Key | 1st
Hourly | Math 1107 | Fall Semester 2010
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) and this copy of the hourly.
Do not share these resources
with anyone else. 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 and Your Toolsheet.
Then hand all of this in to me.
Sign and
Acknowledge: I agree to follow this protocol: (initial)
Sign___________________________Name______________________________Date________________
Show
full work and detail for full credit. Be sure that you have worked all four
cases.
Case One: Random Variables, Pair of Dice
We have a pair of
independently operating dice, whose
probability models are given below:
|
1st
Face Value |
Probability |
|
2nd
Face Value |
Probability |
|
1 |
4/10 |
|
3 |
5/30 |
|
2 |
3/10 |
|
4 |
10/30 |
|
6 |
2/10 |
|
5 |
15/30 |
|
7 |
1/10 |
|
Total |
30/30 |
|
Total |
10/10 |
|
|
|
Each trial of our
experiment involves tossing the pair
of dice: (1st Face Value, 2nd
Face Value), and noting the pair of face values. We compute random
variables from the pair of face values as follows: Define
Max as the
higher of the two face values in the pair, that is,
Max = Maximum of {1st Face
Value, 2nd Face Value}.
Compute the
values and probabilities for Max.
a) List all the
possible pairs of face values, and compute the probability for each pair,
showing full detail.
|
2nd ß
1st Þ |
1(4/10) |
2(3/10) |
6(2/10) |
7(1/10) |
|
3(5/30) |
(1,3) |
(2,3) |
(6,3) |
(7,3) |
|
4(10/30) |
(1,4) |
(2,4) |
(6,4) |
(7,4) |
|
5(15/30) |
(1,5) |
(2,5) |
(6,5) |
(7,5) |
(1,3),(2,3),(6,3),(7,3),(1,4),(2,4),(6,4),(7,4),(1,5),(2,5),(6,5),(7,5)
Pr{(1,3)} = Pr{1 from 1st}*Pr{3
from 2nd } = (4/10)*(5/30) =
20/300
Pr{(2,3)} = Pr{2 from 1st}*Pr{3
from 2nd } = (3/10)*(5/30) = 15/300
Pr{(6,3)} = Pr{6 from 1st}*Pr{3
from 2nd } = (2/10)*(5/30 = 10/300
Pr{(7,3)} = Pr{7 from 1st}*Pr{3
from 2nd } = (1/10)*(5/30) =
5/300
Pr{(1,4)} = Pr{1 from 1st}*Pr{4
from 2nd } = (4/10)*(10/30) =
40/300
Pr{(2,4)} = Pr{2 from 1st}*Pr{4
from 2nd } = (3/10)*(10/30) = 30/300
Pr{(6,4)} = Pr{6 from 1st}*Pr{4
from 2nd } = (2/10)*(10/30 = 20/300
Pr{(7,4)} = Pr{7 from 1st}*Pr{4
from 2nd } = (1/10)*(10/30) =
10/300
Pr{(1,5)} = Pr{1 from 1st}*Pr{5
from 2nd } = (4/10)*(15/30) =
60/300
Pr{(2,5)} = Pr{2 from 1st}*Pr{5
from 2nd } = (3/10)*(15/30) = 45/300
Pr{(6,5)} = Pr{6 from 1st}*Pr{5
from 2nd } = (2/10)*(15/30 = 30/300
Pr{(7,5)} = Pr{7 from 1st}*Pr{5
from 2nd } = (1/10)*(15/30) =
15/300
b) Compute the values and probabilities for
the random variable Max, showing full detail.
Max{(1,3)} = 3 (@20/300)
Max{(2,3)} = 3 (@15/300)
Max{(6,3)} = 6 (@10/300)
Max{(7,3)} = 7 (@5/300)
Max{(1,4)} = 4 (@40/300)
Max{(2,4)} = 4 (@30/300)
Max{(6,4)} = 6 (@20/300)
Max{(7,4)} = 7 (@10/300)
Max{(1,5)} = 5 (@60/300)
Max{(2,5)} = 5 (@45/300)
Max{(6,5)} = 6 (@30/300)
Max{(7,5)} = 7 (@15/300)
Pr{Max=7} =
Pr{One of (7,3),(7,3),(7,5) Shows} =
Pr{(7,3)}+Pr{(7,4)}+Pr{(7,5)} = (5/300) +
(10/300) + (15/300) = 30/300
Pr{Max=6} =
Pr{One of (6,3),(6,3),(6,5) Shows} =
Pr{(6,3)}+Pr{(6,4)}+Pr{(6,5)} = (10/300) +
(20/300) + (30/300) = 60/300
Pr{Max=5} =
Pr{One of (1,5),(2,5) Shows} =
Pr{(1,5)}+Pr{(2,5)} = (60/300) + (45/300) =
105/300
Pr{Max=4} =
Pr{One of (1,4),(2,4) Shows} =
Pr{(1,4)}+Pr{(2,4)} = (40/300) + (30/300) =
70/300
Pr{Max=3} =
Pr{One of (1,3),(2,3) Shows} =
Pr{(1,3)}+Pr{(2,3)} = (20/300) + (15/300) =
35/300
Show full work and detail
for full credit.
Case Two, Long Run Argument and Perfect Samples
Consider the population of
resident live births in the United States in year 2003 – suppose that the
following probability model accurately describes the distribution of year 2003
United States resident live births by birth weight ( in grams ) category:
|
Birthweight
(grams) |
Probability |
|
strictly less than 1500 |
0.00722 |
|
1500-2499 |
0.06520 |
|
2500-3499 |
0.55900 |
|
3500 or more |
0.36800 |
|
Total |
1 |
a) Interpret each probability
using the Long Run Argument.
In long runs of random sampling from year
2003 US resident live births, approximately 0.72% of sampled live births
present birth weights of strictly less than 1500 grams.
In long runs of random sampling from year
2003 US resident live births, approximately 6.52% of sampled live births
present birth weights between 1500 and 2499 grams.
In long runs of random sampling from year
2003 US resident live births, approximately 55.9% of sampled live births
present birth weights between 2500 and 3499 grams.
In long runs of random sampling from year
2003 US resident live births, approximately 36.8% of sampled live births
present birth weights of 3500 or more grams.
b) Compute and discuss
Perfect Samples for n=4000.
|
Birthweight (grams) |
Probability |
Expectedn=4000 |
|
strictly less
than 1500 |
0.00722 |
4000*0.00722 » 28.88 |
|
1500-2499 |
0.0652 |
4000*0.0652 » 260.8 |
|
2500-3499 |
0.559 |
4000*0.559 = 2236 |
|
3500 or more |
0.368 |
4000*0.368 = 1472 |
|
Total |
1 |
4000 |
Expectedn=4000, <1500 = 4000*0.00722 » 28.88
In random samples of 4,000 year 2003 US
resident live births, approximately 28 or 29 of 4,000 sampled births present
birth weights of strictly less than 1500 grams.
Expectedn=4000, [1500,2500) =
4000*0.0652 » 260.8
In random samples of 4,000 year 2003 US
resident live births, approximately 260 or 261 of 4,000 sampled live births present birth weights between
1500 and 2499 grams.
Expectedn=4000, [2500,3500) =
4000*0.559 » 2236
In random samples of 4,000 year 2003 US
resident live births, approximately 2,236 of 4,000 sampled live births present
birth weights between 2500 and 3499 grams.
Expectedn=4000, 3500+ = 4000*0.368 » 1472
In random samples of 4,000 year 2003 US
resident live births, approximately 1,472 of ,4000
sampled live births present birth weights of 3500 or more grams.
Show all work and full detail for full
credit. Provide complete discussion for full credit.
Case Three |
Color Slot Machine | Probability Rules
Here is our slot machine – on
each trial, it produces a color sequence, using the table below:
|
Sequence* |
Probability |
Sequence* |
Probability |
|
BBRRYRRR |
.10 |
BGYRYGYY |
.25 |
|
GGRGBRRB |
.10 |
YYGRRBBY |
.10 |
|
BBYYGGBR |
.15 |
YBYYBGRB |
.10 |
|
GRRYBRGG |
.10 |
GRYYYRGR |
.10 |
*B-Blue, G-Green, R-Red, Y-Yellow
Compute the following conditional
probabilities. Show full work and detail for full credit.
Pr{ Blue Shows Exactly Three Times }
|
Sequence* |
Probability |
Sequence* |
Probability |
|
BBRRYRRR |
.10 |
BGYRYGYY |
.25 |
|
GGRGBRRB |
.10 |
YYGRRBBY |
.10 |
|
BBYYGGBR |
.15 |
YBYYBGRB |
.10 |
|
GRRYBRGG |
.10 |
GRYYYRGR |
.10 |
Pr{ Blue Shows Exactly Three
Times } =
Pr{One of BBYYGGBR or YBYYBGRB Shows} =
Pr{BBYYGGBR} + Pr{YBYYBGRB}
= 0.15 + 0.10 = 0.25
Pr{ “BBY” Shows }
|
Sequence* |
Probability |
Sequence* |
Probability |
|
BBRRYRRR |
.10 |
BGYRYGYY |
.25 |
|
GGRGBRRB |
.10 |
YYGRRBBY |
.10 |
|
BBYYGGBR |
.15 |
YBYYBGRB |
.10 |
|
GRRYBRGG |
.10 |
GRYYYRGR |
.10 |
Pr{ “BBY” Shows } =
Pr{One of BBYYGGBR or YYGRRBBY Shows}
=
Pr{ BBYYGGBR
} + Pr{ YYGRRBBY } = 0.15 + 0.10 = 0.25
Pr{ Blue Shows } – Use the Complementary Rule
|
Sequence* |
Probability |
Sequence* |
Probability |
|
BBRRYRRR |
.10 |
BGYRYGYY |
.25 |
|
GGRGBRRB |
.10 |
YYGRRBBY |
.10 |
|
BBYYGGBR |
.15 |
YBYYBGRB |
.10 |
|
GRRYBRGG |
.10 |
GRYYYRGR |
.10 |
Other Event = OE = “Blue Does Not Show”
Pr{“Blue Does Not Show”} = Pr{ GRYYYRGR} = 0.10
Pr{ Blue Shows } = 1 – Pr{ Blue
Does Not Show } = 1 – 0.10 = 0.90
Case Four |
Conditional Probability | Color Slot Machine
Here is our slot
machine – on each trial, it produces an 8-color sequence, using the table
below:
|
Sequence* |
Probability |
Sequence* |
Probability |
|
RRBBRRYR |
.35 |
YYYRYGYY |
.14 |
|
BYGGYGBR |
.20 |
RYGRRBBY |
.21 |
|
GRGYBRGG |
.10 |
|
|
*B-Blue, G-Green, R-Red, Y-Yellow
Compute the following conditional probabilities. Show full work and detail for full credit.
a) Pr{“RRB” Shows | Green Shows}
|
Sequence* |
Probability |
Sequence* |
Probability |
|
YYYRYGYY |
.14 |
||
|
BYGGYGBR |
.20 |
RYGRRBBY |
.21 |
|
GRGYBRGG |
.10 |
|
|
Pr{Green
Shows} = Pr{One of BYGGYGBR, GRGYBRGG, YYYRYGYY or RYGRRBBY Shows} =
= Pr{BYGGYGBR}+Pr{GRGYBRGG}+Pr{YYYRYGYY}+Pr{RYGRRBBY} =
0.20+0.10+0.14+0.21 = 0.65
|
Sequence* |
Probability |
Sequence* |
Probability |
|
|
|
||
|
RYGRRBBY |
.21 |
||
|
|
|
Pr{“RRB”
and Green Show} = Pr{RYGRRBBY Shows} = 0.21
Pr{“RRB”
Shows | Green Shows} = Pr{“RRB” and Green Show}/Pr{Green Shows} = 0.21/0.65
b) Pr{ “BR” Shows | Green Shows }
|
Sequence* |
Probability |
Sequence* |
Probability |
|
YYYRYGYY |
.14 |
||
|
BYGGYGBR |
.20 |
RYGRRBBY |
.21 |
|
GRGYBRGG |
.10 |
|
|
Pr{Green
Shows} = Pr{One of BYGGYGBR, GRGYBRGG, YYYRYGYY or RYGRRBBY Shows} =
= Pr{BYGGYGBR}+Pr{GRGYBRGG}+Pr{YYYRYGYY}+Pr{RYGRRBBY} =
0.20+0.10+0.14+0.21 = 0.65
|
|
|
||
|
BYGGYGBR |
.20 |
|
|
|
GRGYBRGG |
.10 |
|
|
Pr{“BR”
and Green Show} = Pr{One of BYGGYGBR or, GRGYBRGG Shows} =
= Pr{BYGGYGBR}+Pr{GRGYBRGG} = 0.20+0.10 = 0.30
Pr{“RRB”
Shows | Green Shows} = Pr{“RRB” and Green Show}/Pr{Green Shows} = 0.30/0.65
c) Pr{ Red Shows | Blue Shows}
|
Sequence* |
Probability |
Sequence* |
Probability |
|
RRBBRRYR |
.35 |
|
|
|
BYGGYGBR |
.20 |
RYGRRBBY |
.21 |
|
GRGYBRGG |
.10 |
|
|
Pr{Blue
Shows} = Pr{One of RRBBRRYR, BYGGYGBR, GRGYBRGG or RYGRRBBY Shows} =
= Pr{ RRBBRRYR
}+Pr{ BYGGYGBR }+Pr{ GRGYBRGG }+Pr{ RYGRRBBY } = 0.35 + 0.20+0.10+0.21 = 0.86
|
Sequence* |
Probability |
Sequence* |
Probability |
|
RRBBRRYR |
.35 |
|
|
|
BYGGYGBR |
.20 |
RYGRRBBY |
.21 |
|
GRGYBRGG |
.10 |
|
|
Pr{Red
and Blue Shows} = Pr{One of RRBBRRYR, BYGGYGBR, GRGYBRGG or RYGRRBBY Shows} =
= Pr{ RRBBRRYR
}+Pr{ BYGGYGBR }+Pr{ GRGYBRGG }+Pr{ RYGRRBBY } = 0.35 + 0.20+0.10+0.21 = 0.86
Pr{
Red Shows | Blue Shows} = Pr{ Red and Blue Show}/ Pr{Blue Shows} = 0.86/0.86
Work all four cases.