Instructor Key

1st Hourly

Math 1107

Fall Semester 2002

 

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 indicated 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.

 

 

 

Name (PRINTED)                                Signature                               Date

 

Case One

Random Variables

 

We have a pair of dice – a fair d3 {faces 0,2,4} and a second d3 {faces 1,3,5} – note the probability model for the second d3 below. 

 

Face

Probability

1

0.20

3

0.30

5

0.50

 

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-pair.

 

1.a)       List the possible pairs, and compute a probability for each. Work out at least four pairs in full detail.

 

(1st d3 – fair d3, 2nd d3)

 

Pr{(0,1)}=Pr{0 shows fm 1st }*Pr{ 1 shows fm 2nd } = (1/3)*(2/10) = 2/30

Pr{(0,3)}=Pr{0 shows fm 1st }*Pr{ 3 shows fm 2nd } = (1/3)*(3/10) = 3/30

Pr{(0,5)}=Pr{0 shows fm 1st }*Pr{ 5 shows fm 2nd } = (1/3)*(5/10) = 5/30

 

Pr{(2,1)}=Pr{2 shows fm 1st }*Pr{ 1 shows fm 2nd } = (1/3)*(2/10) = 2/30

Pr{(2,3)}=Pr{2 shows fm 1st }*Pr{ 3 shows fm 2nd } = (1/3)*(3/10) = 3/30

Pr{(2,5)}=Pr{2 shows fm 1st }*Pr{ 5 shows fm 2nd } = (1/3)*(5/10) = 5/30

 

Pr{(4,1)}=Pr{4 shows fm 1st }*Pr{ 1 shows fm 2nd } = (1/3)*(2/10) = 2/30

Pr{(4,3)}=Pr{4 shows fm 1st }*Pr{ 3 shows fm 2nd } = (1/3)*(3/10) = 3/30

Pr{(4,5)}=Pr{4 shows fm 1st }*Pr{ 5 shows fm 2nd } = (1/3)*(5/10) = 5/30

 

 

1.b)      Define the random variable LowTie as either the lower of the two faces when unequal, or the common face value when equal. List the possible values for the variable LowTie, and compute a probability for each.

 

Pr{LT=0} = Pr{(0,1)} + Pr{(0,3)}+ Pr{(0,5)}= (2/30)+(3/30)+(5/30)=10/30

Pr{LT=1} = Pr{(2,1)} + Pr{(4,1)} = (2/30)+(2/30) = 4/30

Pr{LT=2} = Pr{(2,3)} + Pr{(2,5)}=(3/30)+(5/30)=8/30

Pr{LT=3} = Pr{(4,3)} = 3/30

Pr{LT=4} = Pr{(4,5)} = 5/30

 

 

 

1.c)       Define the random variable Sum as the sum of the two faces. List the possible values for the variable Sum, and compute a probability for each.

 

Pr{SUM=1} = Pr{(0,1)}= 2/30

Pr{SUM=2} = 0/30

Pr{SUM=3} = Pr{(0,3)}+ Pr{(2,1)} = (3/30)+ (2/30)=5/30

Pr{SUM=4} = 0/30

Pr{SUM=5} = Pr{(0,5)}+ Pr{(2,3)}+Pr{(4,1) = (5/30)+ (3/30)+(2/30)=10/30

Pr{SUM=6} = 0/30

Pr{SUM=7} = Pr{(2,5)} + Pr{(4,3) = (5/30)+ (3/30)=8/30

Pr{SUM=8} = 0

Pr{SUM=9} = Pr{(4,5)} = 5/30

 

Case Two

Conditional Probability

We have a bowl, as indicated below:

Color

Number in Bowl

Blue

7

Green

5

Red

3

Yellow

1

Total

16

 

Suppose that on each trial of this experiment that we make three (3) draws without replacement from the bowl. Work these directly, without using the usual formula for Pr{A|B}. Show all work and detail for full credit.

2.a)       Compute Pr{ green shows 3rd | red shows 1st and green shows 2nd };

Pr{ green shows 3rd | red shows 1st and green shows 2nd } = (5-1)/(16-1-1) = 4/14

2.b)      Compute Pr{ red shows 3rd | red shows 1st and red shows 2nd };

Pr{ red shows 3rd | red shows 1st and red shows 2nd } = (3-1-1)/(16-1-1) = 1/14

2.c)       Compute Pr{ blue or yellow shows 2nd | yellow shows 1st }.

Pr{ blue or yellow shows 2nd | yellow shows 1st } = (7+1-1)/(16-1) = 7/15

Case Three

Probability Rules

Color Slot Machine

 

Here is our slot machine – on each trial, it produces a 5-color sequence, using the table below:

 

Sequence*

Probability

GBRBY

.25

BBGGB

.10

GBBRG

.25

RGYBG

.10

BBYYG

.10

RRYYB

.10

YBGRY

.10

Total

1.00

 

*          B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered as

           1st to 5th , from left to right: (1st 2nd 3rd 4th 5th)

 

3.a)       Compute Pr{Yellow Shows 3rd} using the Additive Rule. Show full work and detail for full credit.       

 

Pr{Yellow Shows 3rd} = Pr{one of RGYBG,BBYYG,RRYYB,RRYYB shows}=

Pr{ RGYBG}+ Pr{BBYYG}+ Pr{RRYYB}} = .1+.1+.1 = .30 {note the correction}

 

3.b)      Compute Pr{Green Does Not Show} using the Complementary Rule. Show full work and detail for full credit.

 

Pr{Green Shows} = Pr{Green Shows} = Pr{one of  GBRBY, BBGGB, GBBRG,  RGYBG, BBYYG, YBGRY shows}=

Pr{GBRBY} + Pr{BBGGB} + Pr{GBBRG} + Pr{RGYBG} + Pr{BBYYG} + Pr{YBGRY} =

.25+.10+.25+.10+.10+.10 = .90

Pr{Green Does Not Show} = 1 - Pr{Green Shows}= 1 -.90 = .10  {note the correction}

 

 

Case Four

Perfect Samples

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 probabilities  for the population of Kerpuztin’s Syndrome (KS) patients are: 10% Severe, 20% Moderate, 30% Mild and  40% Fatal.

4.a)       Interpret each probability using the Long Run Argument. Be specific and complete for full credit.

[Pr{KS patient is Fatal}=.40] In long runs of draws with replacement from the KS patient population, approximately 40% of sampled KS cases are Fatal.

[Pr{KS patient is Severe}=.10] In long runs of draws with replacement from the KS patient population, approximately 10% of sampled KS cases are Severe.

[Pr{KS patient is Moderate}=.20] In long runs of draws with replacement from the KS patient population, approximately 20% of sampled KS cases are Moderate.

[Pr{KS patient is Mild}=.30] In long runs of draws with replacement from the KS patient population, approximately 30% of sampled KS cases are Mild.

4.b)      Compute the perfect sample of n=150 Kerpuztin’s Syndrome (KS) cases, and describe the relationship of this perfect sample to real samples of Kerpuztin’s Syndrome (KS) cases. Show all work and detail for full credit.

Level    Probability                   Expected Count

Fatal                .40                  e = .40*150 = 60

Severe             .10                  e = .10*150 = 15

Moderate          .20                  e = .20*150 = 30

Mild                .30                  e = .30*150 = 45

Total                1.00                 e = 1.00*150 = 150

In each random sample of 150 KS patients, drawn with replacement from the population of KS

patients, we expect approximately 60 fatal cases, 15 serious cases, 30 moderate cases and 45 mild

cases.

 

Work all four (4) cases.