Key | 1st Hourly | Math 1107 | Spring Semester 2010

 

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) and 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 and Your Toolsheet. Then hand all of this in to me.

 

Sign and Acknowledge:    I agree to follow this protocol: (initial)

 

Sign___________________________Name______________________________Date________________

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

Case One: Random Variables, Pair of Dice

 

We have a pair of independently operating, dice: the first die with face values 1, 2, 3, 4 and the second die with face values 5, 6, 7, 8.  

 

1st Die:

Face

Probability

2nd Die:

Face

Probability

1

1/4

5

1/10

2

1/4

6

2/10

3

1/4

7

3/10

4

1/4

8

4/10

Total

1

Total

1

 

Each trial of our experiment involves tossing the pair of dice, and noting the pair of face values. We compute random variables from the pair of face values as follows: Define SUM as the sum of the face values in the pair, HIGH as the higher of the two face values in the pair and LOW as the lower of the two face values in the pair. Define THING as:

 

Thing = High if Sum is Odd

Thing = Low if Sum is Even.

 

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

 

 

 

1(1/4)

2(1/4)

3(1/4)

4(1/4)

5(1/10)

(1,5)

(2,5)

(3,5)

(4,5)

6(2/10)

(1,6)

(2,6)

(3,6)

(4,6)

7(3/10)

(1,7)

(2,7)

(3,7)

(4,7)

8(4/10)

(1,8)

(2,8)

(3,8)

(4,8)

 

Pr{(1,5)} = Pr{1 from 1st}*Pr{5 from 2nd} = (1/4)*(1/10) = 1/40

Pr{(2,5)} = Pr{2 from 1st}*Pr{5 from 2nd} = (1/4)*(1/10) = 1/40            

Pr{(3,5)} = Pr{3 from 1st}*Pr{5 from 2nd} = (1/4)*(1/10) = 1/40

Pr{(4,5)} = Pr{4 from 1st}*Pr{5 from 2nd} = (1/4)*(1/10) = 1/40

 

Pr{(1,6)} = Pr{1 from 1st}*Pr{6 from 2nd} = (1/4)*(2/10) = 2/40

Pr{(2,6)} = Pr{2 from 1st}*Pr{6 from 2nd} = (1/4)*(2/10) = 2/40            

Pr{(3,6)} = Pr{3 from 1st}*Pr{6 from 2nd} = (1/4)*(2/10) = 2/40

Pr{(4,6)} = Pr{4 from 1st}*Pr{6 from 2nd} = (1/4)*(2/10) = 2/40

 

Pr{(1,7)} = Pr{1 from 1st}*Pr{7 from 2nd} = (1/4)*(3/10) = 3/40

Pr{(2,7)} = Pr{2 from 1st}*Pr{7 from 2nd} = (1/4)*(3/10) = 3/40            

Pr{(3,7)} = Pr{3 from 1st}*Pr{7 from 2nd} = (1/4)*(3/10) = 3/40

Pr{(4,7)} = Pr{4 from 1st}*Pr{7 from 2nd} = (1/4)*(3/10) = 3/40

 

Pr{(1,8)} = Pr{1 from 1st}*Pr{8 from 2nd} = (1/4)*(4/10) = 4/40

Pr{(2,8)} = Pr{2 from 1st}*Pr{8 from 2nd} = (1/4)*(4/10) = 4/40            

Pr{(3,8)} = Pr{3 from 1st}*Pr{8 from 2nd} = (1/4)*(4/10) = 4/40

Pr{(4,8)} = Pr{4 from 1st}*Pr{8 from 2nd} = (1/4)*(4/10) = 4/40

 

 

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

face values.

 

Thing = High if Sum is Odd

Thing = Low if Sum is Even.

 

Thing{(1,5)} = 1, Since Sum=6(Even)

Thing{(2,5)} = 5, Since Sum=7(Odd)

Thing{(3,5)} = 3, Since Sum=8(Even)

Thing{(4,5)} = 5, Since Sum=9(Odd)

Thing{(1,6)} = 6, Since Sum=7(Odd)

Thing{(2,6)} = 2, Since Sum=8(Even)

Thing{(3,6)} = 6, Since Sum=9(Odd)

Thing{(4,6)} = 4, Since Sum=10(Even)

Thing{(1,7)} = 1, Since Sum=8(Even)

Thing{(2,7)} = 7, Since Sum=9(Odd)

Thing{(3,7)} = 3, Since Sum=10(Even)

Thing{(4,7)} = 7, Since Sum=11(Odd)

Thing{(1,8)} = 8, Since Sum=9(Odd)

Thing{(2,8)} = 2, Since Sum=10(Even)

Thing{(3,8)} = 8, Since Sum=11(Odd)

Thing{(4,8)} = 4, Since Sum=12(Even)

 

Compute the probability for each value of THING, showing full detail.

 

Pr{Thing=1} = Pr{One (1,5) or (1,7) of Shows} = Pr{(1,5)} + Pr{(1,7)} = (1/40) + (3/40) = 4/40

Pr{Thing=2} = Pr{One (2,6) or (2,8) of Shows} = Pr{(2,6)} + Pr{(2,8)} = (2/40) + (4/40) = 6/40

Pr{Thing=3} = Pr{One (3,5) or (3,7) of Shows} = Pr{(1,5)} + Pr{(1,7)} = (1/40) + (3/40) = 4/40

Pr{Thing=4} = Pr{One (4,6) or (4,8) of Shows} = Pr{(4,6)} + Pr{(4,8)} = (2/40) + (4/40) = 6/40

Pr{Thing=5} = Pr{One (2,5) or (4,5) of Shows} = Pr{(2,5)} + Pr{(4,5)} = (1/40) + (1/40) = 2/40

Pr{Thing=6} = Pr{One (1,6) or (3,6) of Shows} = Pr{(1,6)} + Pr{(3,6)} = (2/40) + (2/40) = 4/40

Pr{Thing=7} = Pr{One (2,7) or (4,7) of Shows} = Pr{(2,7)} + Pr{(4,7)} = (3/40) + (3/40) = 6/40

Pr{Thing=8} = Pr{One (1,8) or (3,8) of Shows} = Pr{(1,8)} + Pr{(3,8)} = (4/40) + (4/40) = 8/40

 

(4/40)+(6/40)+(4/40)+(6/40)+(2/40)+(4/40)+(6/40)+(8/40)=(4+6+4+6+2+4+6+8)/40 = 40/40 =1

 

Show full work and detail for full credit.

 

Case Two: Probability Rules, Color Slot Machine

 

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

 

Color Sequence*

Color Sequence Probability

GRBB

0.15

BGBG

0.10

BBBB

0.05

RYGB

0.25

YYYY

0.10

BYBY

0.25

YGBR

0.10

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. Briefly interpret each probability using a long run or relative frequency argument.

                 

Pr{Blue Shows} 

 

 

Color Sequence*

Color Sequence Probability

GRBB

0.15

BGBG

0.10

BBBB

0.05

RYGB

0.25

BYBY

0.25

YGBR

0.10

Total

0.90

 

Pr{Blue Shows} = Pr{One of GRBB, BGBG, BBBB, RYGB, BYBY or YGBR Shows} =

Pr{GRBB} + Pr{BGBG} + Pr{BBBB} + Pr{RYGB} + Pr{BYBY} + Pr{YGBR} =

0.15 + 0.10 + 0.05 + 0.25 + 0.25 + 0.10 = 0.90

 

Pr{Blue and Red Both Show}

 

Color Sequence*

Color Sequence Probability

GRBB

0.15

RYGB

0.25

YGBR

0.10

Total

0.50

 

Pr{Blue and Red Both Show} = Pr{One of GRBB, RYGB or YGBR Shows} =

Pr{GRBB} + Pr{RYGB} + Pr{YGBR} = 0.15 + 0.25 + 0.10 = 0.50

 

Pr{Blue or Green Shows} - Use the Complementary Rule

 

Color Sequence*

Color Sequence Probability

YYYY

0.10

Total

0.10

 

Other Event = “Neither Blue Nor Green Shows”

Pr{Other Event} = Pr{YYYY} = 0.10

Pr{Blue or Green Shows} = 1 - Pr{Other Event} = 1 - Pr{YYYY} = 1 - 0.10 = 0.90

 

Check:

 

Color Sequence*

Color Sequence Probability

GRBB

0.15

BGBG

0.10

BBBB

0.05

RYGB

0.25

BYBY

0.25

YGBR

0.10

Total

0.90

 

Pr{Blue or Green Shows} = Pr{One of GRBB, BGBG, BBBB, RYGB, BYBY or YGBR Shows} =

Pr{GRBB} + Pr{BGBG} + Pr{BBBB} + Pr{RYGB} + Pr{BYBY} + Pr{YGBR} =

0.15 + 0.10 + 0.05 + 0.25 + 0.25 + 0.10 = 0.90

 

Show all work and full detail for full credit.

 

Case Three: Long Run Argument, Perfect Samples, Landsteiner's Human Blood Types

 

In the early 20th century, an Austrian scientist named Karl Landsteiner classified blood according to chemical molecular differences – surface antigen proteins. Landsteiner observed two distinct chemical molecules present on the surface of the red blood cells. He labeled one molecule "A" and the other molecule "B." If the red blood cell had only "A" molecules on it, that blood was called type A. If the red blood cell had only "B" molecules on it, that blood was called type B. If the red blood cell had a mixture of both molecules, that blood was called type AB. If the red blood cell had neither molecule, that blood was called type O. The Rh(esus) factor is a blood protein isolated in Rhesus monkeys by K. Landsteiner and A. S. Wiener that is also present in some humans. Humans with this protein are said to be Rh+; those without are said to be Rh-. The introduction of Rh+ blood in Rh- individuals causes a strong immune response, hence individuals who are Rh- cannot safely receive Rh+ donor blood. Safe blood transfusing requires careful matching of ABO and Rh blood types. Suppose that the probabilities for blood types for US residents are noted below:

 

Blood Type (Rh Factor)

Probability

O(+)

0.40

O(-)

0.05

A(+)

0.33

A(-)

0.07

B(+)

0.10

B(-)

0.01

AB(+)

0.025

AB(-)

0.015

Total

1.00

 

Interpret each probability using the Long Run Argument.          

 

In long runs of random sampling of US residents, approximately 40% of sampled US residents present type O+ blood.

 

In long runs of random sampling of US residents, approximately  5% of sampled US residents present type O- blood.

 

In long runs of random sampling of US residents, approximately 33% of sampled US residents present type A+ blood.

 

In long runs of random sampling of US residents, approximately  7% of sampled US residents present type A- blood.

 

In long runs of random sampling of US residents, approximately 10% of sampled US residents present type B+ blood.

 

In long runs of random sampling of US residents, approximately  1% of sampled US residents present type B- blood.

 

In long runs of random sampling of US residents, approximately 2.5% of sampled US residents present type AB+ blood.

 

In long runs of random sampling of US residents, approximately  1.5% of sampled US residents present type AB- blood.

 

Compute the perfect sample of n=3000 US residents, and describe the relationship of this perfect sample to real random samples of US residents.

 

 

Expected Count = 3000*Probability

Expected Count for O+, n=3000 = 3000*0.40 = 1200

Expected Count for O-, n=3000 = 3000*0.05 =   150

Expected Count for A+, n=3000 = 3000*0.33 =   990

Expected Count for A-, n=3000 = 3000*0.07 =   210

Expected Count for B+, n=3000 = 3000*0.10 =   300

Expected Count for B-, n=3000 = 3000*0.01 =     30

Expected Count for AB+, n=3000 = 3000*0.025 = 75

Expected Count for AB-, n=3000 = 3000*0.015 = 45

 

In random samples of 3000 US residents, approximately 1200 of 3000 sampled US residents present type O+ blood.

 

In random samples of 3000 US residents, approximately  150 of 3000 sampled US residents present type O- blood.

 

In random samples of 3000 US residents, approximately 990 of 3000 sampled US residents present type A+ blood.

 

In random samples of 3000 US residents, approximately  210 of 3000 sampled US residents present type A- blood.

 

In random samples of 3000 US residents, approximately 300 of 3000 sampled US residents present type B+ blood.

 

In random samples of 3000 US residents, approximately  30 of 3000 sampled US residents present type B- blood.

 

In random samples of 3000 US residents, approximately 75 of 3000 sampled US residents present type AB+ blood.

 

In random samples of 3000 US residents, approximately  45 of 3000 sampled US residents present type AB- blood.

 

Show all work and full detail for full credit. Provide complete discussion for full credit.

 

Case Four: Color Slot Machine, Conditional Probabilities

 

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

 

Sequence*

Probability

RRYRRR

.10

RGBRRB

.10

BBYGBR

.15

GRRRGG

.10

BGYGYY

.25

RRYYGY

.10

YYGBYY

.20

Total

1.00

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

Compute the following conditional probabilities:

 

Pr{ Yellow Shows Exactly Three Times | Blue Shows}

 

Sequence*

Probability

RGBRRB

.10

BBYGBR

.15

BGYGYY

.25

YYGBYY

.20

Total Blue Shows

.70

 

Pr{Blue Shows} = Pr{One of RGBRRB, BBYGBR, BGYGYY or YYGBYY Shows} =

Pr{RGBRRB} + Pr{BBYGBR} + Pr{BGYGYY} + Pr{YYGBYY} = .10 + .15 + .25 + .20 = .70

 

Sequence*

Probability

BGYGYY

.25

Total Yellow Exactly Three Times and Blue

.25

 

Pr{ Yellow Shows Exactly Three Times and Blue Shows} = Pr{BGYGYY} =.25

 

 

Pr{ Yellow Shows Exactly Three Times | Blue Shows}=

Pr{ Yellow Shows Exactly Three Times and Blue Shows}/ Pr{Blue Shows} = .25/.70

 

Pr{ Yellow Shows | “BR” Shows }

 

Sequence*

Probability

RGBRRB

.10

BBYGBR

.15

Total “BR” Shows

.25

 

Pr{“BR” Shows} = Pr{One of RGBRRB or BBYGBR Shows} = Pr{RGBRRB} + Pr{BBYGBR} = .10 + .15 = .25

 

 

Sequence*

Probability

BBYGBR

.15

Total Yellow and “BR” Shows

.15

 

Pr{Yellow and “BR” Shows} = Pr{BBYGBR} = .15

 

 

Pr{ Yellow Shows | “BR” Shows }= Pr{ Yellow and “BR” Show }/Pr{ “BR” Shows } = .15/.25

 

Pr{ Blue Shows | Yellow Shows}

 

 

Sequence*

Probability

RRYRRR

.10

BBYGBR

.15

BGYGYY

.25

RRYYGY

.10

YYGBYY

.20

Total Yellow Shows

.80

 

Pr{Yellow Shows} = Pr{One of RRYRRR, BBYGBR, BGYGYY, RRYYGY or YYGBYY Shows} =

Pr{RRYRRR} + Pr{BBYGBR} + Pr{BGYGYY} + Pr{RRYYGY} + Pr{YYGBYY} =

.10 + .15 + .25 + .10 + .20 = .80 

 

 

Sequence*

Probability

BBYGBR

.15

BGYGYY

.25

YYGBYY

.20

Total Blue and Yellow Show

.60

 

Pr{Blue and Yellow Show} = Pr{One of BBYGBR, BGYGYY or YYGBYY Shows} =

Pr{BBYGBR} + Pr{BGYGYY} + Pr{YYGBYY} =

.15 + .25 + .20 = .60 

 

Pr{ Blue Shows | Yellow Shows} = Pr{ Blue and Yellow Show}/ Pr{Yellow Shows} = .60/.80

 

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