Question.3210 - General Expectations: (i). For full credit, your written assignments must be accompanied by a narrative explanation/rationale for the process that you used to solve each problem. How did you choose the steps? What is the logic behind the choices that you made? Explain why the problems were solved the way they were solved. Use complete sentences, good English, and proper mathematical notation. (ii). For full credit, your written assignments must include the statement of each problem so the reader knows what you are trying to demonstrate. In cases where the assignment refers you to a book problem, you must also copy the statement of the appropriate problem in the book. Section 6.1: 1. Among 160 math students 45 have taken Discrete Math, 30 have taken History of Math. There are 12 have taken both. (a.) How many students have taken either Discrete Math or History of Math? (b.) How many students have taken only Discrete Math? (c.) How many students have taken only one of Discrete Math or History of Math? (d.) How many students have taken neither course? Section 6.2: 1. A pizza parlor offers whole–wheat, white and gluten–free crust options. The whole–wheat crust comes in three sizes, the white crust comes in four sizes and the gluten–free crust comes in two sizes. (a.) How many crust options does the parlor offer? (b.) Any pizza can be topped with exactly one of the following cheeses: full–fat mozzerella, reduced–fat mozzerella or soy cheese. How many choices does a customer have for a plain cheese pizza? (c.) Suppose that the pizza parlor offers 5 meat toppings and 7 vegetable toppings. How many ways can the customer choose a pizza with 3 distinct toppings? (Remember to factor in the choice of crust and cheese). (d.) How many ways can a customer choose a vegetarian pizza with 3 distinct toppings? (Remember to factor in the choice of crust and cheese). 2. How many numbers in the range 1000–9999 (a.) do not contain the same digit 4 times? (b.) end with an odd digit? (c.) end with an odd digit and have no repeated digits? (d.) have exactly three digits that are 7s? 3. Suppose that a password for a computer system must have exactly 8 characters. (a.) How many passwords are possible that only use lower–case alphabetic characters? (b.) How many passwords are possible that can use either upper– and lower–case characters? (c.) How many passwords are possible that use a combination of upper– and lower–case characters as well as exactly one numeric digit in the last position? (d.) How many passwords are possible that use a combination of upper– and lower–case characters as well as exactly one numeric digit if the position of the numeric digit is variable? 1 Section 6.3: Practice: Do #1 of Section 6.3 in the textbook and check your answer in the back of the book. 1. In a classroom of 30 students, show that there are at least two students with the same last initial. 2. One thousand students are transported on 15 busses. Show that there is a bus with at least 67 students. Each bus can hold a maximum of 80 students. Section 7.1: 1. Seven runners are in a race in which first, second and third place will be awarded. Assuming that there are no ties, how many different outcomes are possible? 2. Forty photographs are entered in a photo competition. How many ways are there to award the top eight places? 3. Three math students and two computer science students come to class late. (a.) There are five students in total. How many ways can the five students walk in through the door late (i.e. 1st, 2nd, 3rd, etc)? (b.) How many ways can the students come in, if a computer science student must be first? Section 7.2: 1. How many subsets with exactly 3 elements does a set with 10 elements have? 2. How many subsets with an odd number of elements does a set with 10 elements have? 3. How many subsets with at most 3 elements does a set with 100 elements have? 4. Do #9 of Section 7.2 in the textbook. Section 7.3: 1. Suppose a die with six sides (1,2,3,4,5,6) is tossed. (a.) What is the probability that the number that appears is a 3? (b.) What is the probability that the number that appears is divisible by 2? (c.) If the die is tossed twice, find the probability that the sum of the numbers appearing is at least 9. 2. Suppose a fair coin is tossed 4 times. (a.) Write out all possible outcomes (i.e. HHTT, etc). (b.) What is the probability of obtaining four heads? (c.) What is the probability of obtaining two heads and two tails? (d.) What is the probability of no heads? (e.) What is the probability of obtaining at least 1 head? 3. A bag contains 3 blue marbles, 2 green marbles, 1 white marble and 6 red marbles. (a.) Suppose a marble is selected from the bag and replaced. What is the probability of receiving a blue marble on the first selection and a blue marble on the second selection? (b.) Suppose a marble is selected from the bag and replaced. What is the probability of receiving a blue marble on the first selection and not a blue marble on the second selection? (c.) Suppose a marble is selected from the bag and replaced. What is the probability of receiving at least one blue marble and a white marble? (d.) Suppose a marble is selected from the bag and not replaced. What is the probability of receiving a blue marble on the first selection and a blue marble on the second selection? 2
Answer Below:
6.1) Total students = 160 Students that have taken maths P(m)= 45 Students that have taken history of maths P(h) = 30 Students that have taken both P(m∩h)= 12 Students that have taken only maths = 45-12 = 33 Students that have taken only history of maths = 30-12 = 18 a) Students that have taken either math or history P(mUh)= P(m)+P(h)-P(m∩h) =45+30-12 =63 b) Students that have taken only maths = 45-12 = 33 c) Students that have only one of discreet math or history of math = P(only discrete math) + P( only history of math) = 33+18 =51 d) Students with neither of course = 160 – P(mUh) =160-63 =97 6.2) 1.a) Types of crust option = Whole wheat, white and gluten free i.e. 3 n number no. of different sizes for whole wheat = 3 no. of different sizes for white crust = 4 no. of different sizes for glunten crust = 2 therefore, no. of different crust option taking size into consideration= 3*4*2= 24 b) Available size for whole wheat = 3 Available size for white crust= 4 Available size for gluten free crust = 2 Now, different types of cheese toppings = 3 No. of ways of selecting a crust = 24 (from a) Now for each of 24 choices , we have 3 types of pizza toppings. Therefore no. of different choice customer has =24*3 = 72 c) meat topping = 5 veg. topping= 7 Total no. of topping = 5+7 = 12 Now. Choosing 3 distinct topping from 12 can be done in C(12,3)= 12*11*10/3*2*1= 35 ways. C(m.n) – no. of ways of choosing ‘n’ distinct item from ‘m’ different items d)No. of veg. topping= 7 now, no. of ways of choosing 3 distinct veg. toppings from 7 is C(7,3)= 7*6*5/3*2*1 = 35ways. 2.a) total numbers in the range 1000 to 9999 = 9000(including both) Numbers containing all the four digits same = 1111,2222,3333,4444,5555,6666,7777,8888,9999 = 9 So, no. of digits not containing all digits similar = 9000- 9 = 8991 b) Every possible combination of three leading digits in this range can end with 0, 2, 4, 6, or 8 (five possible even final digits) or 1, 3, 5, 7, or 9 (five possible odd final digits), so there are just as many ending with odd digits as with even ones. Thus, exactly half the numbers in the range, 9000/2 = 4500, end with an odd digit. C) There are 5 possible odd final digits. For each of those, there are 8 possible first digits (the other four odd numbers, plus 2, 4, 6, and 8). For each possible combination of last and first digit, there are 8 possible choices for each of the middle digits. But we can't choose the same digit for both, so there are only 8 * 7 = 56 choices for the pair of middle digits. Thus, there are 5 * 8 * 56 = 2240numbers in this range that end with an odd digit and have no repeated digits. d) There are 4 possible digit positions for the non-7 digit. If the first digit isn't a 7, there are only 8 other possibilities for it (because it can't be zero or 7). If the non-7 digit isn't first (3 possibilities), there are 9 other possibilities for it. So there are 8 + 3*9 = 35 numbers in this range with exactly 3 digits that are 7s. 3.a) no. of letters in English alphabet’s = 26 Size of password= 8 Each letter in the password can be selected in =26 ways Therefore, no. of different possible passwords = 26^8 = 208,827,064,576 b)taking both the cases into consideration, sample space becomes 52 i.e 26 lower case and 26 upper case alphabets so, no. of ways of selecting each letter in password = 52 since password has 8 letters, total possible passwords = 52^8= 53,459,728,531,456 c) t0tal characters in password = 8 first 7 cahracters are alphabetic. No. of ways of choosing them = 52^7 Now, last character is digit. NO. of digits = 10 Therefore possible passwords that use a combination of upper- and lower-case characters as well as exactly one numeric digit in the last position = 52^7 * 10 = 10,280,717,025,280 d)numeric digit can occupy 1 of 8 spaces i.e. in 8 ways so, Possible passwords that use a combination of upper- and lower-case characters as well as exactly one numeric digit if the position of the numeric digit is variable= 52^7 * 10 * 8 = 82,245,736,202,240 6.3 a) The logic is very simple, since we have only 26 letters in English alphabets and there are 30 students so atleast 2 students share same last initials. b) according to the question, if the bus transports 66 students . Max. number of student that can be carried in 15 bus with a constraint of 66 students = 66*15 = 990 So, there is/are buses with 67 or more students 7.1 1) To arrange k distinct objects from n distinct objects, the formula is given by nPk (0 <= k <= n) where nPk = n! / (n-k)! and where n! = n(n-1)(n-2).....x3x2x1 So for arranging 3 different objects from list of 7 is given by, 7P3 = 7! / (7-3)! = 7!/4! which reduces to 7x6x5 = 210 ways 2) This question is similar to 7.1 and here also we use the permutation formula for permuting 8 things out of 40, the answer is given by:- 40P8 = 40!/32! = 3.10 x 10^12 (to 3 significant figures) 3.a) Three math student and two computer students. No condition is given in the question, so each of the boy can occupy any of 5 positions . One boy can occupy any of 5 position, So next boy can occupy the remaining 4 position. Similarly the third boy can occupy the remaining 3 position and so on. Therefore, 3 math and 2 comp. student can enter the class in 5*4*3*2*1= 120 ways b) for 1 st position we need a computer student as per the question, hence 1 st position can be occupied by choosing 1 out of 2 computer students that turned late. This can be done in C(2,1) = 2 ways Now, the remaining 4 position can be occupied in 4! Ways = 4*3*2*1= 24 ways. No. of ways in which student can enter if computer student has to 1 st = 2*24 = 48 ways. 7.2 1) Choosing 3 elements from a set of 10 can be done in C(10,3)= 10*9*8/ 3*2*1= 120 ways. 2) subset with Odd no. elements out if set of 10 =subsets with 1,3,5,7,9 elements . Now no. of ways of choosing :- 1 element = C(10,1) = 10 3 element = C(10,3) = 120 5 elements= C(10,5)= 252 7 elements = C(10,7)=120 9 elements= C(10,9)=10 waays Now, total ways of subset with odd terms = 10+120+252+120+10 = 512 ways C) At most 3 element means that that set can contain 1,2,3 elements. For a subset with 1 element, possible no. of set = 100 For a subset with 2 elements, possible no. of elements = 100*99 *2 i.e. 1 st element can be chosen in 100 ways , 2 nd element can be chosen in 99 ways and they can be arranged in 2! Ways For A subset with 3 elements, no. of possible subsets = 100*99*98*3! So, total ways of forming subset with atmost 3 element = 100 + 19800 + 5821200 =5841100 7) Sample space for unbiased dice = {1,2,3,4,5,6} = 6 outcomes a) probality of an event is given by possible outcome divided by the sample space P(x) = Probable outcome / sample space Probablity that the no. that appears when a dice is rolled = 1/6 b) 2,4,6 are no. i.e divisible by 2 i.e. 3 cases out of 6 SO, the probality of a no. that appears o dice and is divisible by 2 = 3/6 = ½ c) Possible ways that the sum of no. is greater than or equal to 9 if dice is tossed twice, cases for sum equal 9= {(3,6); (4,5); (5,4) ; (6,3)} = 4 ways sum equal 10 = { (4,6);(5,5);(6,4)} = 3 ways sum equal 11= {(5,6);(6,5)} = 2 ways sum equal 12= {(6,6)} = 1 way Total possible case for a throw of 2 dice and sum atleast 9 = 4+3+2+1 = 10 ways . Sample space for the throw of two dices = 6*6 = 36 as both the dice can give 6 possible outcomes respectively Therefore probability for 2 dices giving sum of atleast 9 = 10/36 = 5/18 d) #9 total newcomers in club = 30 No. of members in committee= 5 Now no. of ways of choosing 5 members from 30 = C(30,5) = 142506 2.a) For a fair coin, possible outcome = head or tail Probability of getting a head= Probablity of getting a tail = ½ All possible outcomes for tossing the coin 4 time are HHHH HHHT,HHTH,HTHH,THHH HHTT,HTHT,,HTTH,THTH,THHT,TTHH HTTT,THTT,TTHT,TTTH TTTT So we have total of 16 possible outcomes b) Probability of obtaining all four head = 1/16 (from ques. a) c) probability of obtaining two head and two tail = 6/16 = 3/8 ( from 3 rd case of ques a) d)Probability of no head= Probability of all tail = 1/16 (from case 5 ques. A) e) possibility of getting at least 1 head = Possibility of getting 1,2,3,4 head Possible ways of getting: 1 head = 4 ways 2 head= 6 ways 3 head= 4 ways 4 head= 1 way So, probability of getting at least 1 head = (4+6+4+1) / 16 = 15/16 3) a)the bag contains 3 blue , 2 green,1 white and 6 red marbles. Total marbles= 12 marbles. No. of ways of getting 1 blue marble = 3 Therefore probability 0f getting blue marble in 1 st attempt= 3/12= ¼ Now, after a draw, if replacement is done, sample space remains the same so, problity of drawing blue ball in second draw is also = 3/12 = ¼ So combined probability of blue marble in 1 st and 2 nd attempt= ¼* ¼ =1/16 b) Total marbles= 12 marbles. No. of ways of getting 1 blue marble = 3 Therefore probability 0f getting blue marble in 1 st attempt= 3/12= ¼ Replacement is done, so the sample space remains the same. No. of ways of selecting a non. Blue marble = 12-3 = 9 So, probability of selecting a non blue marble is = 9/12 = ¾ Now using combined probability for a blue ball in 1 st attempt and a non blue ball in second attempt is given by : ¼* ¾ = 3/16 c) question not clear as no. of selections not mentioned d) Total marbles= 12 marbles. No. of ways of getting 1 blue marble = 3 Therefore probability 0f getting blue marble in 1 st attempt= 3/12= ¼ Now, after a draw, if replacement is not done, So, no. of blue balls become 2 Sample space becomes 11 Now, the probability of drawing 1 blue ball in second draw = 2/11 So combined probability of blue marble in 1 st and 2 nd attempt when replacement is not done = ¼*2/11 =2/44 #9 total newcomers in club = 30 No. of members in committee= 5 Now no. of ways of choosing 5 members from 30 = C(30,5) = 142506 Mod6problem1 In a week no. of days =7 So, in a group of 8 members, max. 7 people can have different days of birth. So the 8 th person neccesarily shares its birthday with some other person.More Articles From Maths
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