Key, Version A
The 1st Hourly
Math 1107
Spring
Semester 2007
Protocol
You will use
only the following resources: Your
individual calculator; individual tool-sheet (single 8.5 by 11 inch sheet); your
writing utensils; blank paper (provided by me) and this copy of the hourly. Do
not share these resources with anyone else.
In each case, show complete detail and
work for full credit. Follow case study solutions and sample hourly keys in presenting
your solutions. Work all four cases. Using only one side of the
blank sheets provided, present your work. Do not write on both sides of the
sheets provided, and present your work only on these sheets. All of your work
goes on one side each of the blank sheets provided. Space out your work.
Do not share information with any other students during this hourly.
Sign and Acknowledge:
I agree to follow this protocol.
Name (PRINTED) Signature Date
Case One
Random Variables
Pair of Dice
We have a pair
of fair dice – a d2 {faces 4, 5} and a d6 {faces 0, 2, 4, 5, 6, 8}. We assume
that the dice operate separately and independently of each other. Suppose that
our experiment consists of tossing the dice, and noting the resulting
face-pair.
a) List the possible face-pairs, and
compute a probability for each.
Writing the pair as (d6 face value, d2 face
value), here are the possible pairs:
|
(0,4) |
(2,4) |
(4,4) |
(5,4) |
(6,4) |
(8,4) |
|
(0,5) |
(2,5) |
(4,5) |
(5,5) |
(6,5) |
(8,5) |
Pr{(0,4)}=Pr{0 from d2}*Pr{4 from d6} =
(1/2)*(1/6) = 1/12
Pr{(2,4)}=Pr{2 from d2}*Pr{4 from d6} = (1/2)*(1/6)
= 1/12
Pr{(4,4)}=Pr{4 from d2}*Pr{4 from d6} =
(1/2)*(1/6) = 1/12
Pr{(5,4)}=Pr{5 from d2}*Pr{4 from d6} =
(1/2)*(1/6) = 1/12
Pr{(6,4)}=Pr{6 from d2}*Pr{4 from d6} =
(1/2)*(1/6) = 1/12
Pr{(8,4)}=Pr{8 from d2}*Pr{4 from d6} =
(1/2)*(1/6) = 1/12
Pr{(0,5)}=Pr{0 from d2}*Pr{5 from d6} =
(1/2)*(1/6) = 1/12
Pr{(2,5)}=Pr{2 from d2}*Pr{5 from d6} =
(1/2)*(1/6) = 1/12
Pr{(4,5)}=Pr{5 from d2}*Pr{5 from d6} =
(1/2)*(1/6) = 1/12
Pr{(5,5)}=Pr{5 from d2}*Pr{5 from d6} =
(1/2)*(1/6) = 1/12
Pr{(6,5)}=Pr{6 from d2}*Pr{5 from d6} =
(1/2)*(1/6) = 1/12
Pr{(8,5)}=Pr{8 from d2}*Pr{5 from d6} =
(1/2)*(1/6) = 1/12
b) Define HIGHTIE as the higher of
the two face values or common value if tied, and LOWTIE as the lower of
the two face values or common value if tied. Consider the random variable MIDDLE
= (HIGHTIE − LOWTIE)/2. Compute the values and probabilities for MIDDLE.
(0,4) ® Hightie=4, Lowtie=0 ®
Middle=(4-0)/2=2
(2,4) ® Hightie=4, Lowtie=2 ®
Middle=(4-2)/2=1
(4,4) ® Hightie=4, Lowtie=4 ®
Middle=(4-4)/2=0
(5,4) ® Hightie=5, Lowtie=4 ®
Middle=(5-4)/2=1/2
(6,4) ® Hightie=6, Lowtie=4 ®
Middle=(6-4)/2=1
(8,4) ® Hightie=8, Lowtie=4 ®
Middle=(8-4)/2=2
(0,5) ® Hightie=5, Lowtie=0 ®
Middle=(5-0)/2=5/2
(2,5) ® Hightie=5, Lowtie=2 ®
Middle=(5-2)/2=3/2
(4,5) ® Hightie=5, Lowtie=4 ®
Middle=(5-4)/2=1/2
(5,5) ® Hightie=5, Lowtie=5 ®
Middle=(5-5)/2=0
(6,5) ® Hightie=6, Lowtie=5 ®
Middle=(6-5)/2=1/2
(8,5) ® Hightie=8, Lowtie=5 ®
Middle=(8-5)/2=3/2
Pr{MIDDLE=0} =
Pr{one of (4,4),(5,5) shows} = Pr{(4,4)}+Pr{(5,5)} = (1/12)+(1/12) = 2/12
Pr{MIDDLE=1/2}
= Pr{one of (5,4),(4,5),(6,5) shows} = Pr{(5,4)}+Pr{(4,5)} +Pr{(6,5)} =
(1/12)+(1/12) +(1/12) = 3/12
Pr{MIDDLE=1} =
Pr{one of (2,4),(6,4) shows} = Pr{(2,4)}+Pr{(6,4)} = (1/12)+(1/12) = 2/12
Pr{MIDDLE=3/2}
= Pr{one of (2,5),(8,5) shows} = Pr{(2,5)}+Pr{(8,5)} = (1/12)+(1/12) = 2/12
Pr{MIDDLE=2} =
Pr{one of (0,4),(8,4) shows} = Pr{(0,4)}+Pr{(8,4)} = (1/12)+(1/12) = 2/12
Pr{MIDDLE=5/2}
= Pr{(0,5) shows} = 1/12
Case Two
Here is our color slot machine – on each
trial, it produces a 4-color sequence, using the table below:
|
Color Sequence* |
Color Sequence Probability |
|
GBRB |
0.150 |
|
BBGG |
0.150 |
|
GBBR |
0.300 |
|
RGYB |
0.060 |
|
YYYY |
0.040 |
|
BBYY |
0.180 |
|
RRYY |
0.020 |
|
YBGR |
0.100 |
|
Total |
1.000 |
*B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered as 1st
to 4th, from left to right: (1st 2nd 3rd
4th)
Compute the
following probabilities. In each of the following, show your intermediate steps
and work. If a rule is specified, you must use that rule for your computation.
a) Pr{Green
Shows} Briefly interpret each probability using a long run or relative
frequency argument.
|
Color Sequence* |
Color Sequence Probability |
|
GBRB |
0.150 |
|
BBGG |
0.150 |
|
GBBR |
0.300 |
|
RGYB |
0.060 |
|
YBGR |
0.100 |
|
Total |
|
Pr{Green Shows}
= Pr{one of GBRB, BBGG,GBBR,RGYB,YBGR shows} = Pr{GBRB}+ Pr{BBGG} + Pr{GBBR} +
Pr{RGYB} + Pr{YBGR shows} = .15+.15+.30+.06+.10 = .76
In long runs of
box-plays, approximately 76% of plays will show green.
b) Pr{Green or
Blue Show} - Use the Complementary Rule. Briefly interpret each probability using a
long run or relative frequency argument.
|
Color Sequence* |
Color Sequence Probability |
|
GBRB |
0.150 |
|
BBGG |
0.150 |
|
GBBR |
0.300 |
|
RGYB |
0.060 |
|
BBYY |
0.180 |
|
YBGR |
0.100 |
|
Color Sequence* |
Color Sequence Probability |
|
YYYY |
0.040 |
|
RRYY |
0.020 |
Pr{Neither Green
nor Blue Shows}=Pr{One of YYYY,YYRR Shows}=Pr{YYYY}+Pr{RRYY}=.04+.02=.06
Pr{Green or
Blue Shows} = 1 - Pr{Neither Green nor Blue Shows}= 1 - .06 =
.94
Pr{Green or
Blue Shows}=Pr{one of GBRB, BBGG,GBBR,RGYB,BBYY,YBGR shows} = = Pr{GBRB}+
Pr{BBGG} + Pr{GBBR} + Pr{RGYB} + +Pr{BBYY} + Pr{YBGR shows}
=.15+.15+.30+.06+.18+.10 = .94
In long runs of
box-plays, approximately 94% of plays will show blue or green.
c) Pr{Green
Shows | Blue Shows} - This is a conditional probability.
|
Color Sequence* |
Color Sequence Probability |
|
GBRB |
0.150 |
|
BBGG |
0.150 |
|
GBBR |
0.300 |
|
RGYB |
0.060 |
|
BBYY |
0.180 |
|
YBGR |
0.100 |
|
Total |
.94 |
Pr{Blue
Shows}=Pr{one of GBRB, BBGG,GBBR,RGYB,BBYY,YBGR shows} = Pr{GBRB}+ Pr{BBGG} +
Pr{GBBR} + Pr{RGYB} + Pr{BBYY} + Pr{YBGR shows} =.15+.15+.30+.06+.18+.10 = .94
Pr{Blue and
Green Show}=Pr{one of GBRB, BBGG,GBBR,RGYB, YBGR shows} = Pr{GBRB}+ Pr{BBGG} +
Pr{GBBR} + Pr{RGYB} + Pr{YBGR shows} =.15+.15+.30+.06+.10 = .76
Pr{Green Shows
| Blue Shows} = Pr{Green and Blue Show}/ Pr{Blue Shows} = .76/.94 ».8085
Case Three
Long Run Argument / Perfect Samples
Prevalence of Viral Subtypes in an
Infected Population
Suppose that we have a virus, say XCV. Suppose further
that XCV has known main types A, B, C and D. Infected individuals may
present one or more types of XCV. Suppose that the probabilities for XCV
types present in infected individuals are noted below:
|
XCV Profile |
Probability |
|
XCV−A |
0.65 |
|
XCV−B |
0.18 |
|
XCV−C |
0.07 |
|
XCV−D |
0.03 |
|
XCV−Multiple Types Present |
0.05 |
|
XCV−Type or Types Unknown |
0.02 |
|
Total |
1.00 |
a) Interpret each probability using the Long
Run Argument. Be specific and complete for full credit.
In long runs of sampling XCV cases with
replacement, approximately 65% of sampled cases present type A, approximately
18% present type B, approximately 7% present type C, approximately 3% present
type D, approximately 5% present multiple types of XCV and approximately 2%
present unknown types of XCV.
b) Compute the perfect sample of n=3,500 XCV-infected
individuals, and describe the relationship of this perfect sample to real random
samples of XCV-infected individuals.
|
XCV Profile |
Probability |
Expected Count n=3500 |
|
XCV−A |
0.65 |
.65*3500=2275 |
|
XCV−B |
0.18 |
.18*3500=630 |
|
XCV−C |
0.07 |
.07*3500=245 |
|
XCV−D |
0.03 |
.03*3500=105 |
|
XCV−Multiple Types Present |
0.05 |
.05*3500=175 |
|
XCV−Type or Types Unknown |
0.02 |
.02*3500=70 |
|
Total |
1 |
3500 |
E3500{XCV−A}
= 3500*Pr{ XCV−A} = 3500*.65 = 2275
E3500{XCV−B}
= 3500*Pr{ XCV−B} = 3500*.18 = 630
E3500{XCV−C}
= 3500*Pr{ XCV−C} = 3500*.07 = 245
E3500{XCV−D}
= 3500*Pr{ XCV−D} = 3500*.03 = 105
E3500{XCV−Multiple}
= 3500*Pr{ XCV−Multiple} = 3500*.05 = 175
E3500{XCV−Unknown}
= 3500*Pr{ XCV−Unknown} = 3500*.02 = 70
Random samples of 3,500 XCV cases, will
contain approximately 2,275 cases presenting type A, approximately 630 cases presenting
type B, approximately 245 cases presenting type C, approximately 105 cases presenting
type D, approximately 175 cases presenting multiple types of XCV and
approximately 70 cases presenting
unknown types of XCV.
Case Four
Three Dice
Probability Model/Random
Variable
Suppose we
have three fair dice: d2(faces 1,2), d2(faces 3,4) and d2(faces 5,6). In our
experiment, we toss this triplet of dice, and note the face value from each
die. For simplicity, we write the outcome as (1st d2 result, 2nd d2 result, 3rd
d2 result). Assume that the dice operate independently and separately. A
triplet is like a pair, only with three items, rather than two.
a) List the
possible face value triplets, and compute a probability for each face value
triplet.
There are
2*2*2=8 possible triplets: (1,3,5), (1,3,6), (1,4,5), (1,4,6), (2,3,5),
(2,3,6), (2,4,5), (2,4,6).
Pr{(1,3,5)} =
Pr{1 from 1st d2}*Pr{3 from 2nd d2}*{5 from 3rd
d2} = (1/2)*(1/2)*(1/2) = 1/8
Pr{(1,3,6)} =
Pr{1 from 1st d2}*Pr{3 from 2nd d2}*{6 from 3rd
d2} = (1/2)*(1/2)*(1/2) = 1/8
Pr{(1,4,5)} =
Pr{1 from 1st d2}*Pr{4 from 2nd d2}*{5 from 3rd
d2} = (1/2)*(1/2)*(1/2) = 1/8
Pr{(1,4,6)} =
Pr{1 from 1st d2}*Pr{4 from 2nd d2}*{6 from 3rd
d2} = (1/2)*(1/2)*(1/2) = 1/8
Pr{(2,3,5)} =
Pr{2 from 1st d2}*Pr{3 from 2nd d2}*{5 from 3rd
d2} = (1/2)*(1/2)*(1/2) = 1/8
Pr{(2,3,6)} =
Pr{2 from 1st d2}*Pr{3 from 2nd d2}*{6 from 3rd
d2} = (1/2)*(1/2)*(1/2) = 1/8
Pr{(2,4,5)} =
Pr{2 from 1st d2}*Pr{4 from 2nd d2}*{5 from 3rd
d2} = (1/2)*(1/2)*(1/2) = 1/8
Pr{(2,4,6)} =
Pr{2 from 1st d2}*Pr{4 from 2nd d2}*{6 from 3rd
d2} = (1/2)*(1/2)*(1/2) = 1/8
b) We define SUM
as the sum of the faces in the face value triplet as SUM = [1st
face] + [2nd face] + [3rd face]. Identify the
possible values of SUM, and compute a probability for each value of SUM.
SUM{(1,3,5)} = 1+3+5 =9
SUM{(1,3,6)} =
1+3+6=10
SUM{(1,4,5)} =
1+4+5=10
SUM{(1,4,6)} =
1+4+6=11
SUM{(2,3,5)} =
2+3+5=10
SUM{(2,3,6)} =
2+3+6=11
SUM{(2,4,5)} =
2+4+5=11
SUM{(2,4,6)} =
2+4+6=12
Pr{SUM=9} =
Pr{(1,3,5)} = 1/8
Pr{SUM=10} =
Pr{one of (1,3,6),(1,4,5),(2,3,5) shows} = Pr{(1,3,6)} + Pr{(1,4,5)} +
Pr{(2,3,5)} = (1/8)+(1/8)+(1/8) = 3/8
Pr{SUM=11} =
Pr{one of (2,3,6),(1,4,6),(2,4,5) shows} = Pr{(1,3,6)} + Pr{(1,4,5)} +
Pr{(2,3,5)} = (1/8)+(1/8)+(1/8) = 3/8
Pr{SUM=12} =
Pr{(2,4,6)} = 1/8
Be certain that you
have worked all four (4) cases. Show full work and detail for full credit.