MATH 1108 R07 MIDTERM EXAM 1 SOLUTION

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1 MATH 1108 R07 MIDTERM EXAM 1 SOLUTION FALL MOON Write your answer neatly and show steps. Except calculators, any electronic devices including laptops and cell phones are not allowed. Do not use the graphing function on your calculator. (1) Quick survey. (a) (1 pt) This class is: Too easy Moderate Too difficult (b) (2 pts) Write any suggestion for improving this class. (For instance, give more examples in class, explain proofs of formulas in detail, give more homework, slow down the tempo,...) (2) Suppose that the universal set U is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 3, 5, 7, 9}, and B = {4, 5, 6, 7}. (a) (2 pts) Find A B. A B = {1, 3, 4, 5, 6, 7, 9} Not using the set notation: 1 pt. (b) (2 pts) Find A. A = {2, 4, 6, 8, 10} (c) (2 pts) Find n(a B). A B = {5, 7} n(a B) = 2 Finding A B = {5, 7}: 1 pt. (d) (2 pts) Find A A. A A = Writing the answer as {}: 1 pt. Date: October 8,

2 MATH 1108 Midterm Exam 1 (3) A cable television company has 8,000 subscribers in a suburban community. The company offers two premium channels: HBO and Showtime. 2,550 subscribers receive HBO, 1,840 receive Showtime, and 5,080 do not receive any premium channel. (a) (4 pts) Sketch a Venn-diagram describing the situation, and indicate the number of subscribers in each region. H: the set of subscribers receiving HBO S: the set of subscribers receiving Showtime U H S n(h) = 2550, n(s) = 1840, n((h S) ) = 5080 n(h S) = n(u) n((h S) ) = = 2920 n(h S) = n(h) + n(s) n(h S) 2920 = n(h S) n(h S) = = 1470 n(h S ) = n(h) n(h S) = = 1080 n(s H ) = n(s) n(h S) = = 370 Sketching an appropriate Venn-diagram: 1 pt. Finding three numbers 1470, 1080, and 370: 1 pt each. (b) (2 pts) How many subscribers receive both HBO and Showtime? 1470 (c) (2 pts) How many subscribers receive HBO but not Showtime?

3 MATH 1108 Midterm Exam 1 (4) A jewelry store chain with 9 stores in Georgia, 12 in Florida, and 8 in Alabama is planning to close 10 of these stores. (a) (3 pts) How many ways can this be done? We have to choose 10 stores from = 29 stores. The order of choices is not important. 29C 10 = 29! 10!(29 10)! = 20, 030, 010 Stating the answer is 29 C 10 : 2 pts. Finding the answer 20, 030, 010: 3 pts. (b) (3 pts) The company decided to close 3 stores in Georgia, 5 in Florida, and 2 in Alabama. In how many ways can this be done? We have to choose 3 from 9 stores in Georgia, 5 from 12 stores in Florida, and 2 from 8 stores in Alabama, at the same time. The order of choices doesn t matter. 9C 3 12 C 5 8C 2 = 9! 3!(9 3)! 12! 5!(12 5)! 8! 2!(8 2)! Stating the answer is 9 C 3 12 C 5 8C 2 : 2 pts. Finding the answer 1, 862, 784: 3 pts. = = 1, 862, 784 (5) Suppose that a single playing card is drawn at random from a standard card deck of 52 cards. (a) (2 pts) Find the probability that the drawn card is a diamond. D: event that the drawn card is a diamond. There are 13 diamond cards among 52 cards. Therefore P (D) = = 1 4. (b) (2 pts) Find the probability that the drawn card is a face card. F : event that the drawn card is a face card. There are 12 face cards among 52 cards. So P (F ) = = (c) (2 pts) Find the probability that the drawn card is a diamond or a face card. There are 3 diamond face cards. So P (D F ) = P (D F ) = P (D) + P (F ) P (D F ) = = = (d) (2 pts) Find the probability that the drawn card is not a diamond. P (D ) = 1 P (D) = = = 3 4 3

4 MATH 1108 Midterm Exam 1 (6) A 4-person committee is selected out of 2 departments, A and B, with 20 and 18 people, respectively. If 4 people are selected at random from the 38 people, what is the probability of selecting (a) (4 pts) all 4 from A? Sample space S: set of combinations of = 38 people taken 4 at a time. n(s) = 38 C 4 = E: event that all 4 were chosen from A n(e) = 20 C 4 = P (E) = n(e) n(s) = Finding an appropriate sample space S and evaluating its number of element 73815: 2 pts. Obtaining the answer : 4 pts. (b) (3 pts) 2 from A and 2 from B? F : event that 2 were chosen from A, and 2 were chosen from B n(f ) = 20 C 2 18 C 2 = = P (F ) = n(f ) n(s) = Describing the correct number of elements 20 C 2 18 C 2 : 2 pts. Getting the probability : 3 pts. (c) (4 pts) at least 3 from A? G: event that at least 3 were chosen from A There are two cases: all of 4 were chosen from A, or 3 of them were chosen from A and the remaining one were chosen from B. n(g) = 20 C C 3 18 C 1 = = P (G) = n(g) n(s) = Describing the correct number of elements 20 C C 3 18 C 1 : 3 pts. Getting the probability : 4 pts. 4

5 MATH 1108 Midterm Exam 1 (7) According to a recent report, 68.1% of men and 64.3% of women in the United States were overweight. It is also known that 49.5% of Americans are men and 50.5% are women. (a) (4 pts) Find the probability that a randomly selected American is an overweight woman. M: event that a chosen person is a man O: event that a chosen person is overweight P (O M)=0.681 O P (M)=0.495 M O P (O M)=0.319 P (M )=0.505 M P (O M )=0.643 M P (O M )=0.357 M P (M O) = P (M ) P (O M ) = Sketching the probability tree: 2 pts. Stating the probability P (M O) we need to find: +1 pt. Getting the answer 0.325: +1 pt. (b) (4 pts) Find the probability that a randomly selected American is overweight. P (O) = P (M O) + P (M O) = P (M)P (O M) + P (M )P (O M ) = Stating the probability P (O) we need to find: 2 pts. Getting the answer 0.662: 4 pts. (c) (2 pts) Is there any relation between gender and overweight? Explain your answer. P (M O) = P (M) P (O) = Because P (M O) P (M) P (O), they are dependent. If you didn t check P (M O) P (M) P (O) (or P (O) P (O M)), you can t get any credit. 5

6 MATH 1108 Midterm Exam 1 (8) (6 pts) A 2009 federal study showed that 63.8% of occupants involved in a fatal car crash wore seat belts. Of those in a fatal crash who wore seat belts, 2% were ejected from the vehicle. For those not wearing seat belts, 36% were ejected from the vehicle. Find the probability that a randomly selected person in a fatal car crash who was ejected from the vehicle was wearing a seat belt. S: event that an occupant involved in a fatal car crash wore a seat belt E: event that the occupant was ejected from the vehicle P (E S)=0.02 E P (S)=0.638 S E P (E S)=0.98 P (S )=0.362 S P (E S )=0.36 E P (E S )=0.64 Want: P (S E) P (S E) = P (S)P (E S) = E P (E) = P (S E) + P (S E) = P (S)P (E S) + P (S )P (E S ) = = P (S E) = P (S E) P (E) Sketching the probability tree: 2 pts. Describing the probability P (S E) we want to find: +1 pt. Finding P (S E) = and P (E) = : +1 pt each. Getting the answer P (E) = : +1 pt. 6

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