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, 3; 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 triplets of face values, and compute the probability for each
triplet,
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), (3,2,3), (3,2,4), (3,3,3),
(3,3,4).
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=3 |
2nd Die=2 |
2nd Die=3 |
|
3rd Die=3 |
(3,2,3) |
(3,3,3) |
|
3rd Die=4 |
(3,2,4) |
(3,3,4) |
Pr{(3,2,3)} = Pr{3 from 1st
Die}*Pr{2 from 2nd Die}*Pr{3 from 3rd Die} =
(1/3)*(1/2)*(1/2) = 1/12
Pr{(3,2,4)} = Pr{3 from 1st
Die}*Pr{2 from 2nd Die}*Pr{4 from 3rd Die} =
(1/3)*(1/2)*(1/2) = 1/12
Pr{(3,3,3)} = Pr{3 from 1st Die}*Pr{3
from 2nd Die}*Pr{3 from 3rd Die} = (1/3)*(1/2)*(1/2) =
1/12
Pr{(3,3,4)} = Pr{3 from 1st
Die}*Pr{3 from 2nd Die}*Pr{4 from 3rd Die} =
(1/3)*(1/2)*(1/2) = 1/12
2. Define LOWTIE as the
lowest of the three face values – if two or more of the face values tie at the same
low value, use the common value. Compute
the values of the random variable LOWTIE, showing in detail how these
values are computed from each triplet of face values.
LOWTIE((1,2,3))=1
LOWTIE((1,2,4))=1
LOWTIE((1,3,3))=1
LOWTIE((1,3,4))=1
LOWTIE((2,2,3))=2
LOWTIE((2,2,4))=2
LOWTIE((2,3,3))=2
LOWTIE((2,3,4))=2
LOWTIE((3,2,3))=2
LOWTIE((3,2,4))=2
LOWTIE((3,3,3))=3
LOWTIE((3,3,4))=3
3. Compute the probability for each value of the random variable LOWTIE, showing full detail – if
you do this correctly, you
should have one probability for each value of LOWTIE.
Pr{LOWTIE=1} = Pr{One of (1,2,3),
(1,2,4), (1,3,3) or (1,3,4) Shows} = Pr{(1,2,3)}+ Pr{(1,2,4)}+ Pr{(1,3,3)}+
Pr{(1,3,4)} = (1/12)+(1/12) (1/12)+(1/12) = 4/12
Pr{LOWTIE=2} = Pr{One of (2,2,3),
(2,2,4), (2,3,3), (2,3,4), (3,2,3) or (3,2,4) Shows} =
Pr{(2,2,3)}+Pr{(2,2,4)}+Pr{(2,3,3)}+Pr{(2,3,4)}+Pr{(3,2,3)}+Pr{(3,2,4)}=
=(1/12)+ (1/12)+ (1/12)+ (1/12)+ (1/12)+
(1/12) = 6/12
Pr{LOWTIE=3} = Pr{One of (3,3,3) or (3,3,4)
Shows} = Pr{(3,3,3)}+Pr{(3,3,4)} = (1/12) + (1/12) = 2/12
Show all work
and full detail for full credit.
Case Two: Long Run Argument and 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. So the blood types are O, A, B, AB. Suppose
that the probabilities for blood types for US residents are noted below:
|
Blood Type |
O |
A |
B |
AB |
|
Probability |
.40 |
.30 |
.10 |
.20 |
1. Interpret each probability
using the Long Run Argument.
In long runs of sampling with replacement,
approximately 40% of sampled US residents present Landsteiner Type O blood.
In long runs of sampling with replacement, approximately
30% of sampled US residents present Landsteiner Type A blood.
In long runs of sampling with replacement,
approximately 10% of sampled US residents present Landsteiner Type B blood.
In long runs of sampling with replacement,
approximately 20% of sampled US residents present Landsteiner Type AB blood.
2. Compute the
perfect sample of n=1500 US residents, and describe the relationship of this
perfect sample to real random samples of US residents.
|
O |
A |
B |
AB |
|
|
Probability |
.40 |
.30 |
.10 |
.20 |
|
Expected Count |
1500*.40=6000 |
1500*.30=4500 |
1500*.10=1500 |
1500*.20=3000 |
Perfect Count for “Landsteiner Type O” = 1500*0.40 = 600
Perfect Count for “Landsteiner Type A” = 1500*0.30 = 450
Perfect Count for “Landsteiner Type B” = 1500*0.10 = 150
Perfect Count for “Landsteiner Type AB” = 1500*0.20 = 300
In random samples of 1,500 US residents, approximately 600 of 1,500
sampled US residents present Landsteiner Type O blood.
In random samples of 1,500 US residents, approximately 450 of 1,500
sampled US residents present Landsteiner Type A blood.
In random samples of 1,500 US residents, approximately 150 of 1,500
sampled US residents present Landsteiner Type B blood.
In random samples of 1,500 US residents, approximately 300 of 1,500
sampled US residents present Landsteiner Type AB blood.
Show all work
and full detail for full credit. Provide complete discussion for full credit.
Case Three: 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 |
|
GGGG |
0.100 |
|
GBBR |
0.250 |
|
BBBB |
0.040 |
|
RGYB |
0.080 |
|
YYYY |
0.030 |
|
BBYY |
0.140 |
|
RRYY |
0.060 |
|
RRRR |
0.010 |
|
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 |
0.250 |
|
BBBB |
0.040 |
|
YYYY |
0.030 |
|
RRYY |
0.060 |
|
Total |
0.580 |
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+0.250+0.040+0.030+0.060 = 0.580
2. Pr{Red and Green both Show} – Use the Complementary Rule
|
Color Sequence* |
Color Sequence Probability |
|
BBGG |
0.075 |
|
GGGG |
0.100 |
|
BBBB |
0.040 |
|
YYYY |
0.030 |
|
BBYY |
0.140 |
|
RRYY |
0.060 |
|
RRRR |
0.010 |
|
Total |
0.455 |
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”
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, GGGG, BBBB, YYYY, BBYY,
RRYY, RRRR Shows} = Pr{BBGG}+ Pr{GGGG}+ Pr{YYYY}+
Pr{BBBB}+Pr{BBYY}+Pr{RRYY}+Pr{RRRR} =
0.075+0.100+0.040+0.030+ 0.140 + 0.060 + 0.01=0.455
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 - 0.455 = 0.5450
|
Color Sequence* |
Color Sequence Probability |
|
GBRB |
0.125 |
|
GBBR |
0.250 |
|
RGYB |
0.080 |
|
YBGR |
0.090 |
|
Total |
0.545 |
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+0.250+0.080+0.090=0.3750+0.170 = 0.5450
Show complete
detail and work for full credit.
Case Four: 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 |
0.30 |
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.