Key
The 1st Hourly
Math 1107
Spring Semester 2008
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 Variable, Fair Dice
We have a triplet of independently
operating fair dice: the first die
has face values 1, 2, 5; the second die has face values 2, 3 and the third die
has face values 3, 4. On each trial of our experiment, we toss the three dice, and note the
triplet of face values.
1. List or otherwise display
all the possible quartets of face values, and compute the probability for each
quartet,
showing full detail. Hint: the
calculation for a pair of dice looks like Pr{(face1, face2)} =
Pr{face1}*Pr(face2} – the calculation for a triplet looks like Pr{(face1,
face2,face3)} = Pr{face1}*Pr(face2}*Pr(face3}.
Triplets are (1,2,3), (1,2,4), (1,3,3),
(1,3,4), (2,2,3), (2,2,4), (2,3,3), (2,3,4), (5,2,3), (5,2,4), (5,3,3),
(5,3,4).
Triplet Probabilities
|
1st Die=1 |
2nd Die=2 |
2nd Die=3 |
|
3rd Die=3 |
(1,2,3) |
(1,3,3) |
|
3rd Die=4 |
(1,2,4) |
(1,3,4) |
Pr{(1,2,3)} = Pr{1 from 1st
Die}*Pr{2 from 2nd Die}*Pr{3 from 3rd Die} =
(1/3)*(1/2)*(1/2) = 1/12
Pr{(1,2,4)} = Pr{1 from 1st
Die}*Pr{2 from 2nd Die}*Pr{4 from 3rd Die} =
(1/3)*(1/2)*(1/2) = 1/12
Pr{(1,3,3)} = Pr{1 from 1st
Die}*Pr{3 from 2nd Die}*Pr{3 from 3rd Die} =
(1/3)*(1/2)*(1/2) = 1/12
Pr{(1,3,4)} = Pr{1 from 1st
Die}*Pr{3 from 2nd Die}*Pr{4 from 3rd Die} =
(1/3)*(1/2)*(1/2) = 1/12
|
1st Die=2 |
2nd Die=2 |
2nd Die=3 |
|
3rd Die=3 |
(2,2,3) |
(2,3,3) |
|
3rd Die=4 |
(2,2,4) |
(2,3,4) |
Pr{(2,2,3)} = Pr{2 from 1st
Die}*Pr{2 from 2nd Die}*Pr{3 from 3rd Die} =
(1/3)*(1/2)*(1/2) = 1/12
Pr{(2,2,4)} = Pr{2 from 1st
Die}*Pr{2 from 2nd Die}*Pr{4 from 3rd Die} =
(1/3)*(1/2)*(1/2) = 1/12
Pr{(2,3,3)} = Pr{2 from 1st
Die}*Pr{3 from 2nd Die}*Pr{3 from 3rd Die} =
(1/3)*(1/2)*(1/2) = 1/12
Pr{(2,3,4)} = Pr{2 from 1st
Die}*Pr{3 from 2nd Die}*Pr{4 from 3rd Die} =
(1/3)*(1/2)*(1/2) = 1/12
|
1st Die=5 |
2nd Die=2 |
2nd Die=3 |
|
3rd Die=3 |
(5,2,3) |
(5,3,3) |
|
3rd Die=4 |
(5,2,4) |
(5,3,4) |
Pr{(5,2,3)} = Pr{5 from 1st
Die}*Pr{2 from 2nd Die}*Pr{3 from 3rd Die} =
(1/3)*(1/2)*(1/2) = 1/12
Pr{(5,2,4)} = Pr{5 from 1st
Die}*Pr{2 from 2nd Die}*Pr{4 from 3rd Die} =
(1/3)*(1/2)*(1/2) = 1/12
Pr{(5,3,3)} = Pr{5 from 1st
Die}*Pr{3 from 2nd Die}*Pr{3 from 3rd Die} =
(1/3)*(1/2)*(1/2) = 1/12
Pr{(5,3,4)} = Pr{5 from 1st
Die}*Pr{3 from 2nd Die}*Pr{4 from 3rd Die} =
(1/3)*(1/2)*(1/2) = 1/12
2. Define HIGHTIE as the highest of the three face values – if two or more of the face
values tie at the
same high value, use the
common value. Compute the values of the
random variable HIGHTIE,
showing in detail how these
values are computed from each triplet of face values.
HIGHTIE((1,2,3))=3
HIGHTIE((1,2,4))=4
HIGHTIE((1,3,3))=3
HIGHTIE((1,3,4))=4
HIGHTIE((2,2,3))=3
HIGHTIE((2,2,4))=4
HIGHTIE((2,3,3))=3
HIGHTIE((2,3,4))=4
HIGHTIE((5,2,3))=5
HIGHTIE((5,2,4))=5
HIGHTIE((5,3,3))=5
HIGHTIE((5,3,4))=5
3. Compute the probability
for each value of the random variable HIGHTIE, showing full detail – if you do this
correctly, you should have one probability for each value of
Pr{HIGHTIE=3}=Pr{One of (1,2,3), (1,3,3),
(2,2,3), (2,3,3) Shows} = Pr{(1,2,3)}+Pr{ (1,3,3) }+Pr{ (2,2,3) }+Pr{ (2,3,3)}=
(1/12)+(1/12)+(1/12)+(1/12) = 4/12 = 1/3 @ .3333
Pr{HIGHTIE=4}=Pr{One of (1,2,4), (1,3,4),
(2,2,4), (2,3,4) Shows} = Pr{(1,2,4)}+Pr{ (1,3,4) }+Pr{ (2,2,4) }+Pr{ (2,3,4)}=
(1/12)+(1/12)+(1/12)+(1/12) = 4/12 = 1/3 @ .3333
Pr{HIGHTIE=5}=Pr{One of (5,2,3), (5,2,4),
(5,3,3), (5,3,4) Shows} = Pr{(5,2,3)}+Pr{ (5,3,3) }+Pr{ (5,2,3) }+Pr{ (5,3,4)}=
(1/12)+(1/12)+(1/12)+(1/12) = 4/12 = 1/3 @ .3333
Show all work
and full detail for full credit.
Case Two: Long Run Argument and Perfect Samples, Consumer
Credit Score
Fair Isaac Corporation developed a consumer credit score, a number
that summarizes the risk present in lending money to a consumer. The consumer
credit score ranges from 300 to 850. Credit bureau scores are often called
“FICO scores” because most credit bureau scores used in the U.S. are produced
from software developed by Fair Isaac and Company. FICO scores are provided to
lenders by the major credit reporting agencies. Suppose that the probabilities for consumer credit scores for US
residents are noted below:
|
Credit Score
Range |
549 and below |
550-599 |
600-699 |
700-799 |
800 and above |
|
Probability |
.07 |
.08 |
.27 |
.45 |
.13 |
1. Interpret each
probability using the Long Run Argument.
In long runs of sampling with replacement, approximately
7% of sampled US residents present a FCIO score of 549 or below.
In long runs of sampling with replacement,
approximately 8% of sampled US residents present a FCIO score between 550 and
599.
In long runs of sampling with replacement,
approximately 27% of sampled US residents present a FCIO score between 600 and
699.
In long runs of sampling with replacement,
approximately 45% of sampled US residents present a FCIO score between 700 and
799.
In long runs of sampling with replacement,
approximately 13% of sampled US residents present a FCIO score between 800 or
above.
2. Compute the
perfect sample of n=3000 US residents, and describe the relationship of this
perfect sample to real random samples of US residents.
Perfect Count for “549 or below” = 3000*.07
= 210
Perfect Count for “between 550 and 599” = 3000*.08 = 240
Perfect Count for “between 600 and 699” = 3000*.27 = 810
Perfect Count for “between 700 and 799” = 3000*.45 = 1350
Perfect Count for “800 or above” = 3000*.13 = 390
In random samples of 3,000 US residents, approximately 210 of 3,000
sampled US residents present FICO scores at or below 549.
In random samples of 3,000 US residents, approximately 240 of 3,000
sampled US residents present a FCIO score between 550 and 599.
In random samples of 3,000 US residents, approximately 810 of 3,000
sampled US residents present a FCIO score between 600 and 699.
In random samples of 3,000 US residents, approximately 1350 of
3,000 sampled US residents present a FCIO score between 700 and 799.
In random samples of 3,000 US residents, approximately 390 of 3,000
sampled US residents present a FCIO score between 800 or above.
Show all work
and full detail for full credit. Provide complete discussion for full credit.
Case Three: Color
Slot Machine, Computation of Conditional Probabilities
Here is our slot machine – on each trial, it produces a 10-color
sequence, using the table below:
|
Sequence* |
Probability |
|
RRBBR RYRRB |
.10 |
|
RRGGRGBRRB |
.10 |
|
BBYYRGYGBR |
.15 |
|
GRRGRGBRGB |
.10 |
|
BGYGYRYGYY |
.25 |
|
RRGYGRRBBB |
.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:
1. Pr{Red Shows
Somewhere in the 1st ─ 4th slots | Yellow Shows
Somewhere in the 7th ─ 10th slots}
Pr{Red Shows in the 1st – 4th
slots|Yellow Shows in the 7th – 10th slots} =
Pr{Red Shows in the 1st – 4th
slots and Yellow Shows in the 7th – 10th slots}/
Pr{ Yellow Shows in the 7th – 10th slots}
|
Sequence* |
Probability |
|
RRBBR RYRRB |
.10 |
|
BBYYRGYGBR |
.15 |
|
BGYGYRYGYY |
.25 |
|
Total |
0.50 |
Pr{ Yellow Shows in the 7th – 10th
slots} = Pr{One of RRBBRRYRRB, BBYYRGYGBR, BGYGYRYGYY shows} =
Pr{RRBBRRYRRB}+ Pr{BBYYRGYGBR}+
Pr{BGYGYRYGYY} = .10+.15+.25 = .50
|
Sequence* |
Probability |
|
RRBBR RYRRB |
.10 |
|
Total |
0.10 |
Pr{ Red Shows in the 1st – 4th
slots and Yellow Shows in the 7th – 10th slots } =
Pr{One of RRBBRRYRRB shows} = .10
Pr{Red Shows in the 1st – 4th
slots|Yellow Shows in the 7th – 10th slots} = .10/.50 =
.20
2. Pr{Green Shows
Anywhere | “RB” Shows Anywhere}
Pr{Green Shows Anywhere|”RB” Shows Anywhere}
=
Pr{ Green Shows Anywhere and ”RB”
Shows Anywhere }/ Pr{”RB” Shows Anywhere}
|
Sequence* |
Probability |
|
RRBBR RYRRB |
.10 |
|
RRGGRGBRRB |
.10 |
|
RRGYGRRBBB |
.10 |
|
Total |
1.00 |
Pr{”RB” Shows Anywhere} = Pr{One of RRBBRRYRRB, RRGGRGBRRB,
RRGYGRRBBB Shows} = Pr{RRBBRRYRRB}+Pr{RRGGRGBRRB}+Pr{RRGYGRRBBB}
=.10+.10+.10 = .30
|
Sequence* |
Probability |
|
RRGGRGBRRB |
.10 |
|
RRGYGRRBBB |
.10 |
|
Total |
0.20 |
Pr{ Green Shows Anywhere and ”RB”
Shows Anywhere } = Pr{One of RRGGRGBRRB,
RRGYGRRBBB Shows} = Pr{RRGGRGBRRB}+Pr{RRGYGRRBBB} =.10+.10
= .20
Pr{Green Shows Anywhere|”RB” Shows Anywhere}
= .20/.30
3. Pr{Yellow
Shows Anywhere | Blue Shows Anywhere}
Pr{Yellow Shows Anywhere | Blue Shows
Anywhere} =
Pr{Yellow Shows Anywhere and Blue
Shows Anywhere}/Pr{ Blue Shows Anywhere}
|
Sequence* |
Probability |
|
RRBBR RYRRB |
.10 |
|
RRGGRGBRRB |
.10 |
|
BBYYRGYGBR |
.15 |
|
GRRGRGBRGB |
.10 |
|
BGYGYRYGYY |
.25 |
|
RRGYGRRBBB |
.10 |
|
YYGBYYBGRR |
.20 |
|
Total |
1.00 |
Pr{ Blue Shows Anywhere} = Pr{one of
RRBBRRYRRB, RRGGRGBRRB, BBYYRGYGBR, GRRGRGBRGB, BGYGYRYGYY, RRGYGRRBBB,
YYGBYYBGRR Shows} =Pr{RRBBRRYRRB}+Pr{RRGGRGBRRB}+Pr{ BBYYRGYGBR}+Pr{GRRGRGBRGB}+Pr{BGYGYRYGYY}+Pr{RRGYGRRBBB}+Pr{YYGBYYBGRR}
= .10+.10+.15+.10+.25+.10+.20 = 1.00
|
Sequence* |
Probability |
|
RRBBR RYRRB |
.10 |
|
BBYYRGYGBR |
.15 |
|
BGYGYRYGYY |
.25 |
|
RRGYGRRBBB |
.10 |
|
YYGBYYBGRR |
.20 |
|
Total |
0.80 |
Pr{Yellow Shows Anywhere and Blue
Shows Anywhere} = Pr{one of RRBBRRYRRB, BBYYRGYGBR, BGYGYRYGYY, RRGYGRRBBB,
YYGBYYBGRR Shows} =Pr{RRBBRRYRRB}+ Pr{
BBYYRGYGBR}+Pr{BGYGYRYGYY}+Pr{RRGYGRRBBB}+Pr{YYGBYYBGRR} = .10+.15+.25+.10+.20
= .80
Pr{Yellow Shows Anywhere | Blue Shows
Anywhere} = .80/1.00 = .80
Show complete
detail and work for full credit. Be sure that you have worked all four cases.
Case Four: 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 |
0.125 |
|
BBGG |
0.075 |
|
GBBR |
0.350 |
|
BBBB |
0.040 |
|
RGYB |
0.080 |
|
YYYY |
0.030 |
|
BBYY |
0.140 |
|
RRYY |
0.070 |
|
YBGR |
0.090 |
|
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.
Pr{Either Blue or Yellow Shows, but not both at the same
time}
|
Color Sequence* |
Color Sequence Probability |
|
GBRB |
0.125 |
|
BBGG |
0.075 |
|
GBBR |
0.350 |
|
BBBB |
0.040 |
|
YYYY |
0.030 |
|
RRYY |
0.070 |
|
Total |
0.690 |
Pr{Either Blue or Yellow Shows, But not at
the Same Time} = Pr{ One of GBRB, BBGG, GBBR, BBBB, YYYY, RRYY Shows } = Pr{GBRB}+
Pr{BBGG}+ Pr{GBBR}+ Pr{BBBB}+ Pr{YYYY}+ Pr{RRYY} =
0.125+0.075+0.350+0.040+0.030+0.070 = 0.690
2. Pr{Red and Green both Show} – Use the Complementary Rule
Event=”Red and Green Both Show”
Other Event = “Either Red Shows and Green Doesn’t,
or Green Shows and Red Doesn’t, or Neither Green nor Red Shows”
|
Color Sequence* |
Color Sequence Probability |
|
n- .BBGG |
0.075 |
|
BBBB |
0.040 |
|
YYYY |
0.030 |
|
BBYY |
0.140 |
|
RRYY |
0.070 |
|
Total |
0.355 |
Pr{ Either Red Shows and Green Doesn't, or
Green Shows and Red Doesn’t, or Neither Green nor Red Shows} =
Pr{One of BBGG, BBBB, YYYY, BBYY, RRYY
Shows} = Pr{BBGG}+Pr{BBBB}+Pr{YYYY}+Pr{BBYY}+Pr{RRYY} = 0.075+0.040+ 0.030+ 0.140 + 0.070 = 0.115+0.170+0.070 =
0.285+0.070=0.355
Pr{ Red and Green Both Show} = 1 -
Pr{ Either Red Shows and Green Doesn’t, or Green Shows and Red Doesn’t, or
Neither Green nor Red Shows } = 1 - 0.355 = 0.645
|
Color Sequence* |
Color Sequence Probability |
|
GBRB |
0.125 |
|
GBBR |
0.350 |
|
RGYB |
0.080 |
|
YBGR |
0.090 |
|
Total |
0.645 |
Check: Pr{Red and Green Both Show} = Pr{One
of GBRB, GBBR, RGYB, YBGR Shows} = Pr{GBRB}+ Pr{GBBR}+ Pr{RGYB}+ Pr{YBGR} =
0.125+0.350+0.080+0.090=0.475+0.170=0.645
Show complete detail
and work for full credit.
Be sure that you have worked all four
cases.