Key
1st Hourly
Math 1107
Fall Semester 2005
Protocol
You will use
only the following resources:
Your individual calculator;
Your individual tool-sheet (one
(1) 8.5 by 11 inch sheet);
Your writing utensils;
Blank Paper (provided by me);
This
copy of the hourly.
Do not share
these resources with
anyone else. Do not collaborate or share information with anyone else
during the test session.
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.
Do not share
information with any other students during this hourly. When you are finished:
Prepare a Cover Sheet: Print only your name on an otherwise blank
sheet of paper. Then stack your stuff as follows: Cover Sheet (Top), Your Work
Sheets, The Test Papers, Your Toolsheet. Then hand all of this in to me.
Sign and
Acknowledge: I agree to follow this protocol.
Name (PRINTED) Signature Date
Case One
Probability Rules
Color Slot
Machine
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 |
0.050 |
|
GGGG |
0.100 |
|
GBBR |
0.250 |
|
BBBB |
0.050 |
|
RGYB |
0.060 |
|
YYYY |
0.040 |
|
BBYY |
0.180 |
|
RRYY |
0.020 |
|
RRRR |
0.025 |
|
YBGR |
0.075 |
|
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.
1.1) Pr{Either Yellow Shows or Green
Shows, but not both}
|
Color
Sequence* |
Color
Sequence Probability |
|
GBRB |
0.150 |
|
BBGG |
0.050 |
|
GGGG |
0.100 |
|
GBBR |
0.250 |
|
BBBB |
0.050 |
|
RGYB |
0.060 |
|
YYYY |
0.040 |
|
BBYY |
0.180 |
|
RRYY |
0.020 |
|
RRRR |
0.025 |
|
YBGR |
0.075 |
|
Total |
1.000 |
Pr{Either
Yellow Shows or Green Shows, but not both}=Pr{One of GBRB, BBGG, GGGG, GBBR,
YYYY, BBYY, RRYY Shows}=
Pr{GBRB} +
Pr{BBGG} + Pr{GGGG} + Pr{GBBR} + Pr{YYYY} + Pr{BBYY} + Pr{RRYY}=
.15+.05+.1+.25+.04+.18+.02 = .79
1.2) Pr{Yellow and Blue both Show} –
Use the Complementary Rule
|
Color
Sequence* |
Color
Sequence Probability |
|
GBRB |
0.150 |
|
BBGG |
0.050 |
|
GGGG |
0.100 |
|
GBBR |
0.250 |
|
BBBB |
0.050 |
|
RGYB |
0.060 |
|
YYYY |
0.040 |
|
BBYY |
0.180 |
|
RRYY |
0.020 |
|
RRRR |
0.025 |
|
YBGR |
0.075 |
|
Total |
1.000 |
Other Event =
OE = “Either Yellow does not Show or Blue does not show or neither Blue nor
Yellow Shows”
Pr{OE}=Pr{ Either
Yellow Shows and Blue does not Show or Blue Shows and Yellow does not Show or
neither Blue nor Yellow Shows }=
Pr{One of GBRB,
BBGG, GGGG, GBBR, BBBB, YYYY, RRYY,RRRR Shows}=
Pr{GBRB} + Pr{BBGG}
+ Pr{GGGG} + Pr{GBBR} + Pr{BBBB} + Pr{YYYY} + Pr{RRYY} + Pr{RRRR}=
=
.15+.05+.1+.25+.05+.04+.02+.025 = .685
Pr{Yellow and
Blue both Show} = 1 – Pr{OE} = 1-.685 = .315
Check: Pr{Yellow and Blue both Show} = Pr{One of
RGYB, BBYY, YBGR Shows} = Pr{RGYB} + Pr{BBYY} + Pr{YBGR} = .06+.18+.075 = .315
1.3) Pr{Red Shows | Green Shows} - This
is a Conditional Probability.
|
Color
Sequence* |
Color
Sequence Probability |
|
GBRB |
0.150 |
|
BBGG |
0.050 |
|
GGGG |
0.100 |
|
GBBR |
0.250 |
|
RGYB |
0.060 |
|
YBGR |
0.075 |
|
Total |
68.50(Model Restricted to “Green Shows”) |
Pr{Green
Shows}=Pr{One of GBRB, BBGG, GGGG, GBBR, RGYB, YBGR Shows}=
Pr{GBRB}+ Pr{BBGG}+
Pr{GGGG}+ Pr{GBBR}+ Pr{RGYB}+ Pr{YBGR}=
.15+.05+.1+.25+.06+.075=.685
Pr{Red Shows
and Green Shows}= Pr{One of GBRB, GBBR, RGYB, YBGR Shows}=
Pr{GBRB}+ Pr{GBBR}+
Pr{RGYB}+ Pr{YBGR}=
.15+.25+.06+.075=.535
Pr{Red Shows |
Green Shows}= Pr{Red Shows and Green Shows}/ Pr{Green Shows}=.535/.685 ≈
.7810
Show complete
detail and work for full credit.
Case Two
Pair of Dice
Random Variable
We have a pair of dice – note the probability models for
the dice below.
|
d4 |
|
d3 |
|
|
Face |
Probability |
Face |
Probability |
|
0 |
1/6 |
0 |
1/4 |
|
1 |
2/6 |
1 |
2/4 |
|
2 |
2/6 |
2 |
1/4 |
|
3 |
1/6 |
|
|
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 pair of faces.
2.1) List the possible pairs, and compute a probability for
each.
Write each pair as (d4,d3).
Then the pairs are (0,0),(0,1),(0,2), (1,0),(1,1),(1,2), (2,0),(2,1),(2,2), (3,0),(3,1),(3,2).
Pr{(0,0)}=Pr{0 from d4}*Pr{0
from d3}=(1/6)*(1/4)=1/24
Pr{(0,1)}=Pr{0 from d4}*Pr{1
from d3}=(1/6)*(2/4)=2/24
Pr{(0,2)}=Pr{0 from d4}*Pr{2
from d3}=(1/6)*(1/4)=1/24
Pr{(1,0)}=Pr{1 from d4}*Pr{0
from d3}=(2/6)*(1/4)=2/24
Pr{(1,1)}=Pr{1 from d4}*Pr{1
from d3}=(2/6)*(2/4)=4/24
Pr{(1,2)}=Pr{1 from d4}*Pr{2
from d3}=(2/6)*(1/4)=2/24
Pr{(2,0)}=Pr{2 from d4}*Pr{0
from d3}=(2/6)*(1/4)=2/24
Pr{(2,1)}=Pr{2 from d4}*Pr{1
from d3}=(2/6)*(2/4)=4/24
Pr{(2,2)}=Pr{2 from d4}*Pr{2
from d3}=(2/6)*(1/4)=2/24
Pr{(3,0)}=Pr{3 from d4}*Pr{0
from d3}=(1/6)*(1/4)=1/24
Pr{(3,1)}=Pr{3 from d4}*Pr{1
from d3}=(1/6)*(2/4)=2/24
Pr{(3,2)}=Pr{3 from d4}*Pr{2
from d3}=(1/6)*(1/4)=1/24
2.2)
Define
HYP as HYP = Ö( (d3 face)2 + (d4 face)2).
List the possible values for HYP, and compute a probability for each value of HYP. “Ö”
denotes the square root function.
HYP{(0,0)} = Ö{02+02}=Ö{0+0}=0
HYP{(0,1)} = Ö{02+12}=Ö{0+1}=1
HYP{(0,2)} = Ö{02+22}=Ö{0+4}=2
HYP{(1,0)} = Ö{12+02}=Ö{1+0}=1
HYP{(1,1)} = Ö{12+12}=Ö{1+1}=Ö2
HYP{(1,2)} = Ö{12+22}=Ö{1+4}=Ö5
HYP{(2,0)} = Ö{22+02}=Ö{4+0}=2
HYP{(2,1)} = Ö{22+12}=Ö{4+1}=Ö5
HYP{(2,2)} = Ö{22+22}=Ö{4+4}=Ö8
HYP{(3,0)} = Ö{32+02}=Ö{9+0}=3
HYP{(3,1)} = Ö{32+12}=Ö{9+1}=Ö10
HYP{(3,2)} =Ö{32+22}=Ö{9+4}=Ö13
Pr{HYP=0}=Pr{(0,0}=1/24
Pr{HYP=1}=Pr{(1,0}+Pr{(0,1)}=(2/24)+(2/24)=4/24
Pr{HYP=Ö2}=Pr{(1,1)}=4/24
Pr{HYP=2}=Pr{(0,2)}+Pr{(2,0)}=(1/24)+(2/24)=3/24
Pr{HYP=Ö5}=Pr{(1,2)}+ Pr{(2,1)}=(2/24)+(4/24)=6/24
Pr{HYP=Ö8}=Pr{(2,2)}=2/24
Pr{HYP=3}=Pr{(3,0)}=1/24
Pr{HYP=Ö10}=Pr{(3,1)}=2/24
Pr{HYP=Ö13}=Pr{(3,2)}=1/24
Show complete detail and work for full credit.
Case Three
Long Run Argument / Perfect
Samples
Landsteiner's Human Blood Types
In
the early 20th century, an Austrian scientist named Karl Landsteiner classified
blood according to chemical molecular differences – surface antigen proteins.
Landsteiner observed two distinct chemical molecules present on the surface of
the red blood cells. He labeled one molecule "A" and the other
molecule "B." If the red blood cell had only "A" molecules
on it, that blood was called type A. If the red blood cell had only
"B" molecules on it, that blood was called type B. If the red blood
cell had a mixture of both molecules, that blood was called type AB. If the red
blood cell had neither molecule, that blood was called type O.
Suppose
that the probabilities for blood types for US residents are noted below:
|
Blood
Type |
Probability: US
White |
Probability: US
Black |
Probability: US
Asian |
|
O |
0.45 |
0.49 |
0.40 |
|
A |
0.40 |
0.27 |
0.28 |
|
B |
0.11 |
0.20 |
0.27 |
|
AB |
0.04 |
0.04 |
0.05 |
|
Total |
1.00 |
1.00 |
1.00 |
|
Blood Type |
Probability: |
Expected |
Probability: |
Expected |
Probability: |
Expected |
|
US White |
Count |
US Black |
Count |
US Asian |
Count |
|
|
O |
0.45 |
675 |
0.49 |
735 |
0.4 |
600 |
|
A |
0.4 |
600 |
0.27 |
405 |
0.28 |
420 |
|
B |
0.11 |
165 |
0.2 |
300 |
0.27 |
405 |
|
AB |
0.04 |
60 |
0.04 |
60 |
0.05 |
75 |
|
Total |
1 |
1500 |
1 |
1500 |
1 |
1500 |
3.1) For
US Whites: Interpret each probability using the Long Run Argument. Be specific
and complete for full credit. Compute the perfect sample of n=1500 US White residents, and
describe the relationship of this perfect sample to real random samples of US White
residents.
|
Blood Type |
Probability: |
Expected |
Probability: |
Expected |
Probability: |
Expected |
|
US White |
Count |
US Black |
Count |
US Asian |
Count |
|
|
O |
0.45 |
675 |
0.49 |
735 |
0.4 |
600 |
|
A |
0.4 |
600 |
0.27 |
405 |
0.28 |
420 |
|
B |
0.11 |
165 |
0.2 |
300 |
0.27 |
405 |
|
AB |
0.04 |
60 |
0.04 |
60 |
0.05 |
75 |
|
Total |
1 |
1500 |
1 |
1500 |
1 |
1500 |
eO,US
White = PO,US White*1500 = .45*1500 = 675
eA,US
White = PA,US White*1500 = .40*1500 = 600
eB,US
White = PB,US White*1500 = .11*1500 = 165
eAB,US
White = PAB,US White*1500 = .04*1500 = 60
In long runs of sampling with replacement
from US White residents, approximately 45% of sampled US White residents will
present type O blood; approximately 40% of sampled US White residents will show
type A blood; approximately 11% of sampled US White residents will show type B
blood and approximately 4% of sampled US White residents will show type AB
blood.
In random samples of 1500 US White
residents, there will be approximately 675 sampled US White residents with type
O blood; approximately 600 sampled US White residents with type A blood;
approximately 165 sampled US White residents with type B blood and
approximately 60 sampled US White residents with type AB blood.
3.2) For
US Blacks: Interpret each probability using the Long Run Argument. Be specific
and complete for full credit. Compute the perfect sample of n=1500 US Black residents, and
describe the relationship of this perfect sample to real random samples of US Black
residents.
|
Blood Type |
Probability: |
Expected |
Probability: |
Expected |
Probability: |
Expected |
|
US White |
Count |
US Black |
Count |
US Asian |
Count |
|
|
O |
0.45 |
675 |
0.49 |
735 |
0.4 |
600 |
|
A |
0.4 |
600 |
0.27 |
405 |
0.28 |
420 |
|
B |
0.11 |
165 |
0.2 |
300 |
0.27 |
405 |
|
AB |
0.04 |
60 |
0.04 |
60 |
0.05 |
75 |
|
Total |
1 |
1500 |
1 |
1500 |
1 |
1500 |
eO,US
Black = PO,US Black*1500 = .49*1500 = 735
eA,US
Black = PA,US Black*1500 = .27*1500 = 405
eB,US
Black = PB,US Black*1500 = .20*1500 = 300
eAB,US
Black = PAB,US Black*1500 = .04*1500 = 60
In long runs of sampling with replacement
from US Black residents, approximately 49% of sampled US Black residents will
present type O blood; approximately 27% of sampled US Black residents will show
type A blood; approximately 20% of sampled US Black residents will show type B
blood and approximately 4% of sampled US Black residents will show type AB
blood.
In random samples of 1500 US Black
residents, there will be approximately 735 sampled US Black residents with type
O blood; approximately 405 sampled US Black residents with type A blood;
approximately 300 sampled US Black residents with type B blood and approximately
60 sampled US Black residents with type AB blood.
3.3) For
US Asians: Interpret eah probability using the Long Run Argument. Be specific
and complete for full credit. Compute the perfect sample of n=1500 US Asian residents, and
describe the relationship of this perfect sample to real random samples of US Asian
residents
|
Blood Type |
Probability: |
Expected |
Probability: |
Expected |
Probability: |
Expected |
|
US White |
Count |
US Black |
Count |
US Asian |
Count |
|
|
O |
0.45 |
675 |
0.49 |
735 |
0.4 |
600 |
|
A |
0.4 |
600 |
0.27 |
405 |
0.28 |
420 |
|
B |
0.11 |
165 |
0.2 |
300 |
0.27 |
405 |
|
AB |
0.04 |
60 |
0.04 |
60 |
0.05 |
75 |
|
Total |
1 |
1500 |
1 |
1500 |
1 |
1500 |
eO,US
Asian = PO,US Asian*1500 = .40*1500 = 600
eA,US
Asian = PA,US Asian*1500 = .28*1500 = 420
eB,US
Asian = PB,US Asian*1500 = .27*1500 = 405
eAB,US
Asian = PAB,US Asian*1500 = .05*1500 = 75
In long runs of sampling with replacement
from US Asian residents, approximately 40% of sampled US Asian residents will
present type O blood; approximately 28% of sampled US Asian residents will show
type A blood; approximately 27% of sampled US Asian residents will show type B
blood and approximately 5% of sampled US Asian residents will show type AB
blood.
In random samples of 1500 US Asian
residents, there will be approximately 600 sampled US Asian residents with type
O blood; approximately 420 sampled US Asian residents with type A blood;
approximately 405 sampled US Asian residents with type B blood and
approximately 75 sampled US Asian residents with type AB blood.
Show complete detail and work for full
credit.
Case Four
Probability Rules
Number Slot
Machine
Here is a number slot machine – on each trial, it
produces a 4-digit sequence, using the table below:
|
Digit
Sequence* |
Digit
Sequence Probability |
|
0121 |
0.150 |
|
1100 |
0.050 |
|
0000 |
0.100 |
|
0112 |
0.250 |
|
1111 |
0.050 |
|
2031 |
0.060 |
|
3333 |
0.040 |
|
1133 |
0.180 |
|
2233 |
0.020 |
|
2222 |
0.025 |
|
3102 |
0.075 |
|
Total |
1.000 |
*Digit slots are 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. Briefly interpret each probability using a long run or relative
frequency argument.
Define SUM14
as the sum of the first and fourth digits in the sequence. Compute and list the
values for SUM14. Compute the probability for each value of SUM14. Show full work and detail for full credit.
|
Digit
Sequence* |
Digit
Sequence Probability |
|
0121 |
0.150 |
|
1100 |
0.050 |
|
0000 |
0.100 |
|
0112 |
0.250 |
|
1111 |
0.050 |
|
2031 |
0.060 |
|
3333 |
0.040 |
|
1133 |
0.180 |
|
2233 |
0.020 |
|
2222 |
0.025 |
|
3102 |
0.075 |
|
Total |
1.000 |
SUM14{0121}=0+1=1
SUM14{1100}=1+0=1
SUM14{0000}=0+0=0
SUM14{0112}=0+2=2
SUM14{1111}=1+1=2
SUM14{2031}=2+1=3
SUM14{3333}=3+3=6
SUM14{1133}=1+3=4
SUM14{2233}=2+3=5
SUM14{2222}=2+2=4
SUM14{3102}=3+2=5
Pr{SUM14=0} =
Pr{0000}=.10
Pr{SUM14=1} =
Pr{One of 0121, 1100 Shows} = Pr{0121}
+ Pr{1100} = .15 + .05 = .20
Pr{SUM14=2} =
Pr{One of 0112, 1111 Shows} = Pr{0112}
+ Pr{1111} = .25 + .05 = .30
Pr{SUM14=3} =
Pr{2031} = .06
Pr{SUM14=4} =
Pr{One of 1133, 2222 Shows} = Pr{1133} + Pr{2222} = .18 + .025 = .205
Pr{SUM14=5} =
Pr{One of 2233, 3102 Shows} = Pr{2233} + Pr{3102} = .02 + .075 = .095
Pr{SUM14=6} =
Pr{3333} = .04
Show complete
detail and work for full credit.
Work all four
cases.