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 THING, where THING = 10 − SUM, where SUM is the sum of the face values in the pair.

 

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

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

3.  Compute the probability for each value of THING 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.

 

THING = 10 − SUM

 

 

1

2

4

5

0

(1,0)

THING=10−(1+0)=9

(2,0)

THING=10−(2+0)=8

(4,0)

THING=10−(4+0)=6

(5,0)

THING=10−(5+0)=5

3

(1,3)

THING=10−(1+3)=6

(2,3)

THING=10−(2+3)=5

(4,3)

THING=10−(4+3)=3

(5,3)

THING=10−(5+3)=2

 

THING{(1,0)}=10 − SUM{(1,0)}= 10 − (1+0) = 10 − 1 = 9  

THING{(2,0)}=10 − SUM{(2,0)}= 10 − (2+0) = 10 − 2 = 8  

THING{(4,0)}=10 − SUM{(4,0)}= 10 − (4+0) = 10 − 4 = 6  

THING{(5,0)}=10 − SUM{(5,0)}= 10 − (5+0) = 10 − 5 = 5 

THING{(1,3)}=10 − SUM{(1,3)}= 10 − (1+3) = 10 − 4 = 6  

THING{(2,3)}=10 − SUM{(2,3)}= 10 − (2+3) = 10 − 5 = 5  

THING{(4,3)}=10 − SUM{(4,3)}= 10 − (4+3) = 10 − 7 = 3  

THING{(5,3)}=10 − SUM{(5,3)}= 10 − (5+3) = 10 − 8 = 2 

 

3.

 

Pr{THING=2}=Pr{(5,3) Shows} = 1/8 = .125

Pr{THING=3}=Pr{(4,3) Shows} = 1/8 = .125

Pr{THING=5}=Pr{One of (5,0), (2,3) Shows} = Pr{(5,0) Shows} + Pr{(2,3) Shows}=(1/8)+(1/8) = 2/8 = 1/4 = .25

Pr{THING=6}=Pr{One of (4,0), (1,3) Shows} = Pr{(4,0) Shows} + Pr{(1,3) Shows}=(1/8)+(1/8) = 2/8 = 1/4 = .25

Pr{THING=8}=Pr{(2,0) Shows} = 1/8 = .125

Pr{THING=9}=Pr{(1,0) Shows} = 1/8 = .125

 

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

 

Severity of cases of Kerpuztin’s Syndrome, Type F (KSF) 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

.35

Severe

.05

Moderate

.40

Mild

.20

Total

1.00

 

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

 

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

population.

2. Describe the perfect sample for 1100 draws with replacement from the KSF patient population. Briefly describe the relationship between this perfect sample and actual samples of 1100 draws with replacement from the KSF patient population.

Show full work and detail for full credit.

 

1.

 

In long runs of sampling with replacement, approximately 35% of sampled patients diagnosed with Kerpuztin’s Syndrome Type F die.  

In long runs of sampling with replacement, approximately 5% of sampled patients diagnosed with Kerpuztin’s Syndrome Type F present severe symptoms.  

In long runs of sampling with replacement, approximately 40% of sampled patients diagnosed with Kerpuztin’s Syndrome Type F present moderate symptoms.  

In long runs of sampling with replacement, approximately 20% of sampled patients diagnosed with Kerpuztin’s Syndrome Type F present mild symptoms.  

 

2.

 

Severity

Probability

Expected Count(n=1100)

Fatal

0.35

1100*.35 = 385

Severe

0.05

1100*.05 = 55

Moderate

0.4

1100*.40 = 440

Mild

0.2

1100*.20 = 220

Total

1

1100*1 = 1100

 

 

 

“Expected Fatal Count for n=1100” = 1100*Pr{Fatal} = 1100*.35 = 385

“Expected Fatal Count for n=1100” = 1100*Pr{Severe} = 1100*.05 = 55

“Expected Fatal Count for n=1100” = 1100*Pr{Moderate} = 1100*.40 = 440

“Expected Fatal Count for n=1100” = 1100*Pr{Mild} = 1100*.20 = 220

 

In random samples of 1100 patients diagnosed with Kerpuztin’s Syndrome Type F with replacement, approximately 385 sampled patients die.

In random samples of 1100 patients diagnosed with Kerpuztin’s Syndrome Type F with replacement, approximately 55 sampled patients present severe symptoms.

In random samples of 1100 patients diagnosed with Kerpuztin’s Syndrome Type F with replacement, approximately 440 sampled patients present moderate symptoms.

In random samples of 1100 patients diagnosed with Kerpuztin’s Syndrome Type F with replacement, approximately 220 sampled patients present mild symptoms.

 

Case Three: Conditional Probability, Color Slot Machine

 

Here is our slot machine – on each trial, it produces an 8-color sequence, using the table below:

 

Sequence*

Probability

RRBBRRYR

.10

RRRGBRRB

.11

BYGGYGBR

.14

GRGYBRGG

.10

YYYRYGYY

.20

RYGRRBBY

.15

YYYYBGRR

.20

Total

1.00

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

 

1. Pr{“BBR” Shows | Yellow Shows}

2. Pr{ “BR” Shows  | Green Shows }

3. Pr{Blue Shows | Red Shows}

 

1.

 

Sequence*

Probability

RRBBRRYR

.10

BYGGYGBR

.14

GRGYBRGG

.10

YYYRYGYY

.20

RYGRRBBY

.15

YYYYBGRR

.20

Total

.89

 

Pr{Yellow Shows} = Pr{One of  RRBBRRYR, BYGGYGBR, GRGYBRGG, YYYRYGYY, RYGRRBBY, YYYYBGRR Shows} =

Pr{RRBBRRYR Shows}+ Pr{BYGGYGBR Shows}+ Pr{GRGYBRGG Shows}+ Pr{YYYRYGYY Shows}+ Pr{RYGRRBBY Shows}+

Pr{YYYYBGRR Shows} = .10 + .14 + .10 + .20 + .15 + .20 = .89

 

 

 

Sequence*

Probability

RRBBRRYR

.10

Total

.10

 

Pr{“BBR” and Yellow Show} = Pr{RRBBRRYR} = .10

 

Pr{“BBR” Shows | Yellow Shows}= Pr{“BBR” and Yellow Show}/ Pr{Yellow Shows} = .10/.89

 

2.

Sequence*

Probability

RRRGBRRB

.11

BYGGYGBR

.14

GRGYBRGG

.10

YYYRYGYY

.20

RYGRRBBY

.15

YYYYBGRR

.20

Total

.90

 

Pr{Green Shows} = Pr{One of RRRGBRRB, BYGGYGBR, GRGYBRGG, YYYRYGYY, RYGRRBBY, YYYYBGRR Shows}=

Pr{RRRGBRRB Shows}+ Pr{BYGGYGBR Shows}+ Pr{GRGYBRGG Shows}+ Pr{YYYRYGYY Shows}+ Pr{RYGRRBBY Shows}+

Pr{YYYYBGRR Shows}= .11 + .14 + .10 + .20 + .15 + .20 = .90

 

 

Sequence*

Probability

RRRGBRRB

.11

BYGGYGBR

.14

GRGYBRGG

.10

Total

.35

 

Pr{“BR” and Green Show} = Pr{One of RRRGBRRB, BYGGYGBR, GRGYBRGG = .11 + .14 + .10 = .35

 

Pr{“BR” Shows  | Green Shows} = Pr{“BR” and Green Show}/Pr{Green Shows} = .35/.90

 

3.

 

Sequence*

Probability

RRBBRRYR

.10

RRRGBRRB

.11

BYGGYGBR

.14

GRGYBRGG

.10

YYYRYGYY

.20

RYGRRBBY

.15

YYYYBGRR

.20

Total

1.00

 

Pr{Red Shows} = Pr{One of RRBBRRYR, RRRGBRRB, BYGGYGBR, GRGYBRGG, YYYRYGYY, RYGRRBBY, YYYYBGRR Shows}=

Pr{ RRBBRRYR Shows} + Pr{RRRGBRRB Shows}+ Pr{BYGGYGBR Shows}+ Pr{GRGYBRGG Shows}+ Pr{YYYRYGYY Shows}+ Pr{RYGRRBBY Shows}+ Pr{YYYYBGRR Shows}= .10 + .11 + .14 + .10 + .20 + .15 + .20 = 1

 

Sequence*

Probability

RRBBRRYR

.10

RRRGBRRB

.11

BYGGYGBR

.14

GRGYBRGG

.10

RYGRRBBY

.15

YYYYBGRR

.20

Total

.80

 

Pr{Blue and Red Show} = Pr{One of RRBBRRYR, RRRGBRRB, BYGGYGBR, GRGYBRGG, RYGRRBBY, YYYYBGRR Shows}=

Pr{ RRBBRRYR Shows} + Pr{RRRGBRRB Shows}+ Pr{BYGGYGBR Shows}+ Pr{GRGYBRGG Shows}+ Pr{YYYRYGYY Shows}+ Pr{RYGRRBBY Shows}+ Pr{YYYYBGRR Shows}= .10 + .11 + .14 + .10  + .15 + .20 = .80

 

 

Pr{Blue Shows | Red Shows} = Pr{Blue and Red Show}/Pr{Red Shows} = .80/1

 

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.

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*

Probability

BGYYYRYGYY

.25

YYGBYYBGRR

.20

Total

.45

 

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

 

 

3.

Sequence*

Probability

RRBBRRYRGG

.10

RRGGRGBRRB

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

BGYYYRYGYY

.25

RRYYGRRBBY

.10

YYGBYYBGRR

.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:

 

Sequence*

Probability

RRBBRRYRGG

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

BGYYYRYGYY

.25

RRYYGRRBBY

.10

YYGBYYBGRR

.20

Total

.90

 

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.