Key

1^st Hourly

Math 1107

Fall Semester 2009

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

Name (PRINTED) Signature Date

Case One | 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
GBRB	0.20
GBBR	0.40
YYYY	0.10
BBYY	0.18
RRYY	0.02
RRRR	0.10
Total	1.000

*B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered as 1^st to 4^th, from left to right.

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.

a) Pr{Either Yellow Shows or Green Shows, but not both}

Color Sequence*	Color Sequence Probability
GBRB	0.20
GBBR	0.40
YYYY	0.10
BBYY	0.18
RRYY	0.02
Total	0.90

Pr{Either Yellow Shows or Green Shows, but not both} =

Pr{ One of GBRB, GBBR, YYYY, BBYY or RRYY Shows} =

Pr{GBRB } + Pr{GBBR } + Pr{ YYYY } + Pr{ BBYY } + Pr{ RRYY } =

0.20 + 0.40 + 0.10 + 0.18 + 0.02 = 0.90

b) Pr{Yellow and Blue both Show} – Use the Complementary Rule

Other Event = “Neither Yellow nor Blue Show, or Yellow Shows but not Blue, or Blue Shows but not Yellow”

Color Sequence*	Color Sequence Probability
GBRB	0.20
GBBR	0.40
YYYY	0.10
RRYY	0.02
RRRR	0.10
Total	0.82

Pr{ Neither Yellow nor Blue Show} = Pr{One of GBRB, GBBR, YYYY, RRYY or RRRR Shows } =

Pr{ GBRB } + Pr{ GBBR } + Pr{ YYYY } + Pr{ RRYY } + Pr{ RRRR } =

0.20 + 0.40 + 0.10 + 0.02 + 0.10 = 0.82

Pr{Yellow and Blue Both Show} = 1 – Pr{ Neither Yellow nor Blue Show} = 1 – 0.82 = .18

Check Pr{Yellow and Blue Both Show} = Pr{ BBYY } = .18

c) Pr{Red Shows | Green Shows} - This is a Conditional Probability.

Color Sequence*	Color Sequence Probability
GBRB	0.20
GBBR	0.40
Total	0.60

Pr{ Green Shows } = Pr{One of GBRB or GBBR Shows} = Pr{ GBRB } + Pr{ GBBR } = 0.20 + 0.40 = 0.60

Color Sequence*	Color Sequence Probability
GBRB	0.20
GBBR	0.40
Total	0.60

Pr{ Red and Green Both Show } = Pr{One of GBRB or GBBR Shows} = Pr{ GBRB } + Pr{ GBBR } = 0.20 + 0.40 = 0.60

Pr{ Red Shows | Green Shows } = Pr{ Red and Green Both Show } = / Pr{ Green Shows } = 0.60/0.60 = 1

Case Two | Random Variable | 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. In each of the following, show your intermediate steps and work.

a) List the possible pairs and compute a probability for each pair.

(d4,d2)	1	2	4	5
0	(1,0)	(2,0)	(4,0)	(5,0)
3	(1,3)	(2,3)	(4,3)	(5,3)

Pr{ (1,0) } = Pr{1 from d4}*Pr{0 from d2} = (1/4)*(1/2) = 1/8

Pr{ (2,0) } = Pr{2 from d4}*Pr{0 from d2} = (1/4)*(1/2) = 1/8

Pr{ (4,0) } = Pr{4 from d4}*Pr{0 from d2} = (1/4)*(1/2) = 1/8

Pr{ (5,0) } = Pr{5 from d4}*Pr{0 from d2} = (1/4)*(1/2) = 1/8

Pr{ (1,3) } = Pr{1 from d4}*Pr{3 from d2} = (1/4)*(1/2) = 1/8

Pr{ (2,3) } = Pr{2 from d4}*Pr{3 from d2} = (1/4)*(1/2) = 1/8

Pr{ (4,3) } = Pr{4 from d4}*Pr{3 from d2} = (1/4)*(1/2) = 1/8

Pr{ (5,3) } = Pr{5 from d4}*Pr{3 from d2} = (1/4)*(1/2) = 1/8

b) List the possible values for THING and compute a probability for each value of THING.

THING { (1,0) } = 10 ─ (1 + 0) = 9

THING { (2,0) } = 10 ─ (2 + 0) = 8

THING { (4,0) } = 10 ─ (4 + 0) = 6

THING { (1,3) } = 10 ─ (1 + 3) = 6

THING { (5,0) } = 10 ─ (5 + 0) = 5

THING { (2,3) } = 10 ─ (2 + 3) = 5

THING { (4,3) } = 10 ─ (4 + 3) = 3

THING { (5,3) } = 10 ─ (5 + 3) =2

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

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

Pr{THING=5} = Pr{(5,0)} + Pr{(2,3)} = (1/8) + (1/8) = 2/8

Pr{THING=6} = Pr{(4,0)} + Pr{(1,3)} = (1/8) + (1/8) = 2/8

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

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

Case Three | Long Run Argument / Perfect Samples | Consumer Credit Score

Fair Isaac Corp., a California-based financial company, developed a consumer credit score, a number that summarizes the risk present in lending money to a consumer. The consumer credit score ranges from 300 to 850. Suppose that the probabilities for consumer credit scores for US residents are noted below:

Credit Score	Probability
499 and below	.02
500-599	.13
600-699	.27
700-799	.45
800 and above	.13
Total	1.00

In each of the following, show your intermediate steps and work.

a) Interpret each probability using the Long Run Argument.

In long runs of random sampling, approximately 2% of sampled US residents have FICO scores of 499 or less.

In long runs of random sampling, approximately 13% of sampled US residents have FICO scores of 500 to 599.

In long runs of random sampling, approximately 27% of sampled US residents have FICO scores of 600 to 699.

In long runs of random sampling, approximately 45% of sampled US residents have FICO scores of 700 to 799.

In long runs of random sampling, approximately 13% of sampled US residents have FICO scores of 800 or or more.

b) Compute the perfect sample of 500 US residents, and describe the relationship of this perfect sample to real random samples of US residents.

Credit Score	Probability	E500
499 and below	.02	500*.02=10
500-599	.13	500*.13=65
600-699	.27	500*.27=135
700-799	.45	500*.45=225
800 and above	.13	500*.13=65
Total	1.00	500*1.00=500

E_{499 and below}= n*P_{499 and below} = 500*.02=10

E_500-599= n*P_500-599 =500*.13=65

E_600-699= n*P_600-699 =500*.27=135

E_700-799= n*P_700-799 =500*.45=225

E_{800 and above}= n*P_{800 and above} =500*.13=65

In random samples of 500, approximately 10 sampled US residents have FICO scores of 499 or less.

In random samples of 500, approximately 65 of sampled US residents have FICO scores of 500 to 599.

In random samples of 500, approximately 135 of sampled US residents have FICO scores of 600 to 699.

In random samples of 500, approximately 225 of sampled US residents have FICO scores of 700 to 799.

In random samples of 500, approximately 65 of sampled US residents have FICO scores of 800 or or more.

Case Four | Conditional Probability | Color Bowl/Draws without Replacement

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

White @ 2, Black @ 3, Blue @ 5, Green @ 5, Red @ 4, Yellow @ 2

Each trial of our experiment consists of five draws without replacement from the bowl. Compute the following conditional probabilities. Compute these directly. In each of the following, show your intermediate steps and work. Compute the following conditional probabilities:

a) Pr{ Yellow shows 3^rd | Blue Shows 2^nd and Yellow shows 1^st}

Before 1^st Draw: White @ 2, Black @ 3, Blue @ 5, Green @ 5, Red @ 4, Yellow @ 2

After Yellow Shows on 1^st Draw: White @ 2, Black @ 3, Blue @ 5, Green @ 5, Red @ 4, Yellow @ 1

After Blue Shows on 2^nd Draw: White @ 2, Black @ 3, Blue @ 4, Green @ 5, Red @ 4, Yellow @ 1

Pr{ Yellow shows 3^rd | Blue Shows 2^nd and Yellow shows 1^st} = 1/(2+3+4+5+4+1) = 1/19

b) Pr{ Blue shows 5^th | Black shows 1^st, Blue shows 2^nd, Blue shows 3^rd, and Green shows 4^th }

Before 1^st Draw: White @ 2, Black @ 3, Blue @ 5, Green @ 5, Red @ 4, Yellow @ 2

After Black Shows on 1^st Draw: White @ 2, Black @ 2, Blue @ 5, Green @ 5, Red @ 4, Yellow @ 2

After Blue Shows on 2^nd Draw: White @ 2, Black @ 2, Blue @ 4, Green @ 5, Red @ 4, Yellow @ 2

After Blue Shows on 3^rd Draw: White @ 2, Black @ 2, Blue @ 3, Green @ 5, Red @ 4, Yellow @ 2

After Green Shows on4^th Draw: White @ 2, Black @ 2, Blue @ 3, Green @ 4, Red @ 4, Yellow @ 2

Pr{ Blue shows 5^th | Black shows 1^st, Blue shows 2^nd, Blue shows 3^rd, and Green shows 4^th }  = 3/(2+2+3+4+4+2) = 3/17

c) Pr{ Blue shows 3^rd | Yellow shows 1^st, Blue shows 2^nd }

Before 1^st Draw: White @ 2, Black @ 3, Blue @ 5, Green @ 5, Red @ 4, Yellow @ 2

After Yellow Shows on 1^st Draw: White @ 2, Black @ 3, Blue @ 5, Green @ 5, Red @ 4, Yellow @ 1

After Blue Shows on 2^nd Draw: White @ 2, Black @ 3, Blue @ 4, Green @ 5, Red @ 4, Yellow @ 1

Pr{ Blue shows 3^rd | Yellow shows 1^st, Blue shows 2^nd } = 4/(2+3+4+5+4+1) = 4/19

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