Key

The 1st Hourly

Math 1107

Fall Semester 2007

 

Protocol: You will use only the following resources:        Your individual calculator; individual tool-sheet (single 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.

 

In each case, 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. All of your work goes on one side each of the blank sheets provided. Space out your work. Do not share information with any other students during this hourly. Sign and Acknowledge:

 

I agree to follow this protocol. 

 

 

Name (PRINTED)                                          Signature                                          Date

 

Case One: Random Variables, Pair of Dice

 

We have a pair of independently operating, fair, dice: the first die with face values 1,2,4,5 and the second die with face values 0,3. Each trial of our experiment involves tossing the pair of dice, and noting the pair of face values. We compute a random variable from the pair of face values as follows: Consider the random variable LOWTIE, where LOWTIE = lower of the two face values or common value if tied.

 

1.  List all the possible pairs of face values, and compute the probability for each pair, showing full detail.

2.  Compute the values of LOWTIE, showing in detail how these values are computed from each pair of face values.

3.  Compute the probability for each value of LOWTIE from step 2, showing full detail.

 

Show full work and detail for full credit.

 

1.

 

 

1

2

4

5

0

(1,0)

(2,0)

(4,0)

(5,0)

3

(1,3)

(2,3)

(4,3)

(5,3)

 

There are eight (8) pairs: (1,0), (2,0), (4,0), (5,0), (1,3), (2,3), (4,3), (5,3).

 

Pr{(1,0) Shows} = Pr{1 from 1st Die Shows}*Pr{0 from 2nd Die Shows} = (1/4)*(1/2) = 1/8

Pr{(2,0) Shows} = Pr{2 from 1st Die Shows}*Pr{0 from 2nd Die Shows} = (1/4)*(1/2) = 1/8

Pr{(4,0) Shows} = Pr{4 from 1st Die Shows}*Pr{0 from 2nd Die Shows} = (1/4)*(1/2) = 1/8

Pr{(5,0) Shows} = Pr{5 from 1st Die Shows}*Pr{0 from 2nd Die Shows} = (1/4)*(1/2) = 1/8

Pr{(1,3) Shows} = Pr{1 from 1st Die Shows}*Pr{3 from 2nd Die Shows} = (1/4)*(1/2) = 1/8

Pr{(2,3) Shows} = Pr{2 from 1st Die Shows}*Pr{3 from 2nd Die Shows} = (1/4)*(1/2) = 1/8

Pr{(4,3) Shows} = Pr{4 from 1st Die Shows}*Pr{3 from 2nd Die Shows} = (1/4)*(1/2) = 1/8

Pr{(5,3) Shows} = Pr{5 from 1st Die Shows}*Pr{3 from 2nd Die Shows} = (1/4)*(1/2) = 1/8

 

2.

 

LT=”Low Tie”

 

 

1

2

4

5

0

(1,0)

LT=0

(2,0)

LT=0

(4,0)

LT=0

(5,0)

LT=0

3

(1,3)

LT=1

(2,3)

LT=2

(4,3)

LT=3

(5,3)

LT=3

 

LT{(1,0)}=0

LT{(2,0)}=0

LT{(4,0)}=0

LT{(5,0)}=0

LT{(1,3)}=1

LT{(2,3)}=2

LT{(4,3)}=3

LT{(5,3)}=3.

 

3.

 

Pr{LT=0}=Pr{One of (1,0), (2,0), (4,0), (5,0) Shows} = Pr{(1,0) Shows} + Pr{(2,0) Shows} + Pr{(3,0) Shows} +Pr{(4,0) Shows} =

(1/8)+ (1/8)+ (1/8)+(1/8) = 4/8 = 2/4 = .50

Pr{LT=1}=Pr{(1,3) Shows} = 1/8 = .125

Pr{LT=2}=Pr{(2,3) Shows} = 1/8 = .125

Pr{LT=3}=Pr{One of (4,3), (5,3) Shows} = Pr{(4,3) Shows} + Pr{(5,3) Shows}=(1/8)+(1/8) = 2/8 = 1/4 = .25

 

Case Two: Long Run Argument, Perfect Samples, Kerpuztin’s Syndrome, Type D

 

Severity of cases of Kerpuztin’s Syndrome, Type D (KSD) 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 distributed as indicated below:

 

Severity

Probability

Fatal

.31

Severe

.09

Moderate

.36

Mild

.24

Total

1.00

 

In our experiment, we draw individual patients (with replacement) from the KS patient population, noting the severity of the case.

 

1. Interpret the probabilities in terms of repeated trials of draws with replacement from the KSD patient

population.

2. Describe the perfect sample for 1200 draws with replacement from the KSD patient population. Briefly describe the relationship between this perfect sample and actual samples of 1200 draws with replacement from the KSD patient population.

Show full work and detail for full credit.

 

1.

 

In long runs of sampling with replacement, approximately 31% of sampled patients diagnosed with Kerpuztin’s Syndrome Type D die.  

In long runs of sampling with replacement, approximately 9% of sampled patients diagnosed with Kerpuztin’s Syndrome Type D present severe symptoms.  

In long runs of sampling with replacement, approximately 36% of sampled patients diagnosed with Kerpuztin’s Syndrome Type D present moderate symptoms.  

In long runs of sampling with replacement, approximately 24% of sampled patients diagnosed with Kerpuztin’s Syndrome Type D present mild symptoms.  

 

2.

 

Severity

Probability

 

Fatal

0.31

1200*.31=372

Severe

0.09

1200*.09=108

Moderate

0.36

1200*.36=432

Mild

0.24

1200*.24=288

Total

1

1200

 

“Expected Fatal Count for n=1200” = 1200*Pr{Fatal} = 1200*.31 = 372

“Expected Fatal Count for n=1200” = 1200*Pr{Severe} = 1200*.09 = 108

“Expected Fatal Count for n=1200” = 1200*Pr{Moderate} = 1200*.36 = 432

“Expected Fatal Count for n=1200” = 1200*Pr{Mild} = 1200*.24 = 288

 

In random samples of 1200 patients diagnosed with Kerpuztin’s Syndrome Type D with replacement, approximately 372 sampled patients die.

In random samples of 1200 patients diagnosed with Kerpuztin’s Syndrome Type D with replacement, approximately 108 sampled patients present severe symptoms.

In random samples of 1200 patients diagnosed with Kerpuztin’s Syndrome Type D with replacement, approximately 432 sampled patients present moderate symptoms.

In random samples of 1200 patients diagnosed with Kerpuztin’s Syndrome Type D with replacement, approximately 288 sampled patients present mild symptoms.

 

Case Three: Conditional Probability, Color Slot Machine

 

Here is our slot machine – on each trial, it produces a 10-color sequence, using the table below:

 

Sequence*

Probability

RRBBRRYRGG

.10

RRGGRGBRRB

.11

BBYYGGYGBR

.14

GRRGGYBRGG

.10

BGYYYRYGYY

.20

RRYYGRRBBY

.15

YYGBYYBGRR

.20

Total

1.00

*B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered as 1st to 10th , from left to right: (1st 2nd 3rd 4th 5th6th7th 8th 9th 10th )

Compute the following conditional probabilities: Show full work and detail for full credit.

 

1. Pr{“GBR” Shows | Yellow Shows}

2. Pr{Green Shows  | “BR” Shows }

3. Pr{Blue Shows | Red Shows}

 

1.

 

Sequence*

Probability

RRBBRRYRGG

.10

BBYYGGYGBR

.14

GRRGGYBRGG

.10

BGYYYRYGYY

.20

RRYYGRRBBY

.15

YYGBYYBGRR

.20

Total

.89

 

Pr{Yellow Shows} = Pr{One of RRBBRRYRGG, BBYYGGYGBR, GRRGGYBRGG, BGYYYRYGYY, RRYYGRRBBY, YYGBYYBGRR Shows} =

Pr{RRBBRRYRGG Shows}+ Pr{BBYYGGYGBR Shows}+ Pr{GRRGGYBRGG Shows}+ Pr{BGYYYRYGYY Shows}+ Pr{RRYYGRRBBY Shows}+ Pr{ YYGBYYBGRR Shows} = .10 + .14 + .10 + .20 + .15 + .20 = .89

 

 

Sequence*

Probability

BBYYGGYGBR

.14

Total

.14

 

Pr{“GBR” and Yellow Show} = Pr{BBYYGGYGBR Shows} = .14

 

Pr{“GBR” Shows | Yellow Shows}= Pr{“GBR” and Yellow Show}/ Pr{Yellow Shows} = .14/.89

 

2.

 

 

Sequence*

Probability

RRBBRRYRGG

.10

RRGGRGBRRB

.11

BBYYGGYGBR

.14

GRRGGYBRGG

.10

Total

.45

 

Pr{“BR” Shows} = Pr{One of RRBBRRYRGG, RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG Shows} =

Pr{RRBBRRYRGG Shows} + Pr{RRGGRGBRRB Shows} + Pr{BBYYGGYGBR Shows} +  Pr{GRRGGYBRGG Shows} =

.10+.11+.14+.10 = .45

 

Sequence*

Probability

RRBBRRYRGG

.10

RRGGRGBRRB

.11

BBYYGGYGBR

.14

GRRGGYBRGG

.10

Total

.45

 

Pr{Green and “BR” Show}= Pr{One of RRBBRRYRGG, RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG Shows} =

Pr{RRBBRRYRGG Shows} + Pr{RRGGRGBRRB Shows} + Pr{BBYYGGYGBR Shows} +  Pr{GRRGGYBRGG Shows} =

.10+.11+.14+.10 = .45

 

Pr{Green Shows  | “BR” Shows } = Pr{Green and “BR” Show} / Pr{Green Shows  | “BR” Shows } = .45/.45 = 1

 

3.

 

Sequence*

Probability

RRBBRRYRGG

.10

RRGGRGBRRB

.11

BBYYGGYGBR

.14

GRRGGYBRGG

.10

BGYYYRYGYY

.20

RRYYGRRBBY

.15

YYGBYYBGRR

.20

Total

1.00

Pr{Red Shows} = Pr{One of RRBBRRYRGG, RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG, BGYYYRYGYY, RRYYGRRBBY, YYGBYYBGRR Shows} = Pr{RRBBRRYRGG Shows} + Pr{RRGGRGBRRB Shows} + Pr{BBYYGGYGBR Shows} + Pr{GRRGGYBRGG Shows} + Pr{BGYYYRYGYY Shows} + Pr{RRYYGRRBBY Shows} + Pr{YYGBYYBGRR Shows} = .10 + .11 + .14 + .10 + .20 + .15 + .20 = 1

 

 

Sequence*

Probability

RRBBRRYRGG

.10

RRGGRGBRRB

.11

BBYYGGYGBR

.14

GRRGGYBRGG

.10

BGYYYRYGYY

.20

RRYYGRRBBY

.15

YYGBYYBGRR

.20

Total

1.00

Pr{Blue and Red Show} = Pr{One of RRBBRRYRGG, RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG, BGYYYRYGYY, RRYYGRRBBY, YYGBYYBGRR Shows} = Pr{RRBBRRYRGG Shows} + Pr{RRGGRGBRRB Shows} + Pr{BBYYGGYGBR Shows} + Pr{GRRGGYBRGG Shows} + Pr{BGYYYRYGYY Shows} + Pr{RRYYGRRBBY Shows} + Pr{YYGBYYBGRR Shows} = .10 + .11 + .14 + .10 + .20 + .15 + .20 = 1

 

Pr{Blue Shows | Red Shows}= Pr{Blue and Red Show}/Pr{Red Shows} = 1/1 = 1

 

 

 

Case Four: Probability Computation Rules, Color Slot Machine

 

Here is our slot machine – on each trial, it produces a 10-color sequence, using the table below:

 

Sequence*

Probability

RRBBRRYRGG

.10

RRGGRGBRRB

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

BGYYYRYGYY

.25

RRYYGRRBBY

.10

YYGBYYBGRR

.20

Total

1.00

 

1. Define the random variable REDCOUNT as the number of times red shows in the sequence. Compute Pr{ REDCOUNT = 3 or 4}.

2. Compute Pr{“BG” Shows}.

3. Compute Pr{“Yellow” Shows}. Use the Complementary rule.

 

 

1.

 

 

Sequence*

REDCOUNT

Probability

RRBBRRYRGG

5

.10

RRGGRGBRRB

5

.10

BBYYGGYGBR

1

.15

GRRGGYBRGG

3

.10

BGYYYRYGYY

1

.25

RRYYGRRBBY

4

.10

YYGBYYBGRR

2

.20

Total

 

1.00

 

Pr{ REDCOUNT = 3 or 4} = Pr{REDCOUNT=3}+ Pr{REDCOUNT=4} =Pr{GRRGGYBRGG Shows} + Pr{RRYYGRRBBY Shows} = .10 + .10 = .20

 

2.

 

 

 

Sequence*

“BG” Shows?

Probability

RRBBRRYRGG

No

.10

RRGGRGBRRB

No

.10

BBYYGGYGBR

No

.15

GRRGGYBRGG

No

.10

BGYYYRYGYY

Yes

.25

RRYYGRRBBY

No

.10

YYGBYYBGRR

Yes

.20

Total

 

1.00

 

Pr{“BG” Shows} = Pr{One of Shows} = Pr{BGYYYRYGYY Shows} + Pr{YYGBYYBGRR Shows} = .25+.20 = .45

 

3.

 

 

 

Sequence*

Yellow Shows?

Probability

RRBBRRYRGG

Yes

.10

RRGGRGBRRB

No

.10

BBYYGGYGBR

Yes

.15

GRRGGYBRGG

Yes

.10

BGYYYRYGYY

Yes

.25

RRYYGRRBBY

Yes

.10

YYGBYYBGRR

Yes

.20

Total

 

1.00

 

Pr{“Yellow” Does Not Show} = Pr{ RRGGRGBRRB Shows} = .10

Pr{“Yellow” Shows} = 1 − Pr{“Yellow” Does Not Show} = 1 − .10 = .90

 

Check

Pr{“Yellow” Shows} = Pr{One of  RRBBRRYRGG, BBYYGGYGBR, GRRGGYBRGG, BGYYYRYGYY, RRYYGRRBBY, YYGBYYBGRR Shows}=

Pr{RRBBRRYRGG Shows}+ Pr{BBYYGGYGBR Shows}+Pr{GRRGGYBRGG Shows}+Pr{BGYYYRYGYY Shows}+Pr{RRYYGRRBBY Shows}+ Pr{YYGBYYBGRR Shows}= .10 +.15 +.10 +.25 +.10 +.20 = .90

 

Show full work and detail for full credit. Be certain that you have worked all four (4) cases. Show full work and detail for full credit.