SALES AND MARKETING Department MATHEMATICS. Combinatorics and probabilities. SOLUTIONS of tutorials and exercises

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1 SALS AND MARKTING Department MATHMATICS 2 nd Semester Combinatorics and probabilities SOLUTIONS of tutorials and exercises Online document : section DUT Maths S2 IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page 1 / 18

2 xercise 1. (Tutorial for lesson page 8) is the set of the inhabitants of a city ; Card() = A is the set of this city s men ; Card(A) = B is the set of this city s retired people ; Card(B) = women are retired. Create, then complete, a contingency table; tell how many men aren t retired; tell how many people are women or retired people. A A B = Card(B) B = Card(B) 1220 = Card(A) 1280 = Card(A ) 2500 = Card() question 1 :{men AND non-retired} = A B. Card( A B) = 950. question 2 :{women OR retired} = A B. - using the union formula : Card( A B) = Card(A ) + Card(B) - Card( A B) = = using a Morgan law : Card( A B) = Card( A B) = 2500 Card( A B) = = 1550 xercise 2. Given A and B, two subsets of, simplify the following expressions: ( A B) ( A B) = A ( B B) = A = A A ( A B) A ( A B) A ( A B) = ( A A) ( A B) = A ( A B) = A ( A B) ( A B) = ( A A ) B = B = B ( A B) ( A B) ( A B) = ( A B) A ( B A) ( B B) = ( A B) A ( B A) = A B A B A = A B B A A = A B B = = = B = ( A B) ( A B) A ( A B) B ( A B) ( A B) = ( A B) ( A B) = ( A A) B = B = B = A B ( A B) = ( A B) ( A B) = ( A B) = A B = A B B = A = A xercise. = {2, 5, 8, 11, 14, 17, 20}. Let A be the subset of the even numbers of and B the one of the multiples of 5. 1) Define the complement of A in. Give its elements. This is the set of every elements of that are not in A, that is to say: the set of the odd numbers of. A = {5, 11, 17}. 2) Give the sets A B and A B. What is their union? A B = 20 ; A B = 2; 8; 14 ; A B A B = A { } { } IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page 2 / 18

3 xercise 4. In a group of 25 students (including 17 women), 20 passed their exam (including 14 women). 1) Build a contingency table dispatching the information above. 2) How many men passed their exam? What is the name of the corresponding set? B passed B didn t pass A men men passed their exam A women This is the set A B xercise 5. Amongst 50 people interviewed for a survey, 244 own a computer with an Internet access, 287 own a smartphone, but inside this last group, 56 don t have a computer with an Internet access. 1) Organize and fill a contingency table using these data. B mob. phone B A comp+inter A ) How many people own a. an Internet access but no smartphone? 1 b. at least one of both? = 00 c. only one of both? = 69 xercise 6. After counting the answers from a survey conducted on a sample of 500 people, it appears that 154 of them go to the cinema at least once a month, 228 buy popcorn when they go to movies, and of those who go to the cinema less than once a month, 11 usually buy popcorn. 1) Let s name A the set of people who go to the cinema at least once a month and B the set of people who buy popcorn when they go to movies; build the corresponding contingency table. B popcorn B A cine >= A ) Answer by naming the adequate subset and justifying the result if not already written in the table: a. Among people who go to the cinema at least once a month, how many buy popcorn? Card(A B) = 97 b. How many people go to the cinema less than once a month? Card(A ) = 46 c. How many people are elements of A or B? Card(A B) = Card(A) + Card(B) - Card(A B) = = 285 IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page / 18

4 xercise 7. (Tutorial for lesson page 9) * by how many ways can we arrange two objects inside three drawers? Two choices (p = 2) have to be made among three drawers (n = ) with possible repetition; once a selection is done, the order is important. ach selection is then a p-list. Number of ways: ² = 9. * how many numbers composed with four figures only contain the figures 1, 2,? Four figures have to be selected (p = 4) among three figures (n = ) with possible repetition (if not, the experiment would be impossible!); once a selection is done, the order is important. ach number is then a p- list. Number of numbers: 4 = 81. * how many words can be written by taking five letters chosen in the set {a, b, e, m, i, r, o}? Five letters have to be selected (p = 5) among seven letters (n = 7) with possible repetition; once a selection is done, the order is important. ach word is then a p-list. Number of words: 7 5 = xercise 8. (Tutorial for lesson page 10) * how many pairs representative/assistant could have been elected from a group of 25 students? Two students must be selected (p = 2) among the 25 (n = 25) without repetition; once a selection is done, the 2 order is important. ach vote is then a permutation. Number of votes: P25 = = 600. * how many ways can blocks be piled, taking them among 10 blocks of different colours? Three blocks have to be selected (p = ) among ten (n = 10) without repetition; once a selection is done, the order is important. ach pile is then a permutation. Number of piles: P10 = = 720. * how many words can be written by taking five different letters chosen in {a, b, e, m, i, r, o}? Five letters have to be selected (p = 5) among seven letters (n = 7) without repetition; once a selection is done, the order is important. ach word is then a permutation. Number of words: P = = xercise 9. (Tutorial for lesson page 12) * How many couples of representatives could be elected from a group of 25 students? Two students must be selected (p = 2) among the 25 (n = 25) without repetition; once a selection is done, the order has no importance. ach vote is then a combination. Number of votes: C25 = = 00. 2! * How many different hands of 8 cards could be given from a deck of 2 playing cards? ight cards have to be selected (p = 8) among the 2 (n = 2) without repetition; once a selection is done, the 8 order has no importance. ach hand is then a combination. Number of hands: C2 = * How many draws of 6 different integers are possible, taking them between 1 and 49? Six cards have to be selected (p = 8) among the 2 (n = 2) without repetition; once a selection is done, the order has no importance. ach hand is then a combination. Number of hands: C = xercise 10. (Tutorial for lesson page 12) This time, the initial set is parted into categories and our purpose is to get a certain number of elements of each category. It is explained, in section 1.2.4, only for situations that lead to combinations, that combinations have to be calculated inside each subset (category) and then multiplied. 1) From a deck of 2 playing cards, how many 8-card hands own exactly spades and 2 hearts? initial set draw spades 8 C8 = 56 2 hearts 8 2 C8 = 28 others 16 C16 = 560 total Number of such hands: C C C = IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page 4 / 18

5 2) In a company, among 20 women and 20 men, 5 women and men have to be chosen at random. How many possibilities are there? initial set draw 5 women 20 5 C20 = men 20 C20 = 1140 total Number of such groups: C C = xercise dice A die is being rolled three times. How many possible outcomes? successive dice rolls : p-lists. 6 = 216 possible outcomes. 1.2 Three dice are being rolled at the same time. How many possible outcomes? simultaneous dice rolls : p-lists. 6 = 216 possible outcomes. 2. numbers and letters 2.1 How many phone numbers of eight figures can theoretically exist? n = 10; p = 8. repetition (of the same figure in a phone number): possible; order (of the figures to create a phone number): essential = phone numbers 2.2 How many ways can six different integers be chosen among [1; 49]? n = 49; p = 6. repetition (of the same integer): no, order (of the six integers once selected): not important. C = How many numbers are composed with three different figures (including 0)? n = 10 available figures; p =. repetition (of the same figure in a number): forbidden, order (of the chosen figures to form a number): essential. P = 720 numbers How many different lists of 4 letters can be created for the vehicles number plates? n = 26 available letters; p = 4. repetition (of the same letter in a list): allowed, order (of the four letters to define a plate): essential = How many anagrams of the word "MATHS" are there? n = 5 available letters, p = 5; repetition (of the same letter into an anagram): no, order (of the letters to create a word): essential. 5! = 120 anagrams 2.6 How many words can be created, taking 4 letters from the word "BRACKT"? n = 7 available letters; p = 4. repetition (of the same letter in a word): possible,order (of the letters to create a word): essential;. 7 4 = words 2.7 How many 10 notes-long melodies can be written, taking notes among A,B,C,D,,F,G? n = 7 available musical notes ; p = 10. repetition (of the same musical note in a melody): authorized, order (of the chosen notes to create a melody): essential; = melodies. arrangements.1 How many ways can 5 objects be arranged into 8 boxes? n = 8 boxes; p = 5 boxes to be selected. repetition (of the same box during the selection): not forbidden, order (of the selected boxes to arrange the different objects): yes. 8 5 = arrangements.2 Same question, but you can t place more than one object per box. n = 8 boxes; p = 5 boxes to be selected. repetition (of the same box during the selection): forbidden, order (of the selected boxes to arrange the different objects): yes. P = 6720 arrangements. Paul drives a team of 5 people. ach month, he evaluates the work of one of them, chosen at random. In a one year period, how many different lists of evaluated people could have been made? n = 5 people ; p = 12 evaluations. repetition (of a person during 12 months): compelling, order (of the 12 names forming a list): yes = lists IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page 5 /

6 4. playing cards 4.1 How many 8 cards hands from a deck of 2 playing cards? n = 2 available cards; p = 8. repetition (of the same card in a hand): impossible, order (of the 8 cards in a hand): no importance. C = hands How many 5 cards hands from a deck of 52 playing cards? n = 52 available cards; p = 5. repetition (of the same card in a hand): impossible, order (of the 5 cards in a hand): no importance. C = hands 5. classifications, elections How many possible different tiercés at the end of a 12 horses race? n = 12 horses; p =. repetition (of the same horse in a tiercé): impossible, order (of the three horses): essential. P = 120 tiercés How many possible different classifications at the end of a 12 horses race? n = 12 horses; p = 12. repetition (of the same horse in a classification): impossible, order (of the twelve horses): essential. 12! = classifications 5. How many ways a delegation of 5 people can be chosen from a group of 40? n = 40 people; p = 5. repetition (of the same person in the delegation): impossible, order (of the five selected people): unimportant. C = delegations pilots are fighting at a formula 1 race. At the end, the only six first will score different numbers of points. How many possible distributions of points are there? n = 16 pilots; p = 6. repetition (of the same pilot into the classification): impossible, order (of the six pilots, for the distribution of points): essential. P = distributions 5.5 How many possible podiums after a race in which 8 runners will compete? n = 8 runners; p =. repetition (of the same runner on the podium): impossible, order (of the three runners on the podium): essential. P = 6 podiums Fifteen people meet. veryone gives a handshake to every other, once. How many handshakes? n = 15 people ; p = 2 people to be selected for a handshake. repetition (of the same person for a handshake): of course not (nobody shakes her/his own hand), order (of both people to form a handshake): certainly not (the handshake A-B is the handshake B-A, it mustn t be counted twice). C = 105 handshakes 2 15 xercise 12. Let consider a group of 1 women and 8 men. 4 people have to be chosen. 1) How many possibilities are there? n = 21 people; p = 4. repetition: no, order: no C21 = ) How many of them contain exactly one man? 1 1 man among 8 and women among 1: C C = 2288 ) 2 men? men? 4 men? no man? 2 man among 8 and 2 women among 1: man among 8 and 1 women among 1: 4 man among 8 and 0 women among 1: 0 man among 8 and 4 women among 1: 8 1 C C = C C = 728 C C = 70 C C = 715 IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page 6 / 18

7 xercise 1. Among the hands of 5 cards taken from a deck of 2 playing cards, how many contain: a. the 4 aces? 4 aces among the 4 aces and 1 other card among the 28 others: C C = b. a square? square of aces, or square of kings, or : 8 possible different squares, giving 8 separate groups of outcomes: the numbers of hands must be added, and are the same for each group = 224. c. exactly spades? spades among 8 and 2 other cards among the 24 others: C C = d. exactly 2 spades and one club? 2 spades among 8 and 1 club among 8 and 2 other cards among the 16 others (heart or diamond): C C C = e. at least one king? The contrary is easier to work on: 0 king among 4 and 5 other cards among 28: C C = The whole number of hands (regardless the number of kings) is: C2 = "At least one king" is then realized in any other hand, whose number is: = f. at least two jacks? reversed: (0 jack among 8 and 5 others among 24) or (1 jack among 8 and 4 others among 24) C C C C C = So, for at least two jacks: 2 ( 4 28 ) ( 4 28 ) g. exactly diamonds and one king? There might be the king of diamonds or not! These are two separated events, whose number of hands have to be added. (1 king of diamond among 1 and 2 diamonds among 7 and 2 other cards among 21 "nor king neither diamond") or (1 king among (kings not diamond) and diamonds among 7 (diamonds not C C C + C C C = = king) and 1 other card among 21) ( ) ( 7 21 ) xercise 14. A jar contains balls: 2 white, green, 5 red. balls are to be taken simultaneously from the jar. Among the possible groups of balls, how many contain a. one single colour? ( red among 5) or ( green among ): C + C = = 11 5 b. the three colours? 1 red among 5 and 1 green among and 1 white among 2: C C C = c. two colours? this is the contrary event of the union of both former events "one colour" and "three colours", hence we have to subtract their numbers of groups from the whole number ( balls among 10): C 11 0 = 79 d. at least two colours? contrary event of "one colour", thus: C10 11 = 109 xercise 15. On loto game (former rules), a player has to select 6 different numbers from the set {1, 2,,, 49}. Then, the official draw is performed, revealing the 6 winner numbers. 1) Calculate the whole number of possible selections. 6 n = 49 available numbers ; p = 6. repetition: no, order: no. C49 = selections 10 IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page 7 / 18

8 2) Among them, how many would contain exactly a. the six winner numbers? 6 numbers selected among the 6 winners: only one possibility b. 5 winner numbers? 5 numbers selected among the 6 winners and 1 loser among the 4 losers: c. 4 winner numbers? 4 numbers selected among the 6 winners and 2 loser among the 4 losers: d. winner numbers? numbers selected among the 6 winners and loser among the 4 losers: C C = C C = C C = e. no winner number? (find two ways ) 1st way: 0 winner number among 6 and 6 losers among C6 C4 = different possible selections 2nd way: let s calculate the number of selections that own 2 winning numbers, the number of selections with only 1 winning number, then if we denote T n the number of selections having n winning numbers (e.g. T 5 = 258) and T the total number of possible selections (198816), one can write: T 0 = T 6 T n n= 1 xercise 16. 1) Independent questions up to you to perform the calculations and find out the result. a. In order to build your team, you have to choose 5 people out of a group of 10 basket players. How many different teams could be made? 5 n = 10 available players; p = 5 players to select. Repetition: no ; order: no. C10 = 252 b. A company has 18 employees and its manager decides to give three awards : best employee, most punctual employee, and less bald employee. How many ways can these awards be given? n = 18 employees; p = to select. Repetition: possible (an employee can be awarded more than once); order: yes (the awards are different and so are people): 18 = 582 c. On a chess board (8 8 tiles), how many ways can you put a king, a queen and a tower? n = 64 available tiles; p = tiles to be chosen. Repetition: no (one tile for one object); order : yes (the objects are different). P64 = It would be better to notice that if we rotate the board 90, then a permutation becomes another... which is the same! Thus, each "storage" of three objects was counted four times in the previous count 2) 20 chips have been put in a bag, numbered from 1 to 20. The chips from #1 to 10 are white; those from #11 to 16 are green; the last ones are red. You have to draw three chips, at random, simultaneously. a. How many different possible draws are there? A simultaneous draw is a combination. Here, n = 20 and p =. Number of combinations:. C20 = 1140 b. How many draws can be made with three white chips? white among 10 white (and 0 other among 10 others): c. How many draws would show three different colours? 1 white among 10 and 1 green among 6 and 1 red among 4: C10 = 120 C C C = d. How many draws would show three chips of the same colour? white among 10 OR green among 6 OR red among 4: = 144 combinations e. How many draws would show three even chips of the same colour? even white among 5 even white OR even green among even green: = 11 combinations IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page 8 / 18

9 xercise 17. (Tutorial for lesson page 16) A random experiment consists in taking one integer, at random, among {1 ; ; 20}. 1) Determine Card(Ω). Ω, set of all possible outcomes, is the set of the integers from 1 to 20 itself. Card(Ω) = 20 2) Let s name some events: A :"get at least 15" and B :"get an even number". Determine : p(a), p(a), p(b), p(a B), p(a B). p(a) = 6/20 = 0, ; p(a) = 1 p(a) = 14/20 = 0,7 ; p(b) = 10/20 = 0,5 A B = {16; 18 ; 20}, hence p(a B) = /20 = 0,15 p(a B) = p(a) +p(b) - p(a B) = 0, + 0,5 0,15 = 0,65 xercise 18. (Tutorial for lesson page 16) A random experiment consists in a simultaneous drawing of letters in our alphabet. 1) Determine Card(Ω) Ω, set of the outcomes, is composed with every combination of three letters. Card(Ω) = C26 = ) b. We set the following events A: "get consonants", B: " get 2 consonants", C: " get 1 consonant" and D: "get vowels" a. Are they mutually exclusive? They are indeed mutually exclusive. e.g. a draw showing 2 consonants (hence belonging to B) doesn t own three consonants (hence does not belong to A), and reciprocally. Same comments for the differences between A and C, A and D, B and C, B and D, C and D. b. Do they represent a partition of Ω? These four subsets do not only have their own elements without sharing a single one, but they also include all the elements of Ω. Hence, the set {A; B; C; D} is a partition of Ω. c. Calculate their cardinal numbers and then their probabilities (writing four significant figures). Finally, check the sum of their cardinal numbers and the sum of their probabilities. alphabet A B C D consonants vowels total 26 0 Card(A) = C C = 1040 p(a) = 1140 / 2600 = 4,85 % Card(B) = C C = 1040 p(b) = 1140 / 2600 = 4,85 % Card(A) = C C = 00 p(c) = 00 / 2600 = 11,54 % Card(A) = C C = 20 p(d) = 20 / 2600 = 0,7692 % xercise 19. (Tutorial for lesson page 16) Random experiment: roll two dice and add both results. 1) What are the different possible sums? {2 ; ; 4 ; 5 ; 6 ; 7 ; 8 ; 9 ; 10 ; 11 ; 12} 2) Are they equally likely? No. e.g.: there are four possibilities for our dice to reach a total of 5: (1, 4) ; (4, 1) ; (2, ) ; (, 2), but only one possibility to make a sum of 2: (1, 1). IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page 9 / 18

10 ) Build a sample space of equally likely outcomes. If each side of a die has the same chances of being obtained than the other, then each pair of sides of two dice has the same chances than any other pair. We can therefore build the following sample space with equally likely pairs: Ω = {(1, 1) ; (1, 2) ; (2, 1) ; (2, 2) ; (1, ) ; (, 1) ; (2, ) ; ; (5, 6) ; (6, 5) ; (6, 6)}, whose cardinal number is 6. 4) We set the following events: A: "the sum equals 10" and B: "the sum is at least 10". Determine p(a) ; p(a) ; p(b). A = {(4, 6) ; (5, 5) ; (6, 4)}. Card(A) =. p(a) = /6 = 1/12. p(a) = 1 p(a) = /6 = 11/12. B = {(4, 6) ; (5, 5) ; (6, 4); (5, 6); (6, 5);(6, 6)}. Card(B) = 6. p(b) = 6/6 = 1/6. xercise 20. (Tutorial for lesson page 17) = {1, 2,,, 10}, A: "even numbers of ", B :"multiples of in " Venn diagram: probabilistic choice tree: 1/ A 8 10 A 1/2 2;4;6;8;10 4/5 1 5 B 7 9 1/2 2/5 contingency table: A A Inside, cardinal numbers of the B 1 2 = Card(B) corresponding intersections must be B 4 7 = Card(B) placed, subtotals ("marginal 5 = Card(A) 5 = Card(A) 10 = Card() frequencies") and the overall total 1) Complete the contingency table. 2) Let s choose a number between 1 and 10, at random, not looking at it. a. What is the probability it would be even? p(a) = 5/10 b. What is the probability it would be a multiple of? p(b) = /10 ) Let s choose a number between 1 and 10, at random, not looking at it, but we re told it s a multiple of. a. What is the probability it would be even? 1/ b. What is the result obtained by the corresponding formula? p ( A) IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page 10 / 18 B /5 ( ) p A B 1 / 10 1 = = = p B / 10 4) Let s choose a number between 1 and 10, at random, not looking at it, but we re told it s even. a. What is the probability it would be a multiple of? 1/5 b. What is the result obtained by the corresponding formula? p ( B) A ( ) p A B 1 / 10 1 = = = p A 5 / 10 5 xercise 21. (Tutorial for lesson page 17) A laboratory has developed a breathalyzer. A reliability test has been done on this product, with a testpopulation on which it's been stated that 2% exceed 0.5 g/l (event ) and so are out of law. veryone exhales into the breathalyzer; the event P refers to a positive result given by this device. The reliability test has given the following results: - 95% of people who really exceed 0.5 g/l got a positive result by the breathalyzer; - 96% of people who don't exceed 0.5 g/l got a negative result by the breathalyzer. A 1;;5;7;9 6B B 2;4;8;10 B;9 B 1;5;7

11 What is then your probability of really exceeding 0.5 g/l, given that your result is positive? The information given above enable us to write: p() = 0.02 ; p (P) = 0.95 ; p ( P ) = The exercise asks us to determine p P (). p( P ) p( P ) pp = = = = = 265. % p P p P + p P xercise 22. (Tutorial for lesson page 17) 1) Taking back exercise 21: are and P independent? there are two ways to decide whether yes or not: * p( P ) = p( P) p? they aren t: * p ( P) p ( P) 2) Taking back exercise 17: are A and B independent? there are two ways to decide whether yes or not: * p( A B) = p( A) p( B)? they are: 0.15 = * p ( B) p ( B) A =? they aren t: A =? they are: /6 = 7/14 xercise 2. dice are rolled together. 1) How many different outcomes are possible? n = 6 available figures; p = obtained. The result of a throw is a p-list. Card(Ω) = 6 = 216 triplets. 2) Calculate the probabilities of the following events: a. A : "get a triple 6" 1 outcome of Ω corresponds: the list (6, 6, 6). Card(A) = 1. p(a) = 1/216. b. B : "get a triple" 6 outcomes of Ω correspond. Card(B) = 6. p(b) = 6/216 = 1/6. c. C : "get a 421" One have to get at the same time a 4, a 2 and a 1, whatever the order. There are! = 6 permutations (orders) of three elements, so: 6 different outcomes in Ω that show 4 2, 1. p(c) = 6/216 = 1/6. d. D : "get at least one 4" The contrary is easier to treat: getting no 4. This happens in case each of the three figures is taken among the figures 1,2,,5,6. The number of such p-lists is 5 = 125. Card(D) = = 91. p(d) = 91/216. e. : "get a sum of 10" The list of the groups of three digits giving a total of 10 must be established; then, we have to tell how many different orders are possible between the three figures, for each group. {1 ; ; 6} 6 orders; {1 ; 4 ; 5} 6 orders; {2 ; 2 ; 6} orders; {2 ; ; 5} 6 orders; {2 ; 4 ; 4} orders; { ; ; 4} orders. All in all, 27 triplets, out of 216, give a sum of 10. p() = 27/216 = 1/8. ) a. Are B and D mutually exclusive? No, they can be realized at the same time with the triplet (4, 4, 4). b. Are C and mutually exclusive? Yes, a 4, a 2 and a 1 don t give a sum of 10. xercise balls lay in a jar: 7 white, 9 red, 2 green. Three balls are simultaneously taken out. 1) What is the probability to see the three colours in your hand? n = 18 available balls; p =. Repetition: no (simultaneous); order: no (simultaneous). Setting A: "One ball for each colour" Card A = C C C = 126. p(a) = 126/ Card Ω = C = IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page 11 / 18

12 2) What is the probability to see only one colour in your hand? The corresponding event, B, is the union of two mutually exclusive events: " white" OR " red", so, we Card B = C + C = 119. p B = 119/ must add their cardinal numbers and their probabilities. ) a. Which third event makes a partition of Ω with both former ones? That s the event C: "two different colours among the three balls", that contain every other outcomes but the ones of A or B. b. Deduce its probability. p(c) = 571/816 4) Calculate the probability to get no white ball or no green ball. We must calculate the cardinal number of two events, not necessarily mutually exclusive (so we can t only add both: we have to use the general formula). Setting D: "0 white" and : "0 green". C + C C = 641. p D = 641/ Card(D ) = Card(D) + Card() Card(D ) = xercise 25. A bag contains 20 coins: n are black and the 20-n others are white. 2 coins have to be picked up together. 1) xpress (with n) the probabilities of the following events: a. A: One black and one white. Card Ω = C = available coins; p = 2 coins to be taken. Order: no, repetition: no. 2 To get an element of A, we need: 1 black from n black and 1 white from 20-n white. 1 1 n( 20 n) n Card( A) = Cn C20 n = n( 20 n). p( A) = 190 b. B: two black To get an element of B, we need 2 black from n black. c. C: two white lement of C: 2 white from 20-n white. ( ) 20 ( ) n n 1 n n Card B = C 2. p( B) 1 n = = n 19 n 20 n 19 n Card C = C20 n =. p( C) = ) Check that the global probability equals 1. The events A, B and C combined are a partition of Ω. Thus, the sum of their probabilities must be 1: n( 20 n) + n( n 1) + ( 20 n)( 19 n) 2n + 40n + n n n 9n 80 p(a) + p(b) + p(c) = = = = ) Determine, by solving an equation, the values of n such that p(c) > n and 19-n are two consecutive integers. Let s test several values: = 12 ; 1 12 = 156 ; 14 1 = 182 ; = n 19 n is more than 190 (that s As long as 20-n is more than or equal to 15 (and then 19-n 14), what we need to get p(c) > 0.5). Therefore, n must be less than 6: n 5. xercise 26. A bank found that 2% of the checks issued by its clients aren't correctly worded (correctly worded : event W). 97% of correctly worded checks are correctly entered by the agent into the bank's data base (event ). When it's not correctly worded, the agent is able to correct the mistakes 5 times out of 100. A check has to be entered into data base. calculate the probabilities of the following events : 1) The agent doesn t enter the check correctly. Here, we're crossing two events and their contraries. We can represent it by a probabilistic choice tree. Given W :"the check is correctly worded" and :"the agent enters it correctly". p = p W + p W = p W p + p W p = = 484 W..... % W IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page 12 / 18

13 0.05 W W ) The check has been correctly worded, given that the agent entered it correctly p( W) p( W) pw p ( W) = = = = = % p 1 - p ) The check has not been correctly worded, given that the agent didn t enter it correctly W p W p W p p ( W) = = = = = 926. % p p One can notice that a little bit less than 60% of the checks for which the agent made a mistake were actually correctly worded by the client. xercise % of people in a city are employees. Among them, 8 people on 10 use their car every day, whereas 0 % of the unemployed do (employee: event ; use car every day: event D). We are to select one person in this city, at random. 1) Display the different categories of people in a probabilistic choice tree D D D D 2) a. What is the probability that this random individual be unemployed? p = 1 p = 0.65 b. What is the probability that this person be an employee who uses his/her car every day? p D = p p D = = 28.. % c. What is the probability that he/she uses his/her car every day? p D = p D + p D = p p D = = 475. % d. Given that this person drives his/her car every day, probability he/she is an employee? p( D) 028. pd = = = % p D e. Are the events and D independent? two ways to answer it: * p( D) = p p( D)? no: * p ( D) p ( D) =? no: IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page 1 / 18

14 xercise 28. (Tutorial for lesson page 16) A lottery is held. 100 tickets are to be sold, 1 each. One ticket is a 0 winner, two are 15 winners, and seven would make the buyer win 1. Considering we want to purchase one ticket, X is the random variable of the net gain (the 1 expense taken into account). 1) Give the probability distribution of X. x i p(x = x i ) or p i ) If we re playing this lottery the same way many times, can we expect to be a long-term winner? (begin by an estimate of what would be likely to occur after a thousand attempts) On 1000 attempts, our expected gain is negative: = -0. This result means that we can expect an average loss of cents per attempt. ven though this accurate result is not likely to occur, the law of large numbers tells us that the more we ll play this lottery, the closer our actual average gain will be to this average expected gain. One cannot expect to be a long-term winner. xercise 29. (Tutorial for lesson page 16) 1) Calculate the expected value and the standard deviation, with the data of exercise 28. Comment. Using the calculator, on stat mode: let s enter the gains on list 1 and their probabilities on list 2. Casio: CALC ST: 1Var X: List 1 and 1Var F: List 2, then XIT, 1VAR. TI: CALC: Stat1Var L 1,L 2. Results: (X) = -0. and σ(x) =.622. This means that in n attempts, with n big enough, the actual average gain is a random variable, normally distributed, centred on -0., with a standard deviation of e.g. in attempts, the standard n deviation of the possible average gain (around -0.) is approximately The fact that its distribution is normal implies for instance that: * there are 68. % chances that the actual average gain would be between and , so between and , and then that we would lose between 2940 and 660; * there are 95.4 % chances that the actual average gain would be between and , so between and , and then that we would lose between 2580 and Globally, on this number of attempts, the standard deviation of the average gain is quite low and makes it virtually impossible to make money. The bigger n is, the lower the standard deviation is, the less likely it is that our loss approaches zero and the greater the chances that our average gain is close to -0.. The calculation of this standard deviation is a numerical translation of the law of large numbers. 2) If the possible gains and loss were doubled (in the initial array), what would these parameters become? Logically, they would be doubled. You may check this thanks to your calculator. ) If the values of X were increased by 0.5, what would these parameters become? The expected value would too, and then would become non negative: However, the standard deviation wouldn t be affected: increase a list of values by the same number only shifts them, but does not change the distances between them. Just check! IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page 14 / 18

15 xercise 0. (Tutorial for lesson page 17) From a jar that contains 7 white balls and black balls, let s draw two balls, one after the other and without putting back the first one. We name X 1 the random variable corresponding to 1 point in case the first ball is black and 0 point in case it s white ; we name X 2 the random variable corresponding to 1 point in case the first ball is black and 0 point in case it s white. a. Give p(x 1 = 0) and p(x 1 = 1). p(x 1 = 0) = 7/10 and p(x 1 = 1) = /10 b. Give p X1 = 0 (X 2 = 0) and p X1 = 0 (X 2 = 1), then p X1 = 1 (X 2 = 0) and p X1 = 1 (X 2 = 1). p X1 = 0 (X 2 = 0) = 6/9 and p X1 = 0 (X 2 = 1) = /9, p X1 = 1 (X 2 = 0) = 7/9 and p X1 = 1 (X 2 = 1) = 2/9 c. Complete the probabilistic choice tree and then the associated probability table. X X 2 P X2 0 42/90 21/90 7/10 7/10 X 1 = 0 6/9 /9 1 21/90 6/90 /10 P X1 7/10 /10 1 /10 X 1 = 1 7/9 2/9 d. Is the knowledge of both marginal distributions sufficient for the knowledge of the joint distribution? Knowing the subtotals of this table doesn t allow us to attribute values to the four empty cells that represent the intersections. e. Compare p(x 1 = 0) p(x 2 = 0) to p((x 1 = 0) (X 2 = 0)). Are the variables X 1 and X 2 independent? 7/10 7/10 42/90. These variables are not independent. xercise 1. A game consists of two identical boxes each having 10 chips numbered 1 through 10. The experiment is to pick a chip in each box. 1) a. Describe one of the possible outcomes. An outcome is an ordered list of two chips. e.g.: (2, 5) ; (10, 8) ; (5, 2), etc. b. xplain why the sample space s cardinal number is 100. These lists consist of two components (p = 2) chosen from 10 (n = 10), with a possible repetition and order to take into account: (5, 2) and (2, 5) are two different results. The sample space is thus a set of p- lists, and Card(Ω) = 10² = 100. c. What is the probability of choosing two even numbers? These p-lists consist of two elements (p = 2) taken from 5 even elements (n = 5). Number of lists: 5² = 25. probability: 25/100 = d. Prove that the probability of two different even numbers is 0.2. These p-lists consist of two elements (p = 2) chosen from 10 (n = 10), without repetition of the same 2 element. Thus, we re here looking for a number of permutations: P5 = 20. One can give another explanation, noticing that on the 25 lists of even numbers, 5 are a double, and then 20 aren t Finally, the requested probability is 20/100 = ) For one game, you have to spend 1. If you get two different even numbers, you win 1; if you get two identical numbers except 1 and 1, you earn 6; 1 and if you get the double one, you win 50; in all other cases, no gain. The random variable X gives the gain at the end of the game, regardless of the initial 1 bet. a. Give the probability distribution of X. x i p(x = x i ) or p i IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page 15 / 18

16 b. Give the expected value of X. (X) = = 1.24 c. Can we expect to win money on playing this game a lot? This expectation is bigger than the bet, with a 0.24 deviation in favour of the player. One can then expect an actual average gain close to 0.24 per game, playing this game a lot (e.g.: 240 in 1000 games). xercise 2. A bag contains 5 white and 10 black balls. You bet 2 for a balls draw together. Get white makes you earn 100; 2 white: 10; 1 white: 2; black: nothing. The random variable X is your gain at the end of a test, once deduced the bet. 1) Give the probability distribution of X. Total number of different possible draws: Card Ω = C = 455 Number of draws that show white: Card A = C C = 10 Number of draws that show 2 white: 2 1 Card B = C C = Number of draws that show 1 white: 1 2 Card C = C C = Number of draws that show 0 white: 0 Probability distribution of X, net gain: Card D = C C = x i p(x = x i ) or p i 10/ / / /455 2) Give the expected value and the standard deviation of X. (X) = 1540/455 =.85. σ(x) = ) If you play a hundred times, what gain is the most likely? 8 xercise. A game consists of a random draw of a letter from the alphabet (which contains 20 consonants and 6 vowels A,, I, O, U, Y). ach letter is assigned a number according to the following table: A B C D F G H I J K L M N O P Q R S T U V W X Y Z We'll set the events C: "the letter is a consonant" and M: "its number is at least 17". Part 1 1) Build a probabilistic choice tree (1st level: C and its contrary; 2nd level: M and its contrary) into which the simple, conditional and intersection probabilities will be placed. 8/20 M 8/26 C 20/26 12/20 M 12/26 6/26 2/6 M 2/26 C 4/6 M 4/26 2) Given that a vowel has been drawn, what s the probability its number is more than 16? p 2 1 C ( M ) = = 6 IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page 16 / 18

17 ) Given that its number is more than 16, what s the probability it s a vowel? p( C M) pm ( C 26 ) = = = = p( M) ) Are the events M and C independent? p( C M) = and p( C) p( M) = These events are not completely independent (slightly dependent, or related). Part 2 The event C M, is awarded a 10 gain and the event C M would make you lose 5; as for the other possibilities: they don t lead to either gain or loss. X is the random variable gain after one draw. 1) Give the probability distribution of X. x i p(x = x i ) or p i 2/26 16/26 8/26 2) Give the expected value of X and its meaning. (X) = -20/26 = This is the average gain expected on a long series of attempts. ) Give the standard deviation of X and its meaning. σ(x).846: average gain fluctuation around (X) at each attempt. xercise 4. Two sales agents A and B of a cooperative work in team for two weeks to obtain orders from potential customers. A is responsible for placing new contracts to existing customers and B is responsible for prospecting new customers. Let s name: X A the random variable measuring the number of contracts obtained by A and X B the random variable measuring the number of contracts obtained by B. It s assumed that X A can only take its values in {0 ; 1 ; 2 ; } and X B in {0 ; 1}. The joint distribution of X A and X B is given through the following table: X B X A ) a. Determine the margin distributions of X A and of X B. By adding the matching crossed probabilities, these distributions are visible: margin distribution of X A : k p(x A = k) margin distribution of X B : k 0 1 p(x B = k) b. Are these variables independent? Let s consider one of the possible crossings: X A = 0 and X B = 0. p((x A = 0) and (X B = 0)) = 0.05 (as seen in the first table) p(x A = 0) p(x B = 0) = = (both intermediate values coming from the 2nd and rd tables) These results are not equal, thus X A and X B are not independent. IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page 17 / 18

18 2) Let s set a new variable, X total number of obtained contracts, by X = X A + X B. a. Give the probability distribution of X. We have to consider every possible crossing between one value of X A and one value of X B, and then give the corresponding value of X: X A X B prob X A + X B Finally, the probability distribution of X is given by consolidating identical values: k p(x = k) b. Calculate (X) and V(X). (X) = = 2 V(X) = (X²) (X)² = 0² ² ² ² ² ² = = 0.9 IUT de Saint-tienne Département TC J.F.Ferraris Math S2 CombProb TxCorr Rev2018 page 18 / 18