"Version 500"

Key

The 1st Hourly

Math 1107

Fall Semester 2006

 

Protocol

 

You will use only the following resources: Your individual calculator, Your individual tool-sheet (single 8.5 by 11 inch sheet), Your writing utensils, Blank Paper (provided by me ) and this copy of the hourly. Share these resources with no-one 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.

 

Sign and Acknowledge: I agree to follow this protocol.

 

________________________________________________________________________

Name (PRINTED)                               Signature                                 Date

 

Case One

Long Run Argument

Perfect Samples

Xyrkztin’s Syndrome

 

Xyrkztin's Syndrome (XS) is a rare, fictitious disease that I invented for this test. Symptoms include:

 

Severity of cases of Xyrkztin's Syndrome (XS) is noted as:  (F)atal, (S)evere, (Mo)derate, or (Mi)ld. Suppose that severity probabilities for the population of XS patients are given below:

 

XS Severity

Probability

Mild

.20

Moderate

.05

Severe

.15

Fatal

.60

Total

1.00

 

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

 

In long runs of sampling with replacement from the population of cases of Xyrkztin's Syndrome, approximately 20% of sampled cases will show mild severity, approximately 5% will show moderate severity, approximately 15% will show severe severity and approximately 60% will show fatal severity.

 

1.b) Compute the perfect sample of n=500 XS cases, and describe the relationship of this perfect sample to real samples of XS cases.

 

ExpectedMild, 500=500*PMild = 500*.20 = 100

ExpectedModerate, 500=500*PModerate = 500*.05 = 25

ExpectedSevere, 500=500*PSevere = 500*.15 = 75

ExpectedFatal, 500=500*PFatal = 500*.60 = 300

 

In random samples of 500 cases of Xyrkztin's Syndrome, approximately 100 sampled cases will show mild severity, approximately 25 will show moderate severity, approximately 75 will show severe severity and approximately 300 will show fatal severity.

 

Case Two

Color Slot Machine

Computation of Conditional Probabilities

 

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

 

Sequence*

Probability

RRBBRRYRRR

.10

RRGGRGBRRB

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

BGYGYRYGYY

.25

RRYYGRRBBY

.10

YYGBYYBGRR

.20

Total

1.00

*B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered as 1st to 6th , from left to right: (1st 2nd 3rd 4th 5th6th7th 8th 9th 10th )

Compute the following conditional probabilities:

 

2.1) Pr{B Shows  1st | Yellow Shows}

 

Sequence*

Probability

RRBBRRYRRR

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

BGYGYRYGYY

.25

RRYYGRRBBY

.10

YYGBYYBGRR

.20

Total

.90

 

Pr{Yellow Shows} = Pr{One of RRBBRRYRRR, BBYYGGYGBR, GRRGGYBRGG, BGYGYRYGYY, RRYYGRRBBY or  YYGBYYBGRR  Shows} = Pr{RRBBRRYRRR} + Pr{BBYYGGYGBR} + Pr{GRRGGYBRGG} + Pr{BGYGYRYGYY} + Pr{RRYYGRRBBY} + Pr{YYGBYYBGRR} =

= .10+ .15 + .10 + .25 + .10 + .20 = .90

 

Pr{B Shows 1st and Yellow Shows } = Pr{ One of  BBYYGGYGBR or BGYGYRYGYY Shows} =

Pr{BBYYGGYGBR}+Pr{ BGYGYRYGYY} = .15+.25 = .40

 

Pr{ B Shows 1st | Yellow Shows} = .40 / .90 » .4444

 

2.2) Pr{ Green Shows  | “BR” Shows }

 

Sequence*

Probability

RRBBRRYRRR

.10

RRGGRGBRRB

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

Total

.45

 

 

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

Pr{RRBBRRYRRR} + Pr{RRGGRGBRRB} + Pr{BBYYGGYGBR} + Pr{GRRGGYBRGG} =

= .10+ .10 + 15 + .10 = .45

 

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

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

=.10 + 15 + .10 = .35

 

Pr{ Green Shows | “BR” Shows} = .35 / .45 » .7778

 

2.3) Pr{ Yellow Shows | Red Shows}

 

Sequence*

Probability

RRBBRRYRRR

.10

RRGGRGBRRB

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

BGYGYRYGYY

.25

RRYYGRRBBY

.10

YYGBYYBGRR

.20

Total

1.00

 

Pr{ Red Shows} = Pr{One of RRBBRRYRRR, RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG, BGYGYRYGYY, RRYYGRRBBY or YYGBYYBGRR Shows} = Pr{RRBBRRYRRR}+Pr{RRGGRGBRRB}+Pr{BBYYGGYGBR}+Pr{GRRGGYBRGG} + Pr{BGYGYRYGYY} + Pr{RRYYGRRBBY} + Pr{YYGBYYBGRR}= .10 + .10 + .15 + .10 + .25 + .10 + .20 = 1

 

Pr{ Red and Yellow Show} = Pr{One of RRBBRRYRRR, BBYYGGYGBR, GRRGGYBRGG, BGYGYRYGYY, RRYYGRRBBY or YYGBYYBGRR Shows} = Pr{RRBBRRYRRR} + Pr{BBYYGGYGBR} + Pr{GRRGGYBRGG} + Pr{BGYGYRYGYY} + Pr{RRYYGRRBBY} + Pr{YYGBYYBGRR}= .10 + .15 + .10 + .25 + .10 + .20 = .90

 

Pr{ Yellow Shows | Red Shows} = Pr{ Red and Yellow Show}/ Pr{ Red Shows} =  .90 / 1 = .90

 

Case Three

Random Variable

Pair of Fair Dice

 

We have a pair of independently operating, fair, dice: the first die with face values 1,2,3,4 and the second die with face values 1,2. 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: First, compute the product of the faces. Then:

 

if product of faces is odd, compute RandomVariable = HighTie

if product of faces is even, compute RandomVariable = LowTie

Where HIGHTIE = higher of the two face values or common value if tied and LOWTIE = lower of the two face values or common value if tied.

 

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

 

 

1

2

3

4

1

(1,1)

(2,1)

(3,1)

(4,1)

2

(1,2)

(2,2)

(3,2)

(4,2)

 

 

Pr{(a,b)} = Pr{a from 1st die}*Pr{b from 2nd die} = (1/4)*(1/4) = 1/16

 

(1,1): Pr{(1,1)} = Pr{1 from 1st d4}*Pr{1 from 2nd d4} = (1/4)*(1/2) = 1/8

(1,2): Pr{(1,2)} = Pr{1 from 1st d4}*Pr{2 from 2nd d4} = (1/4)*(1/2) = 1/8

(2,1): Pr{(2,1)} = Pr{2 from 1st d4}*Pr{1 from 2nd d4} = (1/4)*(1/2) = 1/8

(2,2): Pr{(2,2)} = Pr{2 from 1st d4}*Pr{2 from 2nd d4} = (1/4)*(1/2) = 1/8

(3,1): Pr{(3,1)} = Pr{3 from 1st d4}*Pr{1 from 2nd d4} = (1/4)*(1/2) = 1/8

(3,2): Pr{(3,2)} = Pr{3 from 1st d4}*Pr{2 from 2nd d4} = (1/4)*(1/2) = 1/8

(4,1): Pr{(4,1)} = Pr{4 from 1st d4}*Pr{1 from 2nd d4} = (1/4)*(1/2) = 1/8

(4,2): Pr{(4,2)} = Pr{4 from 1st d4}*Pr{2 from 2nd d4} = (1/4)*(1/2) = 1/8

 

3.2)   Compute the values of the random variable, showing in detail how these values are computed from each pair of face values.

 

RandomVariable = HighTie if Product of Faces is Odd

RandomVariable = LowTie if Product of Faces is Even



Pair

Product

Even/Odd

RV

1

2

3

4

1

(1,1)

1

Odd

1

(2,1)

2

Even

1

(3,1)

3

Odd

3

(4,1)

4

Even

1

2

(1,2)

2

Even

1

(2,2)

4

Even

2

(3,2)

6

Even

2

(4,2)

8

Even

2

 

(1,1) Þ 1*1=1Þ Odd Þ HighTie=1

(1,2) Þ 1*2=2Þ Even Þ LowTie=1

 

(2,1) Þ 2*1=2Þ Even Þ LowTie=1

(2,2) Þ 2*2=4Þ Even Þ LowTie=2

 

(3,1) Þ 3*1=3Þ Odd Þ HighTie=3

(3,2) Þ 3*2=6Þ Even Þ LowTie=2

 

(4,1) Þ 4*1=4Þ Even Þ LowTie=1

(4,2) Þ 4*2=8Þ Even Þ LowTie=2

 

3.3)   Compute the probability for each value of the random variable in 3.2, showing full detail.

 

Pair

Product

Even/Odd

RV

1

2

3

4

1

(1,1)

1

Odd

1

(2,1)

2

Even

1

(3,1)

3

Odd

3

(4,1)

4

Even

1

2

(1,2)

2

Even

1

(2,2)

4

Even

2

(3,2)

6

Even

2

(4,2)

8

Even

2

 

RV

Pair List

1

(1,1), (1,2), (2,1), (4,1)

2

(2,2), (3,2), (4,2)

3

(3,1)

 

 

Pr{Random Variable = 1} = Pr{One of (1,1), (1,2), (2,1) or (4,1) Shows}= Pr{(1,1)} + Pr{(1,2)}+ Pr{(2,1)}+ Pr{(4,1)} = (1/8) + (1/8) + (1/8) + (1/8) = 4/8

 

Pr{Random Variable = 2} = Pr{One of  (2,2), (3,2), (4,2) Shows}= Pr{(2,2)} + Pr{(3,2)}+ Pr{(4,2)} = (1/8) + (1/8) + (1/8) = 3/8

 

Pr{Random Variable = 3} = Pr{(3,1)} =  1/8

 

Case Four

Color Slot Machine

Probability Rules

 

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

 

Sequence*

Probability

RRBBRRYRRR

.10

RRGGRGBRRB

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

BGYGYRYGYY

.25

RRYYGRRBBY

.10

YYGBYYBGRR

.20

Total

1.00

 

*B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered as 1st to 6th , from left to right: (1st 2nd 3rd 4th 5th6th7th 8th 9th 10th )

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.

 

4.1) Pr{“BR” Shows }

 

Pr{“BR” Shows } = Pr{One of RRBBRRYRRR, RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG Shows} = Pr{RRBBRRYRRR } + Pr{RRGGRGBRRB} + Pr{BBYYGGYGBR} + Pr{GRRGGYBRGG} = .10+.10+.15+.10 = .45. In long runs of plays, approximately 45% of observed color sequences contain “BR”.  

 

4.2) Pr{ Green Shows 4th }

 

Pr{ Green Shows 4th }= Pr{One of RRGGRGBRRB,  GRRGGYBRGG,  BGYGYRYGYY Shows} = Pr{ RRGGRGBRRB } + Pr{GRRGGYBRGG } + Pr{ BGYGYRYGYY } = .10+.10 +.25 = .45. In long runs of plays, approximately 45% of observed color sequences contain “G” in the 4th slot.

 

4.3) Pr{ Yellow Shows } – Use the Complementary Rule

 

Event = “Yellow Shows”

Other Event = “Yellow Does Not Show”

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

Pr{“Yellow Shows”} = 1 - Pr{“Yellow Does Not Show”} = 1 - .10 = .90 . In long runs of plays, approximately 90% of observed color sequences contain Yellow.

 

 

Work all four cases. Show full detail and work for full credit.