Instructor Key

1st Hourly

Math 1107

Fall Semester 2003

 

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);

               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

Your Toolsheet

 

Then hand all of this in to me.

 

Sign and Acknowledge:  I agree to follow this protocol.

 

 

 

Name (PRINTED)                                          Signature                                          Date

 

Case One

Conditional Probability

Color Bowl/Draws without Replacement

 

We have a bowl containing the following colors and counts of balls (color @ count):

 

White @ 1, Black @ 2, Blue @ 4, Green @ 4, Red @ 3, Yellow @ 1

 

Each trial of our experiment consists of five draws without replacement from the bowl. Compute the following conditional probabilities. Compute these directly.

 

1.a)         Pr{ Yellow shows 2nd  | Yellow shows 1st} = 0/14

 

Color

Available for 1st Draw

Available for 2nd Draw

Given Yellow on 1st Draw

White

1

1

Black

2

2

Blue

4

4

Green

4

4

Red

3

3

Yellow

1

0

Total

15

14

 

 

1.b)        Pr{ Blue shows 5th | Black shows 1st, Blue shows 2nd, Blue shows 3rd, and Green shows 4th } = 2/11

 

Color

Available for 1st Draw

Available for 2nd Draw

Given Black on 1st Draw

Available for 3rd Draw

Given Blue on 2nd Draw

Available for 4th Draw

Given Blue on 3rd Draw

Available for 5th Draw

Given Green on 4th Draw

White

1

1

1

1

1

Black

2

1

1

1

1

Blue

4

4

3

2

2

Green

4

4

4

4

3

Red

3

3

3

3

3

Yellow

1

1

1

1

1

Total

15

14

13

12

11

 

 

1.c)        Pr{ Blue shows 3rd | Yellow shows 1st, Blue shows 2nd } = 3/13

 

Color

Available for 1st Draw

Available for 2nd Draw

Given Yellow on 1st Draw

Available for 3rd Draw

Given Blue on 2nd Draw

White

1

1

1

Black

2

2

2

Blue

4

4

3

Green

4

4

4

Red

3

3

3

Yellow

1

0

0

Total

15

14

13

 

 

Case Two

Probability Computational Rules

Color Slot Machine

 

Here is our slot machine – on each trial, it produces a 4-color sequence, using the table below:

 

Sequence*

Probability

GBRB

.15

BBGG

.05

GGGG

.10

GBBR

.25

BBBB

.05

RGYB

.05

YYYY

.05

BBYY

.175

RRYY

.025

RRRR

.025

YBGR

.075

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{ Red Shows } = Pr{one  of GBRB,GBBR,RGYB,RRYY,RRRR,YBGR shows} = .15+.25+.05+.025+.025+.075 = .575 or 57.5%

 

Sequence*

Probability

GBRB

.15

GBBR

.25

RGYB

.05

RRYY

.025

RRRR

.025

YBGR

.075

Total

0.575

 

 

2.b)        Pr{ Blue Shows and Red Does Not Show } = Pr{one of BBGG,BBBB,BBYY shows} = .05+.05+.175 = .275 or 27.5%

 

Sequence*

Probability

BBGG

.05

BBBB

.05

BBYY

.175

Total

.275

 

 

2.c)        Pr{Yellow and Green Shows} - Use the Complementary Rule.

 

Two versions were accepted for full credit:

 

Version #1:

 

Pr{neither Yellow nor Green Shows} = Pr{ one of BBBB,RRRR shows} = .05+.025 = .075

 

Sequence*

Probability

BBBB

.05

RRRR

.025

Total

.075

              

Pr{Yellow and Green Shows} = 1 - Pr{neither Yellow nor Green Shows} = 1 - .075 = .925 or 92.5% 

 

Version #2:

 

Pr{either Yellow does not show or Green does not show} = Pr{ one of BBBB,RRRR shows} = .15+.05+.10+.25+.05+.05+.175+.025+.025 = .875

 

Sequence*

Probability

GBRB

.15

BBGG

.05

GGGG

.10

GBBR

.25

BBBB

.05

YYYY

.05

BBYY

.175

RRYY

.025

RRRR

.025

Total

.875

              

Pr{Yellow and Green Shows} = 1 - Pr{either Yellow does not show or Green does not show} = 1 - .875 = .125 or 12.5% 

 

Case Three

Perfect Samples and the Long Run Interpretation

Kerpuztin’s Syndrome

 

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

.001

Severe

.40

Moderate

.10

Mild

.30

No Symptoms

.199

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 KS population, approximately .1% of sampled Kerpuztin's Syndrome patients present Fatal cases.

In long runs of draws with replacement from the KS population, approximately 40% of sampled Kerpuztin's Syndrome patients present Severe cases.

In long runs of draws with replacement from the KS population, approximately 10% of sampled Kerpuztin's Syndrome patients present Moderate cases.

In long runs of draws with replacement from the KS population, approximately 30% of sampled Kerpuztin's Syndrome patients present Mild cases.

In long runs of draws with replacement from the KS population, approximately 19.9% of sampled Kerpuztin's Syndrome patients present no symptoms.

3.b)        Describe the perfect sample for 1500 draws with replacement from the KS patient population. Briefly describe the relationship between this perfect sample and actual samples of 1500 draws with replacement from the KS patient population.

Severity of Case

Probability

Expected Count for Perfect Sample

(for n=1500)

Fatal

.001

.001*1500=1.5

Severe

.40

.40*1500=600

Moderate

.10

.10*1500=150

Mild

.30

.30*1500=450

No Symptoms

.199

.199*1500=398.5

Total

1.00

1.00*1500=1500

 

Random samples of 1500 Kerpuztin's Syndrome patients present approximately: 1.5 fatal cases, 600 severe cases, 150 moderate cases, 450 mild cases and 398.5 asymptomatic cases.

3.c)        Referring to the table, are there any rare events? Briefly explain.

Fatal cases are reasonably rare events. The expected number of fatal cases for n=1500 is 1.5 per sample…in fact, for samples with n<1000, the expected number of fatal cases is less than one.

 

Case Four

Random Variable

THING

 

Our experiment consists of tossing a pair of independently operating, fair dice: d3 with faces  {-1,0,1}; d4 with faces {0,1,2,3}, and observing the pair of face values. A54rygfDefine the random variable THING as follows:

 

THING = 2*(d3 face)2 + (d4 face)3

 

4.a)         List the pairs and compute a probability for each pair.

 

Writing the pairs as (d4 face,d3 face):

 

d3¯d4®

0

1

2

3

-1

(0,-1)

(1,-1)

(2,-1)

(3,-1)

0

(0,0)

(1,0)

(2,0)

(3,0)

1

(0,1)

(1,1)

(2,1)

(3,1)

 

 

Pr{(0,-1)} = Pr{0 from d3}*Pr{-1 from d4}=(1/3)*(1/4)=1/12

Pr{(0,-1)} = Pr{0 from d3}*Pr{-1 from d4}=(1/3)*(1/4)=1/12

Pr{(0,-1)} = Pr{0 from d3}*Pr{-1 from d4}=(1/3)*(1/4)=1/12

Pr{(1,-1)} = Pr{1 from d3}*Pr{-1 from d4}=(1/3)*(1/4)=1/12

Pr{(1,-1)} = Pr{1 from d3}*Pr{-1 from d4}=(1/3)*(1/4)=1/12

Pr{(1,-1)} = Pr{1 from d3}*Pr{-1 from d4}=(1/3)*(1/4)=1/12

Pr{(2,-1)} = Pr{2 from d3}*Pr{-1 from d4}=(1/3)*(1/4)=1/12

Pr{(2,-1)} = Pr{2 from d3}*Pr{-1 from d4}=(1/3)*(1/4)=1/12

Pr{(2,-1)} = Pr{2 from d3}*Pr{-1 from d4}=(1/3)*(1/4)=1/12

Pr{(3,-1)} = Pr{3 from d3}*Pr{-1 from d4}=(1/3)*(1/4)=1/12

Pr{(3,-1)} = Pr{3 from d3}*Pr{-1 from d4}=(1/3)*(1/4)=1/12

Pr{(3,-1)} = Pr{3 from d3}*Pr{-1 from d4}=(1/3)*(1/4)=1/12

 

 

4.b)        List the possible values for THING and compute a probability for each value of THING. 

 

THING = 2*(d3 face)2 + (d4 face)3

 

d3¯d4®

THING

Pr{pair}

0

1

2

3

-1

(0,-1)

2*1+0=2

1/12

(1,-1)

2*1+(1)=3

1/12

(2,-1)

2*1+8=10

1/12

(3,-1)

2*1+27=29

1/12

0

(0,0)

2*0+0=0

1/12

(1,0)

2*0+1=1

1/12

(2,0)

2*0+8=8

1/12

(3,0)

2*0+27=27

1/12

1

(0,1)

2*1+0=2

1/12

(1,1)

2*1+1=3

1/12

(2,1)

2*1+8=10

1/12

(3,1)

2*1+27=29

1/12

 

 

Pr{THING=0} = Pr{ (0,0) } = (1/12) » .0833 or 8.33%

Pr{THING=1} = Pr{ (1,0) } = (1/12) » .0833 or 8.33%

Pr{THING=2} = Pr{ one of (0,-1),(0,1) shows} = (1/12)+(1/12) = 2/12 » .1666 or 16.66%

Pr{THING=3} = Pr{ one of (1,-1),(1,1) shows} = (1/12)+(1/12) = 2/12 » .1666 or 16.66%

Pr{THING=8} = Pr{ (2,0) } = (1/12) » .0833 or 8.33%

Pr{THING=10} = Pr{ one of (2,-1),(2,1) shows } = (1/12) +(1/12) = 2/12 » .1666 or 16.66%

Pr{THING=27} = Pr{ (3,0) } = (1/12) » .0833 or 8.33%

Pr{THING=29} = Pr{ one of (3,-1),(3,1) shows } = (1/12) +(1/12) = 2/12 » .1666 or 16.66%

 

Show full work and detail for full credit. Be sure that you have worked all four cases.

 

Performance Notes

Scores

 

n

mean

p00

p10

p25

p50

p75

p90

p100

88

82.1

15.0

61.0

72.0

86.0

93.5

96.0

100.0

 

There were n=88 scores. The average score was 82.1. The minimum score was 15, and the maximum score was 100 (7 students made this score).

Approximately 10% of the scores were at or below 61, approximately 25% of the scores were at or below 72, approximately 50% of the scores were at or below 86, approximately 75% of the scores were at or below 93.5. Approximately 90% of the scores were at or below 96. Approximately 10% of the scores were at or above 96.

 

score

count

percent

[0,60)

8

9.09

[60,70)

8

9.09

[70,80)

13

14.77

[80,90)

24

27.27

[90,100)

28

31.82

100

7

7.95

 

Do not interpret individual scores as letter grades. However, consistently scoring in the usual ranges will likely yield the traditional letter grade.