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# Mathematics for Business and Economics MATH 1070

UCONN

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This 16 page Class Notes was uploaded by Mary Veum on Thursday September 17, 2015. The Class Notes belongs to MATH 1070 at University of Connecticut taught by Yuriy Shlapak in Fall. Since its upload, it has received 15 views. For similar materials see /class/205821/math-1070-university-of-connecticut in Mathematics (M) at University of Connecticut.

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Date Created: 09/17/15

1 to 03 All You Ever Wanted to Know About Probability Theory but Were Afraid to Ask Theoretical Exercises 1 Let 19 be a uniform probability on a sample space S If S has n elements what is the probability of an element x of S 7 Solution Because 19 is a probability on S we have 1931 1 19sn 1 As 19 is taken a uniform probability all 1931 are equal so 1931 for all elements 31 of S Take the sample space to be S 1 b c d e f and suppose the following are given 1 195 2 190 195 PM 190 19c 005 Calculate all elementary probabilities ie calculate 19a19b19c Solution We know 190 005 so 19a 2 19c 01 Likewise 195 01 Now observe 295 pd 290 1 Ma 7 190 7 296 17 025 075 As 195 Md 19f they each have to be one third of 075 ie 191 19d 19f 025 Take the sample space to be S 1 b c d e f and suppose the following are given Ma 295 290 01 Md 2 196 296 2 190 Calculate 19d 195 and 19f Solution Observe the following pd7e7 f 1 29a7b7 0 1 Ma 7 295 7 296 1 7 03 07 H Furthermore7 notice that pd 2 pe 2 2 pf 4 pf Hence M 7 U M QM p 290 07 i 7190 So pf 01 and thus 195 02 and pd 04 Uniform Probability In a class7 there are 8 boys and 10 girls A class council of 3 pupils is chosen Calculate the probability that the council consists of a 3 boys b exactly 2 boys and 1 girl c at least 1 boy Assume that all pupils are equally popular Solution ways There are 18 pupils in the class 3 are chosen This can be done in C1873 816 a Choosing 3 boys out of a group of 8 and 0 girls out of a group of 10 can be done in C87 3 C1070 ways So C83 C100 C18 3 7 0068627 19443 boys 7 816 b Choosing 2 boys out of a group of 8 and 1 girl out of a group of 10 can be done in C87 2 C107 1 ways So C8 2 C10 1 7 2 b 1 l77 00343137 p Oys g C18 3 816 c Choosing at least 1 boy is equivalent to choosing any combination but 3 girls in other words p at least 1 boy 1 719 3 girls Therefore C10 3 C870 p at least1boy 17 C1873 120 717 0852941 2 In a box there are 9 white pieces of chalk and 1 red one When taking 2 pieces at random7 what is the probability that it is exactly 1 white and 1 red piece Solution The total number of ways in which 2 pieces can be taken out of 10 is C107 2 Taking 1 white piece out of 9 and 1 red piece out of the only red piece available can be done in C971 C171 ways Therefore 1770971100707947 p 1red7 1 white 7 C1072 7 45 7 02 You can also see it as p 1 red7 1 white p rst a red7 then a white 1 p rst a white7 then a red 7 1 9 9 1 7 10 9 10 9 02 Which is7 of course7 the same value One method is no better than the other You should follow whatever line of thought appeals most to you As long as it is correct7 mathematics has no preferences 03 Out of a deck of 52 cards7 we draw a card at random a What is the probability to draw a spade 7 b What is the probability to draw a jack or a king 7 c What is the probability to draw an ace or a hart 7 Solution a Out of 52 cards in a deck7 13 are spades Hence 7 NS 7E7025 icardsi52i 39 290 b Out of 52 cards in a deck7 4 are jacks7 4 are kings None are both7 so we can simply add them In other words7 pJack or King pJack pKing J acks Kings cards cards 4 4 55 01538 c Out of 52 cards in a deck 4 are aces 13 are harts But the ace of harts is both so we have to deduct this one card to avoid double counting ie pA 0r Q7 PAP PA A 1 9 cards cards cards 7 4 13 1 ata a 03076 4 Out of a deck of 52 cards we draw 2 cards at random without replacement a What is the probability to draw a pair of kings 7 b What is the probability to draw a pair 7 c What is the probability that these cards are of consecutive face value but not necessary of the same suit 7 Note that aces are both low and high Solution a A O V There are 4 kings in a deck of cards Of these you choose 2 ie C4 2 The total number of taking 2 cards out of a deck of 52 is C52 2 so the probability is i 1 C4 2 fk 7 000425488 ppa1r o 1ngs C 5239 There are as much pairs of kings as there are pairs ofjacks This is of course true for any face value So the probability is just 13 times the probability of pair of kings ie pany pair 00588235 13 C 4 2 C 52 2 This is the same reasoning as the one used in the calculation of the number of straits in 5 hand poker We proceed as follows the rst card can be any of the 52 in the deck The second card can then be either one higher or one lower so there will be 8 choices for the second card But choosing rst say 40 and then 50 results in the same hand as rst choosing 50 and then 40 so we counted everything twice Therefore we have to divide by 2 528 p2 consecutive cards 052 2 003921568 3 Bernoulli Trials 1 The probability that a certain 10 is faulty is 05 How large is the probability that out of 1000 randomly chosen and tested 10 s that a exactly 10 10 s are faulty b at most 3 10 s are faulty Solution Note that we are dealing with a Bernoulli trial because 0 There are only two possible outcomes faulty or not faulty o The fact that one 10 is faulty is independent of the fact if another one is faulty o The probability of an 10 being faulty is the same for all namely 05 Therefore we have p 1OIC sarefaulty C100010O005100995990 001799 At most 3 10 s are faulty means that either 321 or no 10 s are faulty Hence p at most 3 10 s are faulty p3 10 s are faulty p2 10 s are faulty p1 10 s are faulty 1 p0 10 s are faulty C10003 00053 0995997 C1000 2 00052 0995998 C1000 1 00051 0995 9 C1000 0 00050 09951000 0006654 0033437 0083929 0140303 7 0264323 to A slightly biased coin ptail 051 is thrown 4 times a How large is the probability that tails shows up exactly 2 times 7 b How large is the probability that tails shows up exactly 3 times 7 c Compare a and b with a fair coin Solution Again this is a Bernoulli trial because 0 There are only two possible outcomes heads or tails o The outcome of one coin toss does not in uence the outcome of another one o The probability of heads or tails is the same for each toss Note that the total number of trials is 4 and the success rate is 051 so the failure rate is 049 This allows us to make the following calculations a The probability of throwing 2 times tails and 2 times heads with a biased coin is W Tails biased C4 2 0512 0492 037470006 b The probability of throwing 3 times tails and 1 time heads with a biased coin is p3 Tails biased C4 3 05l3 0491 025999696 c As an unbiased coin has a slightly lower probability to get tails the probability of getting three times tails will be a little lower and that of getting only two times tails will be a little higher lndeed p2 Tails unbiased C4 2 052 052 0375 gt W Tails biased and p3 Tails unbiased C43 053 051 025 lt p3 Tails biased Note that the difference between this biased coin and the unbiased coin is very small This is to be expected as the biased coin is only a 51 to 49 coin as opposed to a straight 5050 ie the coin is only slightly biased 4 Miscellaneous Exercises 1 John Peter Martin and Henry are sitting behind one another in a classroom on the rst second third or fourth row A row can only seat one of them a What is the probability that John is on the rst row 7 b What is the probability that John and Peter are in the middle 7 Solution There are many ways to solve these problems In each case we ll go over two possibilities a You can simply state that there are 4 rows on which John can sit down so the probability that he ll sit on the rst row is 1 out of 4 ie 025 Another way of seeing this is remarking that these 4 guys can take a seat in 4 ways The number of ways in which John is on the rst row ie John on the rst row and let the other three move around is 3 Therefore 31 1 p John on the rst row 4 A T V One way of looking at it is 1 2 1 p John and Peter in the middle 1 3 where g is the probability that John will be in the second or third row and is the probability that Peter is on the third or second row which ever one John is not occupying If you are confused by this you can resort to simply writing out all possibilities So as we argued in case a there are 4 possibilities The only ones of those where John and Peter are in the middle are Therefore 4 p John and Peter in the middle 7 2 An airplane ies over an infantry platoon The platoon res 5000 bullets at the plane A bullet has a chance of only 0001 to hit one of the sensitive areas eg the engine the pilot or the gasoline tank a What is the probability to hit the plane at least once 7 b How many bullets have to be red to have a 50 chance of at least one hit 7 Solution First note that this is a Bernoulli trial because 0 there are only two possible outcomes hit or miss 0 the fact that one bullet hits is independent of the fact if another one does 0 the probability of a bullet hitting is the same for all namely 0001 Furthermore note that at least one hit is the same as everything but no hits at all Therefore p at least one hit 1 7 p all 5000 shots are misses 17 C5000 0 00010 09995000 0993279 where 09995000 is the probability that all 5000 bullets mis Also note that with 5000 bullets you have more than 99 chance of hitting the plane This is to be expected although the probability of hitting with a single bullet is only one out of a 1000 you have 5000 tries for just the one hit Conversely setting the probability of hitting at 50 gives the equation 05 17 0999z where z is the number of bullets red So 705 70999 05 0999m log05 zlog0999 log05 m log0999 m 6928005 So you need 693 bullets to ensure a 50 hit 3 Gary drives a UCONN yellow line bus which has 12 stops From his past record we know Gary to have been on time on 55 of his stops Assume Gary s punctuality neither betters nor worsens a What is the probability that Gary makes every stop on time 7 b What is the probability that Gary makes at least 10 on time 7 Solution Note that we are dealing with a Bernoulli trial because 0 there are only two possible outcomes late or on time o as Gary s punctuality neither betters nor worsens the probability of being on time or not on one bus stop is independent from the previous ones 0 the probability of arriving on time is the same for all stops namely 55 a Simply applying the formula we have p all 12 stops on time C12 12 055 0450 00007662 b At least 10 stops means 1011 or 12 so p at least 10 stops on time p 10 stops on time p 11 stops on time p 12 stops on time C12 10 05510 0452 C12 11 05511 0451 C12 12 05512 0450 0033853 0007523 0000766 0042142 F We toss 5 fair coins What is the probability that the number of heads is higher then the number of tails 7 Solution There cannot be a 5050 because it is impossible to get 25 times tail Therefore we have p more T then H p more H then T 1 Because the probability of Tails equals the probability of Heads the probabilities in the above equation have to be equal Hence p more H then T 3 If you don t see this simply calculate the probabilities as a Bernoulli trial 01 Chevalier de Mere claimed that the probability to throw at least one 1 when throwing one die four times in a row equals the probability of throwing at least one 2 in 24 throws with 2 dice His friend Blaise Pascal proved him wrong How much was he of the mark 7 Solution Note that both experiments are Bernoulli trials for the same reasons as in exercise 32 For throwing one dice we get p at least one 1 in 4 throws 1 7 p no 1 at all in 4 throws 1 0 5 4 17040 7 7 05177469136 a 1 00 For the two dice experiment7 exactly the same reasoning yields p at least one 2 in 24 throws 1 7p no 2 at all in 24 throws 17 0240 2 04914038761 So Chevalier de Mere was about 26 of the mark You pull 13 cards out of a deck of 52 a What is the probability to get 3 aces 7 b What is the probability to get all 13 cards of the same suit 7 Solution The total number of ways to take 13 cards out of 52 is C52713 63501356 1011 a The total number of ways to get exactly 3 aces is C47 3 C39487 107 namely take 3 aces out of the 4 available aces and take 10 cards out of the other 48 non ace cards Therefore C4 3 C48 10 00412 C52 13 p exactly 3 aces b There are only 4 ways in picking out exactly all cards of the same suit namely all harts7 all clubs7 all diamonds7 or all spades So 150511 of the same suit 629907 1012 C52 13 A box contains 75 screws of which 5 are faulty If 25 of these 75 screws are taken out7 what is the probability that exactly 1 screw is faulty Solution The total number of ways in which 25 screws can be taken out of a box of 75 is C757 257 so the probability is C5 1 C70 24 033585 C75 25 p one faulty screw where C571 represents the one bad screw taken out of the 5 bad ones and C707 24 the 24 good screws taken out of the 70 good ones In a jar7 there are 4 bills of 1 and 6 bills of 5 We take out 2 bills at the same time What is the probability of each of the simple outcomes Solution Taking only two bills out of that jar results in either taking out 27 6 or 107 so the state space is s 26 10 a The probability of getting 2 is 7 C42 C00 6 2 M C102 E 15 Where 047 2 stands for taking 2 1 bills and C670 stands for taking no 5 bills b The probability of getting 6 is 0410 61 24 8 W6 lt gt lt gtii C10 2 45 15 Where 047 1 stands for taking 1 1 bill and C671 stands for taking 1 5 bill c The probability of getting 10 is C4 0 C6 2 15 5 10 i i W C10 2 45 15 Where 047 2 stands for taking 2 1 bills and C670 stands for taking no 5 bills Note that adding all the different possibilities does indeed give 1 2 8 5 p2 p6 p10 E E E 1 to In a local lottery 5 prizes are given away and 1000 tickets are sold In a probabilistic mood7 you buy 2 tickets a What is the probability of winning 2 prizes b What is the probability of winning nothing at all Solution The total number of ways in which to buy 2 tickets out of 1000 is C10007 2 a The probability of winning 2 prizes is C5 2 C995 0 000002002 C1000 2 p 2 prizes Where C57 2 stands for buying 2 tickets of the 5 winning ones and where C9957 0 stands for buying none of the 995 loosing tickets b The probability of winning no prizes at all is C5 0 C995 2 099002002 C1000 2 p no prizes Where C57 0 stands for buying no tickets of the 5 winning ones and where C9957 2 stands for buying 2 of the 995 loosing tickets H O An oil company estimates that only 1 oil well out of 20 will yield commercial quantities of oil Assume that successively drilled wells represent independent events If 12 wells are drilled7 nd the probability that a successful well was hit the following number of times 10 a exactly 4 b none at all c at least 2 Solution This is a Bernoulli trial because 0 There are only two possible outcomes viable oil well or not a viable oil well 0 The economic viability of one oil well is independent of the viability of the other ones 0 The probability for an oil well to be viable is the same for all namely 1 out of 20 So we can simply use the formula a Simply stating the formula for n 12 k 4 and p 120 gives 1 1 4 19 8 p 4 good Oil wells C124 E E 0002052 b No oil wells means 0 oil wells and therefore 1 0 19 12 p no good oil wells C12 0 0540360 c At least 2 oil wells means everything but no or only one oil well p at least 2 good oil wells 1 7 p 1 good oil well 7 p no good oil well 1 1 19 11 1 0 19 12 170121 E E 70120lt E 17 0540360 7 0341280 0118359 5 Independence Let A be a subset of the sample space S A is neither the empty set nor the entire sample space S Can A and A0 be independent 7 H Solution Because A O A0 lt1 we know that pA Ac 0 In order for A and Ac to be independent pA Ac pA pAc so pA or pAc would have to be 0 As we have chosen A to be neither 1 nor S this cannot be the case Pull a card out of a deck of 52 Are the events the card is a spade and the card is a jack a queen or a king independent events 7 to Solution The only thing we have to do is to verify pspadeS pJ7Q7K 19SpadeS 0 J7Q7K mimmmpagmmmimmwagomospawgmuu aemnwo Quemsdewwhepmbn v Minimumng g swepmamo g g g ommm mi Manda mam Minion Thav ue smut so no goo mi a Conditional Probability You gm no Mom Ema p new me who am m Pam zommz 0 32am znr zow camMe no lmms ME 02 M 4 mm a mm a mu m Soiqu mm mm Dugnm 5 m mm m we ms ammo w w dam dams what we no dams dugam m i mm way no Emmy ms i M2 L397FEOF omoozois ommoamamosa 7 Theorem of Total Pmtability was m Mmmm emu Hangva New Da MHvdaatmion m mmm 0mm m a Datum cammm ha 5 g m mama to cum gammme Mumwmdgwmm MthW MAME m we wan mm mm m mammth am We Ema mm m promumy mm m cmw mm mm m 5 mm mm m mainly 4mg mu m NW mm Wm g v Ms M 112 mm 5quot ma Wham m m 5mm amt mama mm mm x nm I N aw Dallu mm m Kanpur mum m mama a a a a o n 5 Msmmsme sm mew 5a m wawmumwm mg g xnumpmmmsumnsmm thsuhEqu mammammwxmx mama N am wmgmm m at mum um n5 ma 7 mmbmmoxmgamwn mm um the arswdam was5 m 7 59mm 1n ms am the mummy quota mg n m WW w mwmws v 39 f I v 39299 n Amifaazi szaz 53 a awms s as 2 33 3 33 2 E a 22 z i zqmgagm aa ig E 3523 2257 a 533 a 5122 23quot 5 Ease a f a wamza gao x a 321 3 s aaamiz a 532me 2 3 3 SEE 9 j amgam g u igiaini u E oo 305 m m 5 a 2 m 33 a 89 m A E v 3 2 am 3 ma a a a s a a 35 E3 i 53 a z 25 a 2 dis a 33 g 332 3 as 32 3a 323 a E s a 5 352 a 3 3 a 42 x 5 3 a 233 mi 2amp3 a w gn i5 51 3 m3 5 a m m s a e 3 a 2 922 5 353 922 5 31 922 E y E 3 3 535 a 3 SEE a as E a E 3 in 53 a 3 3332 923 may a m 33 Ea ns 3 3332 a mama mhmahssnu Whamamg mmaa wwbdwmzlampusaimuhem mum quelmamdewm Km and Lnbnmbuh am mam sp 4 25 and ms of the mm mm m Pacmuse Mmmepms 39 mm 2 292m 4 2mm ammmmmgmsmm Manny mm mm magma m K b7 4 WM 5 the mummy um my m pm sh p 5 50nd and mm m Km 5mm 59mm mummy use 5 ms pmblmhas the dlnwmg shape WM 9 m whammy a sued W m be mind by add me pmbzbdmg a 53m Sand whomn5 m sp Kismgm mmammmmm msbmls dmmw wde um mm a m pmmmw mm mania z mend mu lawmmgimegm WW whom bum mm WW whom bum um Ms Mamas 0mm we ows b ummmmmmmm m muonth 035 003 Pm b dm hoas DDZHMS DDZDZD DNZDET A m whammy Bums my wrung in ma mm m m momma mm n is mania as oawzows 3 A company has plants in Berlin Hamburg Munich and Cologne which resp produce 15 20 30 and 35 of the company s cathode ray tubes The probability that there is something wrong with the cathode ray tube produced by these plants is resp 5 4 3 and 2 A cathode ray tube turns out to be faulty What is the probability that it was made in resp Berlin Hamburg Munich and Cologne Solution First apply the theorem of total probability pa cathode ray tube is faulty pfaultleerlin pBerlin pfaultylHamburg pHamburg pfaultylMunich pMunich pfaultleologne pCologne 005 015 004 02 003 03 002 035 00315 Then apply Bayes7 theorem for each of the 4 possibilities pfaultleerlin pBerlin 7 005 015 a pBerlinlfaulty pfaulty 00315 0238095 b pHamburglfaulty flemagg gfmambmg 0832 0253968 C MM i E u pfaultyleirgclliEb Munich 0285714 d pltc E u pfaultyleiE ji aCOlOgne 039 H

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