FALL 2015 - 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
1 2 3 4 5 6 7
(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}, andB ={4,5,6,7}.
(a) (2 pts) FindA∪B.
A∪B ={1,3,4,5,6,7,9}
• Not using the set notation: 1 pt. (b) (2 pts) FindA0. A0 ={2,4,6,8,10} (c) (2 pts) Findn(A∩B). A∩B ={5,7}⇒n(A∩B) =2 • FindingA∩B ={5,7}: 1 pt. (d) (2 pts) FindA∩A0. A∩A0 =∅
(3) A cable television company has 8,000 subscribers in a suburban community. The company offers two premium channels: HBO and Showtime. 2,550 sub-scribers receive HBO, 1,840 receive Showtime, and 5,080 do not receive any pre-mium 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
H S U 5080 1080 1470 370 n(H) = 2550, n(S) = 1840, n((H∪S)0) = 5080 n(H∪S) = n(U)−n((H∪S)0) = 8000−5080 = 2920 n(H∪S) =n(H) +n(S)−n(H∩S)⇒2920 = 2550 + 1840−n(H∩S) ⇒n(H∩S) = 2550 + 1840−2920 =1470 n(H∩S0) = n(H)−n(H∩S) = 2550−1470 =1080 n(S∩H0) =n(S)−n(H∩S) = 1840−1470 =370
• Sketching an appropriate Venn-diagram: 1 pt.
• Finding three numbers1470,1080, and370: 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?
(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 choose10stores from9 + 12 + 8 = 29stores. The order of choices is not important.
29C10 =
29!
10!(29−10)! =20,030,010
• Stating the answer is29C10: 2 pts.
• Finding the answer20,030,010: 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. 9C3·12C5·8C2 = 9! 3!(9−3)! · 12! 5!(12−5)! · 8! 2!(8−2)! = 84·792·28 =1,862,784
• Stating the answer is9C3 ·12C5·8C2: 2 pts.
• Finding the answer1,862,784: 3 pts.
(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. ThereforeP(D) = 13 52 =
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. SoP(F) = 12 52 =
3 13.
(c) (2 pts) Find the probability that the drawn card is a diamond or a face card. There are 3 diamond face cards. SoP(D∩F) = 3
52. P(D∪F) =P(D) +P(F)−P(D∩F) = 13 52 + 12 52− 3 52 = 22 52 = 11 26
(d) (2 pts) Find the probability that the drawn card is not a diamond. P(D0) = 1−P(D) = 1− 13 52 = 39 52 = 3 4
(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 spaceS: set of combinations of 20 + 18 = 38people taken 4 at a time.
n(S) =38C4 =73815.
E: event that all 4 were chosen from A n(E) =20C4 = 4845.
P(E) = n(E)
n(S) = 4845
73815 ≈0.0656
• Finding an appropriate sample spaceSand evaluating its number of element73815: 2 pts.
• Obtaining the answer0.0656: 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) = 20C2×18C2 = 190·153 = 29070
P(F) = n(F)
n(S) =
29070
73815 ≈0.3938
• Describing the correct number of elements20C2×18C2: 2 pts.
• Getting the probability0.3938: 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) = 20C4+20C3×18C1 = 4845 + 1140·18 = 25365
P(G) = n(G)
n(S) = 25365
73815 ≈0.3436
• Describing the correct number of elements20C4+20C3×18C1: 3 pts.
(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 over-weight woman.
M: event that a chosen person is a man O: event that a chosen person is overweight
O M P(O|M)=0.681 7 7 P(O0|M)=0.319 / /O0 P(M)=0.495 >> P(M0)=0.505 M0P(O|M 0)=0.643 / / P(O0|M0)=0.357 ' ' M M0 P(M0∩O)=P(M0)·P(O|M0) = 0.505·0.643≈0.325
• Sketching the probability tree: 2 pts.
• Stating the probabilityP(M0∩O)we need to find: +1 pt. • Getting the answer0.325: +1 pt.
(b) (4 pts) Find the probability that a randomly selected American is over-weight.
P(O)=P(M ∩O) +P(M0∩O) =P(M)P(O|M) +P(M0)P(O|M0)
= 0.495·0.681 + 0.505·0.643≈0.662
• Stating the probabilityP(O)we need to find: 2 pts. • Getting the answer0.662: 4 pts.
(c) (2 pts) Is there any relation between gender and overweight? Explain your answer.
P(M ∩O) = 0.495·0.681 ≈0.337
P(M)·P(O) = 0.495·0.662 ≈0.328
BecauseP(M ∩O)6=P(M)·P(O), they aredependent.
• If you didn’t checkP(M ∩O) 6= P(M)·P(O)(orP(O) 6= P(O|M)), you can’t get any credit.
(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
E S P(E|S)=0.02 8 8 P(E0|S)=0.98 / / E0 P(S)=0.638 ?? P(S0)=0.362 S0 P(E|S 0)=0.36 / / P(E0|S0)=0.64 & & E E0 Want: P(S|E) P(S∩E) =P(S)P(E|S) = 0.638·0.02≈0.01276 P(E) =P(S∩E) +P(S0∩E) =P(S)P(E|S) +P(S0)P(E|S0) = 0.638·0.02 + 0.362·0.36 = 0.14308 P(S|E) = P(S∩E) P(E) ≈0.0892
• Sketching the probability tree: 2 pts.
• Describing the probabilityP(S|E)we want to find: +1 pt. • FindingP(S∩E) = 0.01276andP(E) = 0.14308: +1 pt each. • Getting the answerP(E) =0.0892: +1 pt.
