Solutions to Unit 1 - The Burns Home Page

Solutions to Unit 4.21a) 7 of diamonds 1b) ace of spades, ace of diamonds, ace of hearts, ace of clubs 1c) 2, 3, 4, 5, 6, 7, 8, 9,10all of clubs1d) 2, 4, 6, 8,10 of clubs, diamonds, hearts, and spades3a) There are 5 possible outcomes3b) Pblack3c) Predn35n25nnblackall possibleredall possible3d) There are no possible outcomes therefore the probability is 04a) P joker4b) Pred jokern jokernall possible cards254127n red jokernall possible cards154

4c) Pqueen4d) Pblack4e) P

5c) Pred5d) PB75e) P

6. a) P A48n An Snnb) P Ac) P A7. a) P Aodd numberany number nn An Sn2814n An Sdivisible by 4any numbern

)P A 1 P AnA1nSn1nred blocksnumber blocks3112348. a) P Ab) P Ac) P A nn An SS'sn letters111 nn An SM'sn letters211411 nn An Snvowelsletters

10 Possible values {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}nA10. a) P An S10 b) P A10 c) P A12.n all headsnpossible values1878 nn An Sn381 tail, 2 tail, 3 tail nn An Sn12 a) nS 36possible valuesHHT, HTH, THHpossible valuesDie 1 2 3 4 5 61 1, 1 2, 1 3, 1 4, 1 5, 1 6, 12 1, 2 2, 2 3, 2 4, 2 5, 2 6, 23 1, 3 2, 3 3, 3 4, 3 5, 3 6, 34 1, 4 2, 4 3, 4 4, 4 5, 4 6, 45 1, 5 2, 5 3, 5 4, 5 5, 5 6, 56 1, 6 2, 6 3, 6 4, 6 5, 6 6, 6

12 b) P An An Sntotals 7npossible values6361612 c) 1P AnA1nSn1nP Atotals 7possible values61365614.4444% this will be equal to the probability ratio we need100let the number of caffeine-free diets be x,44 n diet100 nall drinks44 16 x100 44 x1936 44x1600 100x336 56xx 6

Unit 4.3 Solutions1. a) {6, 9}b) {5, 2, 6, 9, 12, 10}c) {9,10}d) {4, 2, 9, 10, 6, 1}e) f) {9}2a) P A B P A PB P AB0.7 0.5 0.2 PAB0.7 0.7 PABPAB0SinceP AB=0, mutually exclusive.2b) P A B P A PB P AB 0.7 0.2 0.15 0.75SinceP AB=0.15, not mutually exclusive2c) P A B P A PB P AB0.9 0.3 PB 00.9 0.3PBPB 0.6SinceP AB=0, mutually exclusive3a) remember P A and B P ABSince mutually exclusive P AB 03b) P A B P A PB P AB 0.2 0.30 0.5

3c Since all probabilities must add up to 1, and A and B are mutually exclusive1=P(A)+P(B)+P(C)1=0.2+0.3+P©P(C)=0.55. Let A represent the event that it is a femaleLet B represent the event that it is a tetrafemale or tetra P A PB P ABnAnBnA B nS nS nSP P A B26 36 14 56 56 564856676. a)SCFT10 73b) PCnCn S102012

c) PCnTn S320d) P F 1P FnF1nS71 201320 8.SHAC14 16 812The number combined in both sets is 50-12=38Let HA represent Head aches only, B represent both, and C represent coldsTherefore HA+B=30C+B=24HA+B+C=38We have three equations and three variables, therefore we can do this.Rewrite the first two equations in terms of BHA=30-BC=24-BPlug these into the third equation.(30-B)+B+(24-B)=3854-B=38B=16

Therefore HA 30 1614C 24 16 8b i) 14 students have just a headacheP HA1450725n HAn Sb ii) 14+8+16=38 have headache or coldP HA or Cn HA or CnS38501925b iii) 12+14=26 have no cold symptomPnone26501325n nonen S10. Let A represent the event of drawing an aceLet B represent the event of drawing a club

nAnBnA B nS nS nSP A B P A P B P A B4 13 1 52 52 52165241314 a) Let A represent the event of being taller than 185 cmLet B represent the event of having dark hair PA BnS nS nSP A B P A P B P A Bn P B n A n BP A40 19 28 PAB45 45 457B4514 b) Let A represent the event of being taller than 185 cmLet B represent the event of having dark hairFrom above P AB40458914 c) Let A represent the event of being taller than 185 cm 1P AP A191452645

Unit 4.4 Solutions1a) Let A represent the event that the student is male.Let B represent the event that the student likes schoolP B AP AP ABP An An SP AB193312P B A 3319331219n ABnS1233The probability of the event that a student likes school given that the student is maleis 1219 .1b) Let A represent the event that the student dislikes school.Let B represent the event that the student is female.P B AP AP ABP An An S833

P ABP B A n ABnS33333383338The probability of the event that a student is female given that the student dislikesschool is 3 8 .5a) Let A represent the event that the patient is a smoker.Let B represent the event that the patient is dying from lung cancer.P B AP AP ABP ABP An An S250010000n ABnS61010000610P B A 10000250010000610250061250

The probability of the event that a patient will die from lung cancer given that the61patient is a smoker is250 .5b) Let A represent the event that the patient is a non-smoker.Let B represent the event that the patient is dying from lung cancer.P B AP AP ABP ABP An An S750010000n ABnS11010000110P B A 10000750010000110750011750The probability of the event that a patient will die from lung cancer given that thepatient is a non-smoker is 11750 .

6. Let A represent the event that first marble is red.Let B represent the event that the second marble is red.P A B P A BP AP A38n An SP B An B AnS272 3PA B 7 83289a) Let A represent the event that first draw is a joker.Let B represent the event that the second draw is an ace.P A B P A BP AP An An SP B A254n B AnS4534 2PA B 53 5441431

9b) Let A represent the event that first draw is a numbered card.Let B represent the event that the second draw is an red joker.P A B P A BP AP An An SP B A3654n B AnS1531 36PA B 53 5421599b) Let A represent the event that first draw is a queen.Let B represent the event that the second draw is an queen.P A B P A BP AP An An SP B A454n B AnS3533 4PA B 53 542477

9d) Let A represent the event that first draw is a black card.Let B represent the event that the second draw is a black card.P A B P A BP AP An An SP B A2754n B AnS265326 27PA B 53 5413539e) Let A represent the event that first draw is a numbered card below 10.Let B represent the event that the second draw is the same numbered card.P A B P A BP AP An An SP B A3254n B AnS3533 32PA B 53 5416477

9f) Let A represent the event that first draw is a red joker or red ace.Let B represent the event that the second draw is an red joker or ace.It is easier to find complementary because of the either draw aspect.1P AB51 501 54 535247719a) Let A represent the event that first draw is a red marble.Let B represent the event that the second draw is green marble.P A BP AP ABP An An S6104 6PA B 9 104154P A B 1561049

19b) Let A represent the event that the first marble is redLet B represent the event that the second marble is red.P A B P A BP AP AP B An An S610n B AnS595 6PA B 9 101319c) Let A represent the event that the first marble is green.Let B represent the event that the second marble is green.P A B P A BP AP AP B An An S410n B AnS393 4PA B 9 10215

19b) Let A represent the event that the first marble is redLet B represent the event that the second marble is red.We have two situationsred red green red red red red red greengreenP P P P P PUnit 4.5 Solutions5 6 4 4 9 10 9 1023452a) TTFTTFFTTTFFTFFFTFTFTFTFTFTFTF

2b) Pperfectn perfectnS116or 1 1 1 12 2 2 22c) P n 3 correct n 1 incorrect3 correct orn Sn S41614we can look at it this way1 1 1 1 4 , we multiply by 4 because there are 4 ways of having 3 correct.2 2 2 43a) (7D7D)3b) (AS7D)3c) (2C2C, 2C3C, 2C4C…10C9C, 10C10C)3d) (AS2C, AS4C, AS6C, AD10C)4a) 6x6=364b) The only way to obtain a sum of 2 is 1 white and 1 red4c) 1R2W1W2R

5a) nS ncards ndice406 240There are 240 possible outcomes5b) The elements of the sample space (card number and suit, number on dice)5ci)Let A represent the event of drawing an even cardLet B represent the event of an even number on dieeven and even P AP BnAnB nS nSP P A B20 3 40 6141 25cii) Let A represent the event of drawing an even cardLet B represent the event of an odd number on dieeven and odd P APBnAnB nS nSP P A B20 3 40 6141 2

5ciii) Let A represent the event of drawing a 3 cardLet B represent the event of an number less than or equal to 3 number on die3 and 3 P AP BnAnB nS nSP P A B4 340 61201 25civ) Let A represent the event of drawing a suit of cardLet B represent the event of number on dieArrangements are {(1,6); (2,6); (3,4); (4,3); (5,2); (6,1)}sum=7 P A PBnAnSnBnSP P A B 6 61 24 1 640 6110arrangements5civ) Let A represent the event of drawing a suit of cardLet B represent the event of number on dieArrangements are {(5,6); (6,5); (7,4); (8,3); (9,2); (10,1)}sum=7 P A PBnAnSnBnSP P A B 6 61 24 1 640 6110arrangements

6a) Let A represent the event of drawing a joker firstLet B represent the event of drawing an ace on the second P A B P A P B2 4 54 542729This is not conditional because we do not require the finding of the probability ofdrawing an ace given that the joker was drawn.6b) Let A represent the event of drawing a numbered card firstLet B represent the event of drawing a red joker on the second P A B P A P B36 1 54 541816c) Let A represent the event of drawing a queen card firstLet B represent the event of drawing a queen on the second P A B P A P B4 4 54 5447296d) Let A represent the event of drawing a black card firstLet B represent the event of drawing a black on the second P A B P A P B27 27 54 5414

6e) Let A represent the event of drawing a numbered card < 10 firstLet B represent the event of drawing the same on the second P A B P A P B32 4 54 54327296f) Let A represent the event of drawing a red joker or red ace card firstLet B represent the event of drawing a red joker or red ace on the secondEasier to do complement. 1 P A B 1 P A P B 51 51154 54353248a) Let A represent the event of being a union memberP An An S510128b) Let A represent the event of being a union memberLet B represent the event of being a union memberNotice the name is not replaced nAnB nS nSP A B P A P B 5 410 9291 2

8c) Let A represent the event of being a union memberLet B represent the event of being a union memberLet C represent the event of being a union memberLet C represent the event of being a union memberNotice the name is not replaced nAnBnCnD nS nS nS nSP A B C D P A P B P C P D1 2 3 45 4 3 2 10 9 8 71428d) We have 4 possible situationsNUUU, UNUU, UUNU, UUUNWe want P NUUU UNUU UUNU UUUN P NUUU PUNUU PUUNU PUUUN 5 5 4 3 5 5 4 3 5 4 5 3 5 4 3 5 10 9 8 7 10 9 8 7 10 9 8 7 10 9 8 75219a) Let A represent the event of drawing a chocolate bar firstLet B represent the event of drawing a chocolate bar second P A B P A P B10 10 20 20149b) Let A represent the event of drawing a chocolate bar firstLet B represent the event of drawing a chocolate bar secondLet C represent the event of drawing a fruit bar firstLet D represent the event of drawing a fruit bar secondLet E represent the event of drawing a toffee bar firstLet F represent the event of drawing a toffee bar second

P A PB PC PD PE P FP A B P C D P E F 10 10 7 7 3 3 20 20 20 20 20 20792001 keep any candy791200121200P spending $3000 P 1st month P 2nd month P 3rd month1 1 1 2 2 2189c) P 10a) 10b) The two situations are spending $5000 or $60001 1 1P6000 2 2 2181 1 1P5000 32 2 238we spend 2000, 2000, 2000we need to multiply by 3 because of three casesthree cases (1000, 2000, 2000); (2000, 1000, 2000); (2000, 2000, 1000)we can look at it this way1000,2000,2000 2000,1000,2000 2000,2000,1000P P P1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 21 1 1 8 8 838

more 5000 6000P P P3 1 8 8121 spending $30001187810c) P 11a) nS 101010100011b) Pending with a 511c) Pfirst is 1 or 2n last is 5nS101011010101001000110n 1 or 2nS21010101010200100015