Key
The 1st Hourly
Math 1107
Fall 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
independently operating, fair, dice: the first die with face values 1,2,4,5 and the second die with face values 0,3. Each trial of our experiment
involves tossing the pair of dice, and noting the pair of face values. We
compute a random variable from the pair of face values as follows: Consider
the random variable HIGHTIE, where HIGHTIE = higher of the two face values or
common value if tied.
1. List all the possible pairs of face values,
and compute the probability for each pair, showing full detail.
2.
Compute the values of HIGHTIE, showing in detail how these values are
computed from each pair of face values.
3. Compute the probability for each value of HIGHTIE from step 2,
showing full detail.
Show
full work and detail for full credit.
1.
|
|
1 |
2 |
4 |
5 |
|
0 |
(1,0) |
(2,0) |
(4,0) |
(5,0) |
|
3 |
(1,3) |
(2,3) |
(4,3) |
(5,3) |
There are eight (8) pairs: (1,0), (2,0), (4,0),
(5,0), (1,3), (2,3), (4,3), (5,3).
Pr{(1,0) Shows} = Pr{1 from 1st
Die Shows}*Pr{0 from 2nd Die Shows} = (1/4)*(1/2) = 1/8
Pr{(2,0) Shows} = Pr{2 from 1st
Die Shows}*Pr{0 from 2nd Die Shows} = (1/4)*(1/2) = 1/8
Pr{(4,0) Shows} = Pr{4 from 1st
Die Shows}*Pr{0 from 2nd Die Shows} = (1/4)*(1/2) = 1/8
Pr{(5,0) Shows} = Pr{5 from 1st
Die Shows}*Pr{0 from 2nd Die Shows} = (1/4)*(1/2) = 1/8
Pr{(1,3) Shows} = Pr{1 from 1st
Die Shows}*Pr{3 from 2nd Die Shows} = (1/4)*(1/2) = 1/8
Pr{(2,3) Shows} = Pr{2 from 1st
Die Shows}*Pr{3 from 2nd Die Shows} = (1/4)*(1/2) = 1/8
Pr{(4,3) Shows} = Pr{4 from 1st
Die Shows}*Pr{3 from 2nd Die Shows} = (1/4)*(1/2) = 1/8
Pr{(5,3) Shows} = Pr{5 from 1st
Die Shows}*Pr{3 from 2nd Die Shows} = (1/4)*(1/2) = 1/8
2.
HT=”High Tie”
|
|
1 |
2 |
4 |
5 |
|
0 |
(1,0) HT=1 |
(2,0) HT=2 |
(4,0) HT=4 |
(5,0) HT=5 |
|
3 |
(1,3) HT=3 |
(2,3) HT=3 |
(4,3) HT=4 |
(5,3) HT=5 |
HT{(1,0)}=1
HT{(2,0)}=2
HT{(4,0)}=4
HT{(5,0)}=5
HT{(1,3)}=3
HT{(2,3)}=3
HT{(4,3)}=4
HT{(5,3)}=5.
3.
Pr{HT=1}=Pr{(1,0) Shows} = 1/8 = .125
Pr{HT=2}=Pr{(2,0) Shows} = 1/8 = .125
Pr{HT=3}=Pr{One of (1,3), (2,3) Shows} =
Pr{(1,3) Shows} + Pr{(2,3) Shows}=(1/8)+(1/8) = 2/8 = 1/4 = .25
Pr{HT=4}=Pr{One of (4,0), (4,3) Shows} =
Pr{(4,0) Shows} + Pr{(4,3) Shows}=(1/8)+(1/8) = 2/8 = 1/4 = .25
Pr{HT=5}=Pr{One of (5,0), (5,3) Shows} =
Pr{(5,0) Shows} + Pr{(5,3) Shows}=(1/8)+(1/8) = 2/8 = 1/4 = .25
Case Two: Conditional
Probability, Color Slot Machine
Here is our slot
machine – on each trial, it produces a 10-color sequence, using the table
below:
|
Sequence* |
Probability |
|
RRBBRRYRGG |
.10 |
|
RRGGRGBRRB |
.10 |
|
BBYYGGYGBR |
.15 |
|
GRRGGYBRGG |
.10 |
|
BGYYYRYGYY |
.25 |
|
RRYYGRRBBY |
.10 |
|
YYGBYYBGRR |
.20 |
|
Total |
1.00 |
*B-Blue, G-Green, R-Red, Y-Yellow,
Sequence is numbered as 1st to 6th , from left to right:
(1st 2nd 3rd 4th 5th6th7th
8th 9th 10th )
Compute the following conditional
probabilities: Show full work and detail
for full credit.
1. Pr{“GBR” Shows | Yellow Shows}
2. Pr{Green Shows | “BR” Shows }
3. Pr{Blue Shows | Red Shows}
1.
Pr{Yellow Shows}
|
Sequence* |
Probability |
|
RRBBRRYRGG |
.10 |
|
BBYYGGYGBR |
.15 |
|
GRRGGYBRGG |
.10 |
|
BGYYYRYGYY |
.25 |
|
RRYYGRRBBY |
.10 |
|
YYGBYYBGRR |
.20 |
|
Total |
.90 |
Pr{Yellow
Shows} = Pr{One of
RRBBRRYRGG,BBYYGGYGBR,GRRGGYBRGG,BGYYYRYGYY,RRYYGRRBBY,YYGBYYBGRR Shows} =
Pr{RRBBRRYRGG
Shows}+ Pr{BBYYGGYGBR Shows}+ Pr{GRRGGYBRGG Shows}+ Pr{BGYYYRYGYY Shows}+
Pr{RRYYGRRBBY Shows}+ Pr{YYGBYYBGRR Shows} = .10+.15+.10+.25+.10+.20 = .90
Pr{“GBR” and
Yellow Show}
|
Sequence* |
Probability |
|
BBYYGGYGBR |
.15 |
|
Total |
.15 |
Pr{“GBR” and
Yellow Show} = Pr{BBYYGGYGBR Shows} =.15
Pr{“GBR” Shows
| Yellow Shows}= Pr{“GBR” and Yellow Show}/ Pr{Yellow Shows} = .15/.90 = 1/6 » .1667
2.
Pr{“BR” Shows}
|
Sequence* |
Probability |
|
RRBBRRYRGG |
.10 |
|
RRGGRGBRRB |
.10 |
|
BBYYGGYGBR |
.15 |
|
GRRGGYBRGG |
.10 |
|
Total |
.45 |
Pr{“BR”
Shows}=Pr{One of RRBBRRYRGG, RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG Shows} =
Pr{RRBBRRYRGG Shows} +
Pr{RRGGRGBRRB
Shows}+Pr{BBYYGGYGBR Shows}+ Pr{GRRGGYBRGG Shows} = .10+.10+.15+.10 =.45
Pr{Green and
“BR” Show}
|
Sequence* |
Probability |
|
RRBBRRYRGG |
.10 |
|
RRGGRGBRRB |
.10 |
|
BBYYGGYGBR |
.15 |
|
GRRGGYBRGG |
.10 |
|
Total |
.45 |
Pr{Green and
“BR” Show}=Pr{One of RRBBRRYRGG, RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG Shows} =
Pr{RRBBRRYRGG Shows} +
Pr{RRGGRGBRRB
Shows}+Pr{BBYYGGYGBR Shows}+ Pr{GRRGGYBRGG Shows} = .10+.10+.15+.10 =.45
Pr{Green
Shows | “BR” Shows } = Pr{Green and “BR” Show} / Pr{ “BR” Shows } =
.45/.45 = 1
3.
Pr{Red Shows}
|
Sequence* |
Probability |
|
RRBBRRYRGG |
.10 |
|
RRGGRGBRRB |
.10 |
|
BBYYGGYGBR |
.15 |
|
GRRGGYBRGG |
.10 |
|
BGYYYRYGYY |
.25 |
|
RRYYGRRBBY |
.10 |
|
YYGBYYBGRR |
.20 |
|
Total |
1.00 |
Pr{Red Shows} =
Pr{ One of
RRBBRRYRGG,RRGGRGBRRB,BBYYGGYGBR,GRRGGYBRGG,BGYYYRYGYY,RRYYGRRBBY,YYGBYYBGRR
Shows} =
Pr{RRBBRRYRGG Shows}+Pr{RRGGRGBRRB
Shows}+Pr{BBYYGGYGBR Shows}+Pr{GRRGGYBRGG Shows}+
Pr{BGYYYRYGYY Shows}+Pr{RRYYGRRBBY Shows}+Pr{YYGBYYBGRR
Shows} = .10+.10+.15+.10+.25+.10+.20=1.00
Pr{Blue and Red
Show}
|
Sequence* |
Probability |
|
RRBBRRYRGG |
.10 |
|
RRGGRGBRRB |
.10 |
|
BBYYGGYGBR |
.15 |
|
GRRGGYBRGG |
.10 |
|
BGYYYRYGYY |
.25 |
|
RRYYGRRBBY |
.10 |
|
YYGBYYBGRR |
.20 |
|
Total |
1.00 |
Pr{Blue and Red
Show} =
Pr{ One of
RRBBRRYRGG,RRGGRGBRRB,BBYYGGYGBR,GRRGGYBRGG,BGYYYRYGYY,RRYYGRRBBY,YYGBYYBGRR
Shows} =
Pr{RRBBRRYRGG Shows}+Pr{RRGGRGBRRB
Shows}+Pr{BBYYGGYGBR Shows}+Pr{GRRGGYBRGG Shows}+
Pr{BGYYYRYGYY Shows}+Pr{RRYYGRRBBY
Shows}+Pr{YYGBYYBGRR Shows} = .10+.10+.15+.10+.25+.10+.20=1.00
Pr{Blue Shows |
Red Shows} = Pr{Blue and Red Show} / Pr{Red Shows} = 1/1 = 1
Case Three: Long Run Argument, Perfect
Samples, Kerpuztin’s Syndrome, Type C
Severity of cases
of Kerpuztin’s Syndrome, Type C (KSC) is noted as: (F)atal, (S)evere,
(Mo)derate, or (Mi)ld. Suppose that severity levels for the population of
Kerpuztin’s Syndrome (KS) patients are distributed as indicated below:
|
Severity of Case |
Probability |
|
Fatal |
.30 |
|
Severe |
.10 |
|
Moderate |
.35 |
|
Mild |
.25 |
|
Total |
1.00 |
In our experiment, we draw individual patients (with replacement)
from the KS patient population, noting the severity of the case.
1. Interpret the probabilities in terms of
repeated trials of draws with replacement from the KSC patient population.
2. Describe the perfect sample for 800
draws with replacement from the KSC patient population. Briefly describe the
relationship between this perfect sample and actual samples of 800 draws with
replacement from the KSC patient population.
1.
In long runs of sampling with replacement,
approximately 30% of sampled patients diagnosed with Kerpuztin’s Syndrome Type
C die.
In long runs of sampling with replacement,
approximately 10% of sampled patients diagnosed with Kerpuztin’s Syndrome Type
C present severe symptoms.
In long runs of sampling with replacement,
approximately 35% of sampled patients diagnosed with Kerpuztin’s Syndrome Type
C present moderate symptoms.
In long runs of sampling with replacement,
approximately 25% of sampled patients diagnosed with Kerpuztin’s Syndrome Type
C present mild symptoms.
2.
|
Severity of Case |
Probability |
Perfect Count n=800 |
|
Fatal |
.30 |
800*.30=240 |
|
Severe |
.10 |
800*.10=80 |
|
Moderate |
.35 |
800*.35=280 |
|
Mild |
.25 |
800*.25=200 |
|
Total |
1.00 |
800*1=800 |
“Expected Fatal Count for n=800” = 800*Pr{Fatal}
= 800*.30 = 240
“Expected Fatal Count for n=800” =
800*Pr{Severe} = 800*.10 = 80
“Expected Fatal Count for n=800” =
800*Pr{Moderate} = 800*.35 = 280
“Expected Fatal Count for n=800” =
800*Pr{Mild} = 800*.25 = 200
In random samples of 800 patients diagnosed
with Kerpuztin’s Syndrome Type C with replacement, approximately 240 sampled
patients die.
In random samples of 800 patients diagnosed
with Kerpuztin’s Syndrome Type C with replacement, approximately 80 sampled
patients present severe symptoms.
In random samples of 800 patients diagnosed
with Kerpuztin’s Syndrome Type C with replacement, approximately 280 sampled
patients present moderate symptoms.
In random samples of 800 patients diagnosed
with Kerpuztin’s Syndrome Type C with replacement, approximately 200 sampled
patients present mild symptoms.
Show
full work and detail for full credit.
Case Four: Probability
Computation Rules, Color Slot Machine
Here is our slot
machine – on each trial, it produces a 10-color sequence, using the table
below:
|
Sequence* |
Probability |
|
RRBBRRYRGG |
.10 |
|
RRGGRGBRRB |
.10 |
|
BBYYGGYGBR |
.15 |
|
GRRGGYBRGG |
.10 |
|
BGYYYRYGYY |
.25 |
|
RRYYGRRBBY |
.10 |
|
YYGBYYBGRR |
.20 |
|
Total |
1.00 |
1. Define the random variable REDCOUNT as the number of times red shows
in the sequence. Compute Pr{ REDCOUNT = 3 or 4}.
2. Compute Pr{“BG” Shows}.
3. Compute Pr{“Yellow” Shows}. Use the Complementary rule.
1.
|
Sequence* |
REDCOUNT |
Probability |
|
RRBBRRYRGG |
5 |
.10 |
|
RRGGRGBRRB |
5 |
.10 |
|
BBYYGGYGBR |
1 |
.15 |
|
GRRGGYBRGG |
3 |
.10 |
|
BGYYYRYGYY |
1 |
.25 |
|
RRYYGRRBBY |
4 |
.10 |
|
YYGBYYBGRR |
2 |
.20 |
|
Total |
|
1.00 |
Pr{
REDCOUNT = 3 or 4} = Pr{
REDCOUNT = 3} + Pr{ REDCOUNT =
4} = Pr{ GRRGGYBRGG Shows} + Pr{ RRYYGRRBBY Shows}
= .10 + .10 = .20
2.
|
Sequence* |
“BG” Shows? |
Probability |
|
RRBBRRYRGG |
No |
.10 |
|
RRGGRGBRRB |
No |
.10 |
|
BBYYGGYGBR |
No |
.15 |
|
GRRGGYBRGG |
No |
.10 |
|
BGYYYRYGYY |
Yes |
.25 |
|
RRYYGRRBBY |
No |
.10 |
|
YYGBYYBGRR |
Yes |
.20 |
|
Total |
|
1.00 |
Pr{“BG” Shows}
= Pr{One of BGYYYRYGYY, YYGBYYBGRR Shows} = Pr{BGYYYRYGYY
Shows}+Pr{YYGBYYBGRR Shows} = .25+.20 = .45
3.
|
Sequence* |
Yellow Shows? |
Probability |
|
RRBBRRYRGG |
Yes |
.10 |
|
RRGGRGBRRB |
No |
.10 |
|
BBYYGGYGBR |
Yes |
.15 |
|
GRRGGYBRGG |
Yes |
.10 |
|
BGYYYRYGYY |
Yes |
.25 |
|
RRYYGRRBBY |
Yes |
.10 |
|
YYGBYYBGRR |
Yes |
.20 |
|
Total |
|
1.00 |
Pr{Yellow Does
Not Show} = Pr{ RRGGRGBRRB Shows} = .10
Pr{Yellow
Shows} = 1 − Pr{Yellow Does Not Show} = 1 − .10 = .90
Check:
Pr{Yellow Shows} = Pr{One of RRBBRRYRGG,
BBYYGGYGBR, GRRGGYBRGG, BGYYYRYGYY, RRYYGRRBBY, YYGBYYBGRR Shows} =
Pr{RRBBRRYRGG Shows}+Pr{BBYYGGYGBR
Shows}+Pr{GRRGGYBRGG Shows}+Pr{BGYYYRYGYY Shows}+
Pr{RRYYGRRBBY Shows}+Pr{YYGBYYBGRR Shows} =
.10 + .15 + .10 + .25 + .10 + .20 = .90
Show
full work and detail for full credit. Be
certain that you have worked all four (4) cases. Show full work and detail for full
credit.