, We have P(E0) = P(E0|sevenonthefirstroll)P(sevenonthefirstroll) +P(E0|evenonthefirstroll)P(evenonthefirstroll) +P(E0|neithersevennorevenonthefirstroll)P(neithersevennorevenonthefirstroll) = 1P(A) + 0P(B) + P(E0)P(C) so P(E0) = P(A) 1 −P(C) = 1 4 For j > 0 we have P(Ej) = P(Ej|sevenonthefirstroll)P(sevenonthefirstroll) +P(Ej) = P(Ej|evenonthefirstroll)P(evenonthefirstroll) +P(Ej) = P(Ej|neithersevennorevenonthefirstroll)P(neithersevennorevenonthefirstroll) = 0P(A) + P(Ej=1)P(B) + P(Ej)P(C) which implies that (1 −P(C))P(Ej) = P(Ej−1)P(B) so P(Ej) = ( P(B) 1 −P(C) )j,P(E0) = ( 3 4 )j 1 4 So the desired probability is 5∑ i=0 i∑ j=0 ( 3 4 )i=j 1 4 ( 3 4 )j 1 4 = ( 1 4 )2 i∑ j=0 (i + 1)( 3 4 )i =' 0.5550 The initial identity in red is incorrect as stated. The presentation needs to be improved. If you correct your solution and improve your presentation, send me a message and I will take another look and give you some credit 0/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Set 5, problem 4 An urn contains 12 balls, of which 4 are white. Three players (A, B, and C) successively draw from the urn in that order repeatedly. The winner is the first one to draw a white ball. Find the probability of winning for each player if a. each ball is replaced after it is drawn. b. the balls that are withdrawn are not replaced. Solution: a. There are 4 white and 8 non-white balls, The probability of winning for with replacement 4 12 = 1 3 and probability that A does not win is 8 12 = 2 3 . Now, probability that A does not win in first draw is 2 3 . If it happens then B is exactly in the same situation as A was at beginning. SO now chance 2 of B to win is 2 3 A. Similarly if player A and B does not win then C is in the same situation so probability of C wins is ( 2 3 )2A. The sum of three probabilities is 1. A + ( 2 3 )A + ( 2 3 )2A = 1 A(1 + ( 2 3 ) + ( 2 3 )2) = 1 A(( 5 3 ) + ( 4 9 )) = 1 A = 9 19 So P(Awins) = 0.4737 P(Bwins) = ( 2 3 )( 9 19 ) = 0.3158 P(Cwins) = ( 2 3 )2( 9 19 ) = 0.2105 Your solution has some merit since your explanation is reasonable. Your presentation, however, is very poor and does not make explicit use of the formulæ you use. Again, if you give a cleaner proof, I can give you credit. b. The balls are withdrawn without replacement. The probabilities that 1st white ball is drawn at 1st, 2nd, 3rd, 4th, ......, 9th draw is 1st = 1 3 2nd = 8 12 4 11 = 8 33 3rd = 8 12 7 11 4 10 = 28 165 4th = 8 12 7 11 6 10 4 9 = 56 495 5th = 8 12 7 11 6 10 5 9 4 8 = 7 99 6th = 8 12 7 11 6 10 5 9 4 8 4 7 = 4 99 7th = 8 12 7 11 6 10 5 9 4 8 3 7 4 6 = 2 99 8th = 8 12 7 11 6 10 5 9 4 8 3 7 2 6 4 5 = 4 495 9th = 8 12 7 11 6 10 5 9 4 8 3 7 2 6 1 5 4 4 = 1 495 Now we know balls are drawn by first A, then B, then C, then A and so on. P(Awins) = P(whiteball1stdraw) + P(whiteballat4thdraw) + P(whiteballat7thdraw) = 1 3 + 56 495 + 2 99 P(Awins) = 0.4667 P(Bwins) = 8 33 + 7 99 + 4 495 = 53 165 P(Bwins) = 0.3212 3 P(Cwins) = 28 165 + 4 99 + 1 495 = 7 33 P(Cwins) = 0.2121 I gave you full credit for part b. 2.5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Set 6, problem 2 Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed n times. What are the possible values of X ? Solution: A coin is tossed six times and X represents the difference between the number of heads and the number of tails. Let H be the event of heads, and T be the event of tails. then, the following are the possible values: X(6H, 0T) = |6 − 0| = 6 X(5H, 1T) = |5 − 1| = 4 X(4H, 2T) = |4 − 2| = 2 X(3H, 3T) = |3 − 3| = 0 X(2H, 4T) = |2 − 4| = 2 X(1H, 5T) = |1 − 5| = 4 X(0H, 6T) = |0 − 6| = 6 Thus, the possible values of X are 0, 2, 4, 6 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Set 6, problem 3 Suppose that two teams play a series of games that ends when one of them has won i games. Suppose that each game played is, independently, won by team A with probability p. Find the expected number of games that are played when i = 2 and when i = 3. Also, show that this number is maximized when p = 0.5 in both cases. Solution: Let X denote the number of games played a. When i =2 E(X) = 2(p2 + (1 −p)2) + 3(2p2(1 −p) + 2p(1 −p)2) = −2p2 + 2p + 2 By letting d dp EX = 0 we can get p = 1 2 at which EX is maximized.( d 2 dp2 EX|p= 1 2 < 0) b. When i = 3 E(X) = 3(p3 + (1 −p)3) + 4(3p3(1 −p) + 3p(1 −p)3) + 5((6p3(1 −p)2) + 6p2(1 −p)3) = 6p4 − 12p3 + 3p2 + 3p + 3 4 By letting d dp EX = 0 we can get p = 1 2 at which EX is maximized.( d 2 dp2 EX|p= 1 2 < 0) 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Set 6, problem 6 You have $1000, and a certain commodity presently sells for $2 per ounce. Suppose that after one week the commodity will sell for either $1 or $4 an ounce, with these two possibilities being equally likely. a. If your objective is to maximize the expected amount of money that you possess at the end of the week, what strategy should you employ? b. If your objective is to maximize the expected amount of the commodity that you possess at the end of the week, what strategy should you employ? Solution: Suppose that you buy x ounces during the first week, and then you will have 1000 − 2x dollars until next week a. Let M denote the amount of money that you possess at the end of the week. EM = (1000 − 2x + x) 1 2 + (1000 − 2x + 4x) 1 2 = 1000 + 1 2 x The expected amount of money would be larger if you bought more during the first week. Therefore, the best strategy would be to use your all money to get 500 ounces of the commodity and sell them after one week. b. Let N denote the amount of the commodity EN = (x + 1000 − 2x) 1 2 + (x + 1000 − 2x 4 1 2 ) = − 1 4 x + 625 The expected amount of the commodity would be smaller if you bought more during the first week. Therefore, the best strategy would be not to buy anything during the first week, and spend all the money after one week. 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Set 6, problem 8 If X has distribution function FX, what is the distribution function FY of Y = e X? Solution: Let Y = eX. Since we know that X has a distribution function F, FY (y) = P(Y ≤ y) = P(eX ≤ y) = P(X ≤ ln(y)) = F(ln(y)) 5/5 5 Homework Assignment 4 Set 7, problem 1 If E[X] = 1 and V ar(X) = 5 find a. E[(2 + X)2] b. V …