Draft Key | The 1st Hourly |
Math 1107 | Spring Semester 2011
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) 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. Write on one side each of the sheets provided. 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 Variable, Pair of Dice
We
have a pair of dice– note the probability models for the dice
below.
|
1st d4 |
2nd d4 |
||
|
Face |
Probability |
Face |
Probability |
|
1 |
4/10 |
3 |
1/10 |
|
2 |
4/10 |
4 |
2/10 |
|
7 |
1/10 |
5 |
3/10 |
|
8 |
1/10 |
6 |
4/10 |
|
Total |
10/10=1 |
Total |
10/10=1 |
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 face-value-pair.
List
the possible pairs of face values, and compute a probability for each pair of
face values.
|
|
1(4/10) |
2(4/10) |
7(1/10) |
8(1/10) |
|
3(1/10) |
(1,3) |
(2,3) |
(7,3) |
(8,3) |
|
4(2/10) |
(1,4) |
(2,4) |
(7,4) |
(8,4) |
|
5(3/10) |
(1,5) |
(2,5) |
(7,5) |
(8,5) |
|
6(4/100 |
(1,6) |
(2,6) |
(7,6) |
(8,6) |
(1,3),(2,3),(7,3),(8,3),(1,4),(2,4),(7,4),(8,4),(1,5),(2,5),(7,5),(8,5),(1,6),(2,6),(7,6),(8,6)
Pr{(1,3)} = Pr{1 from 1st}*Pr{3
from 2nd} = (4/10)*(1/10) = 4/100
Pr{(1,4)} = Pr{1 from 1st}*Pr{4
from 2nd} = (4/10)*(2/10) = 8/100
Pr{(1,5)} = Pr{1 from 1st}*Pr{5
from 2nd} = (4/10)*(3/10) = 12/100
Pr{(1,6)} = Pr{1 from 1st}*Pr{6
from 2nd} = (4/10)*(4/10) = 16/100
Pr{(2,3)} = Pr{2 from 1st}*Pr{3
from 2nd} = (4/10)*(1/10) = 4/100
Pr{(2,4)} = Pr{2 from 1st}*Pr{4
from 2nd} = (4/10)*(2/10) = 8/100
Pr{(2,5)} = Pr{2 from 1st}*Pr{5
from 2nd} = (4/10)*(3/10) = 12/100
Pr{(2,6)} = Pr{2 from 1st}*Pr{6
from 2nd} = (4/10)*(4/10) = 16/100
Pr{(7,3)} = Pr{7 from 1st}*Pr{3
from 2nd} = (1/10)*(1/10) = 1/100
Pr{(7,4)} = Pr{7 from 1st}*Pr{4
from 2nd} = (1/10)*(2/10) = 2/100
Pr{(7,5)} = Pr{7 from 1st}*Pr{5
from 2nd} = (1/10)*(3/10) = 3/100
Pr{(7,6)} = Pr{7 from 1st}*Pr{6
from 2nd} = (1/10)*(4/10) = 4/100
Pr{(8,3)} = Pr{8 from 1st}*Pr{3
from 2nd} = (1/10)*(1/10) = 1/100
Pr{(8,4)} = Pr{8 from 1st}*Pr{4
from 2nd} = (1/10)*(2/10) = 2/100
Pr{(8,5)} = Pr{8 from 1st}*Pr{5
from 2nd} = (1/10)*(3/10) = 3/100
Pr{(8,6)} = Pr{8 from 1st}*Pr{6
from 2nd} = (1/10)*(4/10) = 4/100
Define HIGH as the highest of the face values
in the pair. Define LOW as the lowest of the face values in the pair. Define
DIFFCU =( (HIGH -
LOW) )3.
Compute and list the possible values for the
variable DIFFCU
DIFFCU{(1,3)} = ( (3 - 1) )3 = 23
= 8 (@ 4/100)
DIFFCU{(1,4)} = ( (4 - 1) )3 = 33
= 27 (@8/100)
DIFFCU{(1,5)} = ( (5 - 1) )3 = 43
= 64 (@12/100)
DIFFCU{(1,6)} = ( (6 - 1) )3 = 53
= 125 (@16/100)
DIFFCU{(2,3)} = ( (3 -
2) )3 = 13 = 1 (@4/100)
DIFFCU{(2,4)} = ( (4 -
2) )3 = 23 = 8 (@8/100)
DIFFCU{(2,5)} = ( (5 -
2) )3 = 33 = 27 (@12/100)
DIFFCU{(2,6)} = ( (6 -
2) )3 = 43 = 64 (@16/100)
DIFFCU{(7,3)} = ( (7 -
3) )3 = 43 = 64 (@ 1/100)
DIFFCU{(7,4)} = ( (7 -
4) )3 = 33 = 27 (@2/100)
DIFFCU{(7,5)} = ( (7 -
5) )3 = 23 = 8 (@3/100)
DIFFCU{(7,6)} = ( (7 -
6) )3 = 13 = 1 (@4/100)
DIFFCU{(8,3)} = ( (8 -
3) )3 = 53 = 125 (@1/100)
DIFFCU{(8,4)} = ( (8 -
4) )3 = 43 = 64 (@2/100)
DIFFCU{(8,5)} = ( (8 -
5) )3 = 33 = 27 (@ 3/100)
DIFFCU{(8,6)} = ( (8 -
6) )3 = 23 = 8 (@ 4/100)
DIFFCU{(2,3)} = ( (3 -
2) )3 = 13 = 1 (@4/100)
DIFFCU{(7,6)} = ( (7 -
6) )3 = 13 = 1 (@4/100)
DIFFCU{(1,3)} = ( (3 - 1) )3 = 23
= 8 (@ 4/100)
DIFFCU{(2,4)} = ( (4 - 2) )3 = 23
= 8 (@8/100)
DIFFCU{(7,5)} = ( (7 - 5) )3 = 23
= 8 (@3/100)
DIFFCU{(8,6)} = ( (8 - 6) )3 = 23
= 8 (@ 4/100)
DIFFCU{(1,4)} = ( (4 - 1) )3 = 33
= 27 (@8/100)
DIFFCU{(2,5)} = ( (5 - 2) )3 = 33
= 27 (@12/100)
DIFFCU{(7,4)} = ( (7 - 4) )3 = 33
= 27 (@2/100)
DIFFCU{(8,5)} = ( (8 - 5) )3 = 33
= 27 (@ 3/100)
DIFFCU{(1,5)} = ( (5 - 1) )3 = 43
= 64 (@12/100)
DIFFCU{(2,6)} = ( (6 - 2) )3 = 43
= 64 (@16/100)
DIFFCU{(7,3)} = ( (7 - 3) )3 = 43
= 64 (@ 1/100)
DIFFCU{(8,4)} = ( (8 - 4) )3 = 43
= 64 (@2/100)
DIFFCU{(1,6)} = ( (6 - 1) )3 = 53
= 125 (@16/100)
DIFFCU{(8,3)} = ( (8 -
3) )3 = 53 = 125 (@1/100)
Pr{DIFFCU = 1} = Pr{(7,6)} +
Pr{(2,3)} = (4/100) + (4/100) = 8/100
Pr{DIFFCU = 8} = Pr{One of (1,3), (2,4), (7,5) or (8,6) Shows} =
Pr{(1,3)}+Pr{(2,4)}+Pr{(7,5)}+Pr{(8,6)} =
(4/100)+(8/100)+(3/100)+(4/100)=19/100
Pr{DIFFCU = 27} = Pr{One of (1,4),(2,5),(7,4) or (8,5) Shows} =
Pr{(1,4)}+Pr{(2,5)}+Pr{(7,4)}+Pr{(8,5)} =
(8/100)+(12/100)+(2/100)+(3/100)=25/100
Pr{DIFFCU = 64} = Pr{One of (1,5),(2,6),(7,3),(8,4) Shows} =
Pr{(1,5)}+Pr{(2,6)}+Pr{(7,3)}+Pr{(8,4)} =
(12/100)+(16/100)+(1/100)+(2/100)=31/100
Pr{DIFFCU = 125} = Pr{One of (1,6), (8,3) Shows} = Pr{(1,6)}+
Pr{(8,3)} = (16/100)+(1/100) =17/100
Show full work and detail
for full credit
Case
Two | Long Run Argument and Perfect Samples | Plurality in US Resident
Pregnancies
Plurality
is the number of siblings born as the result
of a single pregnancy. Singleton pregnancies yield one born infant, twin
pregnancies yield two infants and triplet pregnancies yield three infants. Suppose that the probabilities tabled below apply to
pregnancies to US resident mothers with pregnancies yielding one or more live
births:
|
Plurality |
Probability |
|
Singleton
(1) |
0.9660 |
|
Twins
(2) |
0.0330 |
|
Triplets+
(3 or more) |
0.0010 |
|
Total |
1.00 |
Interpret each probability using the
Long Run Argument.
In long runs of random sampling, approximately 96.6% of sampled US
resident pregnancies were singleton.
In long runs of random sampling, approximately 3.3% of sampled US
resident pregnancies were twins.
In long runs of random sampling, approximately 0.1% of sampled US
resident pregnancies were triplets or of higher plurality.
Compute and discuss Perfect Samples
for n=5,000.
|
Plurality |
Probability |
E5000 |
|
Singleton (1) |
0.966 |
4830 |
|
Twins (2) |
0.033 |
165 |
|
Triplets+ (3 or more) |
0.001 |
5 |
|
Total |
1 |
5000 |
Expected Count Singleton = 5000*PSingleton = 5000*0.9660 = 4830
Expected Count Twins = 5000*PTwins = 5000*0.0330 = 165
Expected Count Triplets+ = 5000*PTriplets+ =
5000*0.001 = 5
In random samples of 5000 US resident pregnancies, approximately
4830 of 5000 of sampled US resident pregnancies were singleton.
In random samples of 5000 US resident pregnancies, approximately
165 of 5000 of sampled US resident pregnancies were twins.
In random samples of 5000 US resident pregnancies, approximately 5
of 5000 of sampled US resident pregnancies were triplets or of higher
plurality.
Show full work and detail
for full credit.
Case Three | Conditional
Probability | Color Slot Machine
Here is our slot machine
– on each trial, it produces an color sequence, using
the table below:
|
Sequence* |
Probability |
|
BBRRYR |
.20 |
|
RRBRRB |
.11 |
|
BYGGBR |
.24 |
|
GRGYBR |
.10 |
|
RYRBBY |
.15 |
|
YYYYRR |
.20 |
|
Total |
1.00 |
Compute the following conditional probabilities: Show full work and detail for full credit.
Pr{ “BR” Shows | Yellow Shows }
Prior Event = “Yellow
Shows”
|
Sequence* |
Probability |
|
BBRRYR |
.20 |
|
BYGGBR |
.24 |
|
GRGYBR |
.10 |
|
RYRBBY |
.15 |
|
YYYYRR |
.20 |
|
Total |
0.89 |
Pr{Yellow Shows} = Pr{One of BBRRYR, BYGGBR, GRGYBR,
RYRBBY or YYYYRR Shows} =
Pr{BBRRYR}+Pr{BYGGBR}+Pr{GRGYBR}+Pr{RYRBBY}+Pr{YYYYRR}
= .20 + .24 + .10 + .15 + .20 = 0.89
Joint Event = “BR” and
Yellow Show
|
Sequence* |
Probability |
|
BBRRYR |
.20 |
|
BYGGBR |
.24 |
|
GRGYBR |
.10 |
|
Total |
0.54 |
Pr{“BR” and Yellow Show} = Pr{One of BBRRYR, BYGGBR
or GRGYBR Shows} =
Pr{BBRRYR}+Pr{BYGGBR}+Pr{GRGYBR} = .20 + .24 +
.10 = 0.54
Pr{ “BR” Shows | Yellow Shows } = Pr{ “BR” and Yellow
Shows }/ Pr{ Yellow Shows } = 0.54/0.89
Pr{ “RBB” Shows | Yellow Shows }
Prior Event = “Yellow
Shows”
|
Sequence* |
Probability |
|
BBRRYR |
.20 |
|
BYGGBR |
.24 |
|
GRGYBR |
.10 |
|
RYRBBY |
.15 |
|
YYYYRR |
.20 |
|
Total |
0.89 |
Pr{Yellow Shows} = Pr{One of BBRRYR, BYGGBR, GRGYBR,
RYRBBY or YYYYRR Shows} =
Pr{BBRRYR}+Pr{BYGGBR}+Pr{GRGYBR}+Pr{RYRBBY}+Pr{YYYYRR}
= .20 + .24 + .10 + .15 + .20 = 0.89
Joint Event = “RBB” and
Yellow Show
|
Sequence* |
Probability |
|
RYRBBY |
.15 |
|
Total |
0.15 |
Pr{“RBB” and Yellow Show} = Pr{ RYRBBY} = .15
Pr{ “RBB” Shows |
Yellow Shows } = Pr{ “RBB” and Yellow Show }/ Pr{ Yellow Shows } = 0.15/0.89
Pr{ Red Shows | Blue Shows }
Prior Event = “Blue
Shows”
|
Sequence* |
Probability |
|
BBRRYR |
.20 |
|
RRBRRB |
.11 |
|
BYGGBR |
.24 |
|
GRGYBR |
.10 |
|
RYRBBY |
.15 |
|
Total |
0.80 |
Pr{Blue Shows} = Pr[One of
BBRRYR, RRBRRB, BYGGBR, GRGYBR or RYRBBY Shows} =
Pr[BBRRYR}+Pr{RRBRRB}+Pr{BYGGBR}+Pr{GRGYBR}+Pr{RYRBBY}
= .20+.11+.24+.10+.15 = 0.80
Joint Event = “Blue and Red Show”
|
Sequence* |
Probability |
|
BBRRYR |
.20 |
|
RRBRRB |
.11 |
|
BYGGBR |
.24 |
|
GRGYBR |
.10 |
|
RYRBBY |
.15 |
|
Total |
0.80 |
Pr{Blue and Red Show} =
Pr[One of BBRRYR, RRBRRB, BYGGBR, GRGYBR or RYRBBY Shows} =
Pr[BBRRYR}+Pr{RRBRRB}+Pr{BYGGBR}+Pr{GRGYBR}+Pr{RYRBBY}
= .20+.11+.24+.10+.15 = 0.80
Pr{ Red Shows | Blue Shows } = Pr{ Red and Blue
Show}/Pr{Blue Shows } = 0.80/0.80 = 1
Show full work and detail
for full credit
Case Four | Color Slot Machine |
Probability Rules
Using the color slot machine from
Case Three, compute the following probabilities. If a rule is specified, you
must use that rule.
Pr{“GR” Shows }
|
Sequence* |
Probability |
|
BBRRYR |
.20 |
|
RRBRRB |
.11 |
|
BYGGBR |
.24 |
|
GRGYBR |
.10 |
|
RYRBBY |
.15 |
|
YYYYRR |
.20 |
|
Total |
1.00 |
Pr{“GR” Shows } =
Pr{GRGYBR} = 0.10
Pr{ Blue Shows 3rd or 4th }
|
Sequence* |
Probability |
|
BBRRYR |
.20 |
|
RRBRRB |
.11 |
|
BYGGBR |
.24 |
|
GRGYBR |
.10 |
|
RYRBBY |
.15 |
|
YYYYRR |
.20 |
|
Total |
1.00 |
Pr{ Blue Shows 3rd
or 4th } = Pr{One of RRBRRB or RYRBBY Shows} = Pr{RRBRRB} +
Pr{RYRBBY} = 0.11+0.15 = 0.26
Pr{ Green Shows } – Use the Complementary Rule
Other Event = “Green Does Not Show”
|
Sequence* |
Probability |
|
BBRRYR |
.20 |
|
RRBRRB |
.11 |
|
RYRBBY |
.15 |
|
YYYYRR |
.20 |
|
Total |
0.66 |
Pr{Green Does Not Show} =
Pr[One of BBRRYR, RRBRRB, RYRBBY or YYYYRR Shows} =
Pr{BBRRYR}+Pr{RRBRRB}+Pr{RYRBBY}+Pr{YYYYRR}
= .20+.11+.15+.20=0.66
Pr{ Green Shows } = 1 – Pr{
Green Does Not Show } = 1 – 0.66 = 0.34
Show full work and detail
for full credit. Work all four cases.