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: Long Run Argument, Perfect Samples, Prevalence of Viral Subtypes

 

Suppose that we have a virus, say XCV. Suppose further that XCV has known main types A, B, C and D. Infected individuals may present one or more types of XCV. Suppose that the probabilities for XCV types present in infected individuals are noted below:

 

XCV Profile

Probability

XCV−A

0.70

XCV−B

0.13

XCV−C

0.08

XCV−D

0.02

XCV−Multiple Types Present

0.04

XCV−Type or Types Unknown

0.03

Total

1.00

 

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

 

In long runs of random sampling XCV-infected individuals, approximately 70% of sampled individuals present XCV type A.

 

In long runs of random sampling XCV-infected individuals, approximately 13% of sampled individuals present XCV type B.

 

In long runs of random sampling XCV-infected individuals, approximately 8% of sampled individuals present XCV type C.

 

In long runs of random sampling XCV-infected individuals, approximately 2% of sampled individuals present XCV type D.

 

In long runs of random sampling XCV-infected individuals, approximately 4% of sampled individuals present XCV multiple types.

 

In long runs of random sampling XCV-infected individuals, approximately 3% of sampled individuals present XCV type or types unknown.

 

Compute the perfect sample of n=450 XCV-infected individuals, and describe the relationship of this perfect sample to real random samples of XCV-infected individuals.

 

Expected Count (for n=450) = 450*Probability

Expected Count (for XCV−A, n=450) = 450*0.70 = 315

Expected Count (for XCV−B, n=450) = 450*0.13 = 58.5

Expected Count (for XCV−C, n=450) = 450*0.08 = 36

Expected Count (for XCV−D, n=450) = 450*0.02 =  9

Expected Count (for XCV−Multiple Types, n=450) = 450*0.04 = 18

Expected Count (for XCV−Type or Types Unknown, n=450) = 450*0.03 = 13.5

 

In random samples of 450 XCV-infected individuals, approximately 315 of 450 sampled individuals present XCV type A.

 

In random samples of 450 XCV-infected individuals, approximately 58 or 59 of 450 sampled individuals present XCV type B.

 

In random samples of 450 XCV-infected individuals, approximately 36 of 450 sampled individuals present XCV type C.

 

In random samples of 450 XCV-infected individuals, approximately 9 of 450 sampled individuals present XCV type D.

 

In random samples of 450 XCV-infected individuals, approximately 18 of 450 sampled individuals present XCV multiple types.

 

In random samples of 450 XCV-infected individuals, approximately 13 or 14 of 450 sampled individuals present XCV type or types unknown.

 

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

0.90

 

Pr{Blue Shows} = Pr{One of  GRBB, BGBG, BBBB, RYGB, BYBY, 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 Green Both Show}

 

Color Sequence*

Color Sequence Probability

GRBB

0.15

BGBG

0.10

RYGB

0.25

YGBR

0.10

Total Blue and Green Show

0.60

 

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

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

0.15 + 0.10 + 0.25 + 0.10 = 0.60

 

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

 

 

Color Sequence*

Color Sequence Probability

YYYY

0.10

Total Other Event

0.10

 

Other Event = “Neither Blue Nor Green Show”

 

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 Blue or Green Shows

0.90

 

Pr{Blue or Green Shows} = Pr{One of  GRBB, BGBG, BBBB, RYGB, BYBY, 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 full work and detail for full credit.

 

Case Three, Conditional Probability, Color Bowl, Draws without Replacement

 

We have a bowl containing the following colors and counts of balls (color@count):

 

Blue @ 4, Green @ 3, Red @ 2, Yellow @ 1

 

Each trial of our experiment consists of four draws without replacement from the bowl. Compute the following conditional probabilities. Compute these directly.

 

Pr{Blue shows 4th | Blue shows 1st, Red shows 2nd , Green  Shows 3rd} 

 

Before 1st Draw: Blue @ 4, Green @ 3, Red @ 2, Yellow @ 1

After 1st Draw: Blue @ 3, Green @ 3, Red @ 2, Yellow @ 1

After 2nd Draw: Blue @ 3, Green @ 3, Red @ 1, Yellow @ 1

After 3rd Draw: Blue @ 3, Green @ 2, Red @ 1, Yellow @ 1

 

Pr{Blue shows 4th | Blue shows 1st, Red shows 2nd , Green  Shows 3rd} = 3/(3+2+1+1) = 3/7 

 

Pr{Green shows 3rd | Green shows 1st, Green shows 2nd, Blue shows 3rd} 

 

This is a typo, but taken as is, the answer is simple:

Pr{Green shows 3rd | Green shows 1st, Green shows 2nd, Blue shows 3rd} = 0, since we are given that Blue, not Green, shows/showed 3rd.

 

In class, I indicated that this was the intended probability:

 

Pr{Green shows 4th | Green shows 1st, Green shows 2nd, Blue shows 3rd} 

 

Before 1st Draw: Blue @ 4, Green @ 3, Red @ 2, Yellow @ 1

After 1st Draw: Blue @ 4, Green @ 2, Red @ 2, Yellow @ 1

After 2nd Draw: Blue @ 4, Green @ 1, Red @ 2, Yellow @ 1

After 3rd Draw: Blue @ 3, Green @ 1, Red @ 2, Yellow @ 1

 

Pr{Green shows 4th | Green shows 1st, Green shows 2nd, Blue shows 3rd} = 1/(3+1+2+1) = 1/7 

 

Pr{Yellow 2nd  | Yellow shows 1st}

 

Before 1st Draw: Blue @ 4, Green @ 3, Red @ 2, Yellow @ 1

After 1st Draw: Blue @ 4, Green @ 3, Red @ 2, Yellow @ 0

 

Pr{Yellow shows 2nd  | Yellow shows 1st} = 0/(4+3+2+0) = 0/9

 

Show all work and full detail for full credit.

 

Case Four: 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

4/10

5

1/4

2

3/10

6

1/4

3

2/10

7

1/4

4

1/10

8

1/4

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 Even

Thing = Low if Sum is Odd.

 

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

 

 

1(4/10)

2(3/10)

3(2/10)

4(1/10)

5(1/4)

(1,5)

(2,5)

(3,5)

(4,5)

6(1/4)

(1,6)

(2,6)

(3,6)

(4,6)

7(1/4)

(1,7)

(2,7)

(3,7)

(4,7)

8(1/4)

(1,8)

(2,8)

(3,8)

(4,8)

 

(1,5), (2,5), (3,5), (4,5), (1,6), (2,6), (3,6), (4,6), (1,7), (2,7), (3,7), (4,7), (1,8), (2,8), (3,8), (4,8)

 

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

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

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

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

 

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

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

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

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

 

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

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

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

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

 

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

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

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

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

Thing = Low if Sum is Odd.

 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 

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

 

Pr{Thing=1} = Pr{One of (1,6), (1,8) Shows} = Pr{(1,6)} + Pr{(1,8)} = (4/40) + (4/40) = 8/40

Pr{Thing=2} = Pr{One of (2,5), (2,7) Shows} = Pr{(2,5)} + Pr{(2,7)} = (3/40) + (3/40) = 6/40

Pr{Thing=3} = Pr{One of (3,6), (3,8) Shows} = Pr{(3,6)} + Pr{(3,8)} = (2/40) + (2/40) = 4/40

Pr{Thing=4} = Pr{One of (4,5), (4,7) Shows} = Pr{(4,5)} + Pr{(4,7)} = (1/40) + (1/40) = 2/40

Pr{Thing=5} = Pr{One of (1,5), (3,5) Shows} = Pr{(1,5)} + Pr{(3,5)} = (4/40) + (2/40) = 6/40

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

Pr{Thing=7} = Pr{One of (1,7), (3,7) Shows} = Pr{(1,7)} + Pr{(3,7)} = (4/40) + (2/40) = 6/40

Pr{Thing=8} = Pr{One of (2,8), (4,8) Shows} = Pr{(2,8)} + Pr{(4,8)} = (3/40) + (1/40) = 4/40

 

(8/40)+(6/40)+(4/400+(2/40)+6/40)+(4/40)+(6/40)+(4/40) = (8+6+4+2+6+4+6+4))/40 =40/40=1

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