Instructor Key
1st Hourly
Math 1107
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);
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 your name on an otherwise
blank sheet of paper. Then stack your stuff as follows:
Cover Sheet (Top)
Your
Work Sheets
The
Test Papers
Your
Toolsheet
Then
hand all of this in to me.
Sign and Acknowledge: I agree
to follow this protocol.
Case One
Conditional Probability
Modification - Bonus Points Only
Consider three dice, all fair:
d2 with faces {1,2}
d3 with faces {1,2,3}
d4 with faces {1,2,3,4}.
Consider a two-step experiment:
first, select a die at random, then toss that die. Suppose that each of the
dice has an equal chance of being selected for each toss. Assume that the dice
are fair.
List the possible face values that can result from this
experiment. Compute the probability for each face value. Show full detail and
work for full credit.
Die selection probabilities:
Pr{d2 is selected}= 1/3, Pr{d3 is selected}= 1/3, Pr{d4 is
selected}= 1/3
Conditional Probabilities:
Pr{face from d2 | d2 is selected} = 1/2 for faces 1,2
Pr{face from d3 | d3 is selected} = 1/3 for faces 1,2,3
Pr{face from d4 | d4 is selected} = 1/2 for faces 1,2,3,4
Pr{ 1 shows } = Pr{1 shows from
d2}+Pr{1 shows from d3}+Pr{1 shows from d4}=
(1/3)*(1/2)+(1/3)*(1/3)+(1/3)*(1/4)
= 1/6 + 1/9 + 1/12 = (9*12+6*12+6*9)/(6*9*12) @
Pr{ 2 shows } = Pr{2 shows from
d2}+Pr{2 shows from d3}+Pr{2 shows from d4}=
(1/3)*(1/2)+(1/3)*(1/3)+(1/3)*(1/4)
= 1/6 + 1/9 + 1/12 = (9*12+6*12+6*9)/(6*9*12) @
Pr{ 3 shows } = Pr{3 shows from
d2}+Pr{3 shows from d3}+Pr{3 shows from d4}=
(1/3)*0+(1/3)*(1/3)+(1/3)*(1/4)
= 1/9 + 1/12 = (9+12)/(9*12) @
Pr{ 4 shows } = Pr{4 shows from
d2}+Pr{4 shows from d3}+Pr{4 shows from d4}=
(1/3)*0+(1/3)*0+(1/3)*(1/4) =
1/12 @
Case Two
34
Points Maximum
Color Slot Machine –
Computation of Conditional Probabilities.
Here is our slot machine – on each
trial, it produces a 4-color sequence, using the table below:
|
Sequence* |
Probability |
|
GBRB |
.25 |
|
BBGG |
.10 |
|
GBBR |
.25 |
|
RGYB |
.10 |
|
BBYY |
.10 |
|
RRYY |
.10 |
|
YBGR |
.10 |
|
Total |
1.00 |
* 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.
2.a) Pr{ Blue Shows }
Pr{ Blue Shows } = Pr{ one of GBRB,BBGG,GBBR,RGYB,BBYY,YBGR
shows} =
Pr{ GBRB } + Pr( BBGG } + Pr{ GBBR } + Pr{ RGYB } + Pr{ BBYY
} + Pr{ YBGR } =
.25 + .10 + .25 + .10 + .10 + .10 =
.90
2.b) Pr{ Blue Shows and Green Does Not Show }
Pr{ Blue Shows and Green Does Not Show} = Pr{ BBYY shows} =
.10
2.c) Pr{ Yellow Shows} - Use the Complementary Rule.
Pr{ Yellow Does Not Show } = Pr{ one of GBRB, BBGG, GBBR
shows } =
Pr{GBRB} + Pr{BBGG}
+ Pr{GBBR shows } = .25 + .10 + .25 =
.60
Pr{ Yellow Shows } = 1 - Pr{ Yellow Does Not Show } = 1
- .60 = .40
Case Three
Kerpuztin’s Syndrome
33 Points Maximum
Severity of cases of Kerpuztin’s Syndrome (KS) 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
given as: Fatal, Severe, Moderate, Mild and No Symptoms. Suppose that these
probabilities for these severity levels are given in the table below:
|
Severity of
Case |
Probability |
|
Fatal |
.01 |
|
Severe |
.09 |
|
Moderate |
.10 |
|
Mild |
.30 |
|
No Symptoms |
.50 |
|
Total |
1.00 |
In our experiment,
we draw individual patients (with replacement) from the KS patient population,
noting the severity of the case.
3.a) Interpret
the probabilities in terms of repeated trials of draws with replacement from
the KS patient population.
In long runs of draws with replacement from
the population of KS cases, approximately 1% of sampled cases will show as
fatal.
In long runs of draws with replacement from
the population of KS cases, approximately 9% of sampled cases will show as
severe.
In long runs of draws with replacement from
the population of KS cases, approximately 10% of sampled cases will show as
moderate.
In long runs of draws with replacement from
the population of KS cases, approximately 30% of sampled cases will show as
mild.
In long runs of draws with replacement from
the population of KS cases, approximately 50% of sampled cases will show no
symptoms.
3.b) Describe the perfect sample for
120 draws with replacement from the KS patient population. Briefly describe the
relationship between this perfect sample and actual samples of 120 draws with
replacement from the KS patient population.
|
Severity of
Case |
Probability |
Expected
or Perfect Count (n=120) |
|
Fatal |
.01 |
.01*120
= 1.2 cases out of 120 |
|
Severe |
.09 |
.09*120
= 10.8 cases out of 120 |
|
Moderate |
.10 |
.10*120
= 12 cases out of 120 |
|
Mild |
.30 |
.30*120
= 36 cases out of 120 |
|
No Symptoms |
.50 |
.50*120
= 60 cases out of 120 |
|
Total |
1.00 |
1.00*120
= 120 cases out of 120 |
In random samples of 120 draws with
replacement from the population of KS cases, approximately 1.2 cases will show
as fatal, 10.8 cases will show as severe, 12 cases will show as moderate, 36
cases will show as mild and 60 cases will show no symptoms of KS.
Case Four
Random Variable
33 Points Maximum
Our experiment consists of tossing
a pair of independently operating, fair dice:
d2 with faces {1,2}
d4 with faces {1,2,3,4},
and observing the pair of face
values. Define the random variable SUM as the sum of the faces in the pair.
4.a) List the pairs and compute a probability for each pair.
Write the pairs as (face from d2,face from d4). The pairs
are: (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3),
(2,4). Compute the probabilities as:
Pr{(1,1)} = Pr{1 from d2}*{1 from d4} = (1/2)*(1/4) = 1/8
Pr{(1,2)} = Pr{1 from d2}*{2 from d4} = (1/2)*(1/4) = 1/8
Pr{(1,3)} = Pr{1 from d2}*{3 from d4} = (1/2)*(1/4) = 1/8
Pr{(1,4)} = Pr{1 from d2}*{4 from d4} = (1/2)*(1/4) = 1/8
Pr{(2,1)} = Pr{2 from d2}*{1 from d4} = (1/2)*(1/4) = 1/8
Pr{(2,2)} = Pr{2 from d2}*{2 from d4} = (1/2)*(1/4) = 1/8
Pr{(2,3)} = Pr{2 from d2}*{3 from d4} = (1/2)*(1/4) = 1/8
Pr{(2,4)} = Pr{2 from d2}*{4 from d4} = (1/2)*(1/4) = 1/8
4.b) List the possible values for SUM and compute a probability
for each value of SUM.
The sums are: 2, 3, 4, 5, 6. Compute the probabilities as:
Pr{SUM=2} = Pr{(1,1)} = 1/8
Pr{SUM=3} = Pr{(1,2), (2,1) } = 1/8 + 1/8 = 2/8
Pr{SUM=4} = Pr{(1,3), (2,2)} = 1/8 + 1/8 = 2/8
Pr{SUM=5} = Pr{(1,4), (2,3)} = 1/8 + 1/8 = 2/8
Pr{SUM=6} = Pr{(2,4)} = 1/8
Show full work and
detail for full credit.
Be sure that you have
worked all four cases.