MA 114 Introduction to Finite Mathematics Midterm Exam 1 Solution June 2, 2010 Instructions: Show all work and justify all your steps relevant to the solution of each problem. No texts, notes, or other aids are permitted. Calculators are not permitted. Please turn off your cellphones. You have 90 minutes to complete this exam. There are seven problems which carry a total 100 points. Good luck! (10 pts) Problem 1. Let U = {1, 2, 3, 4, 5, 6}, A = {1, 2, 3, 4}, B = {3, 4, 5}, and C = {3, 6}. Compute A (B C) c (*) B C = {3, 4, 5, 6} (B C) c = {1, 2} A (B C) c = {1, 2} (10 pts) Problem 2. In a sack I have three Oreos and two Keebler elves. I will draw a cookie one at a time until I have drawn two Oreos. Write the tree diagram illustrating all possible outcomes, and tell me the number of outcomes which result in at least one Keebler elf being drawn. (*) 5 of the 6 possible paths involve at least one Keebler elf. 1
(15 pts) Problem 3. Watsonville has 1000 households. 400 have children in elementary school 300 have children in middle school 350 have children in high school 150 have children in both elementary and middle school 150 have children in both middle school and high school 100 have children in both elementary school and high school 50 have children in all three schools a. Create a Venn Diagram of the above information (*) The diagram is below. b. How many households have children in at least two schools? (*) 300 households have children in at least two schools. (Add the numbers in the center four partitions where either two or all three circles overlap) c. How many households have children in middle school, but not high school? (*) 150 households have children in middle school, but not high school. (Add the numbers in the M partition and subtract any also in the H partition) (10 pts) Problem 4. A coffee shop offers three choices of coffee, three choices of bagels, and two choices of cream cheese. How many coffee-bagel-cheese combinations can be purchased? (*) This is an exercise using the multiplication principle. I am making three decisions. The first, coffee, I have three choices. The second, bagels, I also have three choices. For the third, cream cheese, I have two choices. 3 3 2 = 18 ways I can make the three decisions. 2
(16 pts) Problem 5. Mara is enrolling in the local university. She has found five English courses, and four philosophy courses that she likes. a. If Mara can enroll in four courses, how many possible schedules can she have? (*) Mara must choose from 9 courses 4 of them. She can have C(9, 4) schedules. Order does not matter because she can only select which courses to take, not the time that the courses meet. b. If at least one course must be an English course and one must be a philosophy course, how many possible schedules can Mara have? (*) There are three ways Mara can select courses in this manner. She can have one English course and three philosophy, two English courses and two philosophy, or one English course and three philosophy. If Mara enrolls in one English course and three philosophy courses, there are 5 ways to choose the English course, and C(4, 3) ways to choose the philosophy course. If Mara enrolls in two English courses and two philosophy, there are C(5, 2) ways to choose two English courses, and C(4, 2) ways to choose two philosophy courses. If Mara enrolls in three English courses and one philosophy, there are C(5, 3) ways to choose the three English courses, and 4 ways to choose the one philosophy course. Add the three values to find the total number of options Mara has: 5 C(4, 3) + C(5, 2) C(4, 3) + C(5, 3) 4 c. If Mara selects courses at random, what is the probability two are English courses and two are philosophy courses? (*) Let E be the event Mara enrolls in two English courses and two philosophy courses. As shown above, there are C(5, 2) C(4, 2) ways Mara can select two English courses and two philosophy courses. Then n(e) = C(5, 2) C(4, 2). From part (a) there are C(9, 4) ways in total she can select courses. This is n(s). Then P(E) = C(5,2) C(4,2) C(9,4) 3
d. Mara wants a new name. How many ways can she make up a new name for herself by re-arranging the same four letters? (*) Like with the APPLE problem and the WOODCHUCK problems from class, the decisions being made here are how to place the letters. There three distinct letters, and four positions to fill. We can work through the letters in any orders, but the process is the same. With all blanks to fill, there are 4 ways to place the M. Now three blanks remain. We need to place an A in two of them. There are C(3, 2) ways to select two of the three blanks. There is only one blank remaining now. The R must go there. Hence, there are 12 ways to re-arrange the words in MARA. This answer is acceptable for full credit, but it is not the correct answer! The careful reader could earn a bonus by noting that the question was in how many ways can a new name be formed. The original combination ( MARA ) must not be counted. Hence, there are 11 ways to make a new name. (24 pts) Problem 6. Marcus is one of nine men attending a council meeting. In attendance are also four women. a. The thirteen people in attendance are to sit at a round table. In how many ways can they be seated? (*) We did a similar example in class. The issue here is that the people are being sat relative to the first person placed. Once we place the first person, there are 12! ways to fill in the remaining seats. b. The thirteen people are to sit in a single row. In how many ways can they be seated? (*) There are 13! ways to fill the blanks. c. The thirteen people are to sit in a single row. In how many ways can they be seated if the men and women are seated together? (*) There are 9! ways to seat nine men in a group, and 4! ways to seat four women in a group. There are also two ways to place the groups (either women to the left or men to the left). Hence, the total number of options is 2 9! 4! d. One member is to be selected at random to speak. What is the probability it is a man? (*) Formally, let E be the event a man is selected. n(e) is 9. The sample space, S, contains 13 elements. Then the probability is 9 13. 4
e. What is the probability Marcus will speak given that a man is selected? (*) Let A be the event a man is chosen, and B the event Marcus is chosen. We want to determine P(B A). If Marcus is selected then a man is selected. So P(A B) = P(B) = 1 13. From before, P(A) = 9 13. Then P(B A) = P(A B) P(A) = 1 13 9 13 = 1 9. f. Three members will be selected at random to speak. What is the probability Marcus will be included in the group of speakers? (*) Let E be the event Marcus is included in the group of three speakers. We want to count the number of ways this can happen. If we have three seats and place Marcus into one of them, two seats remain and there are twelve members to choose from. Hence, there are C(12, 2) ways Marcus can be included in a group of three speakers. Hence, n(e) = C(12, 2). From before, n(s) = C(13, 3). Then P(E) = C(12,2) C(13,3). 5
(15 pts) Problem 7. Let P(A) =.30, P(B) =.40, P(A B) =.10. Find each of the following. a. P(A B c ) (*) A helpful Venn diagram is below. P(A B c ) = P(A Bc ) P(B c ) =.20 1 P(B) =.20.60 = 2 6 b. P(A B c ) (*) P(A B c ) = P(A) + P(B c ) P(A B c ) =.30 +.60.20 =.70 c. P((A B) c ) (*) P(A B) = P(A B) P(B) =.10.40 = 1 4. P((A B) c ) = 1 P(A B) = 3 4 6