Key

The 1st Hourly

Math 1107

Spring Semester 2009

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. Do not use any external resources during this hourly.

Sign and Acknowledge: I agree to follow this protocol. 

 

Name (PRINTED)                                             Signature                                             Date

Case One | Probability Rules | Color Slot Machine

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

 

Sequence

Probability

RBRRYRGG

.10

RRGGBRRB

.10

BBYYYGBR

.15

GRRGGRGG

.10

BGYYYRYY

.25

RRYYGRRB

.10

YYGBYYBR

.20

Total

1.00

 

1. Define BLUECOUNT as the number of times that blue shows in the color sequence.

Compute Pr{BLUECOUNT = 2 or 3}.

 

Sequence

Blue Count

Probability

RBRRYRGG

1

.10

RRGGBRRB

2

.10

BBYYYGBR

3

.15

GRRGGRGG

0

.10

BGYYYRYY

1

.25

RRYYGRRB

1

.10

YYGBYYBR

2

.20

Total

 

1.00

 

Here are the full computations for BLUECOUNT:

 

Pr{BLUECOUNT = 0} = Pr{GRRGGRGG} = .10

 

Pr{BLUECOUNT = 1} = Pr{One of  RBRRYRGG, BGYYYRYY or RRYYGRRB Shows } =

Pr{ RBRRYRGG} + Pr{ BGYYYRYY} + Pr{ RRYYGRRB } = .10 + .25 + .10 = .45

 

Pr{BLUECOUNT = 2} = Pr{One of  RRGGBRRB, YYGBYYBR  Shows } = Pr{ RRGGBRRB } +

Pr{ YYGBYYBR } = .10 + .20 = .30

 

Pr{BLUECOUNT = 3} = Pr{ BBYYYGBR } = .15

 

Here is the required calculation:

 

Pr{BLUECOUNT = 2 or 3} = Pr{One of  RRGGBRRB, YYGBYYBR or BBYYYGBR  Shows } =

Pr{ RRGGBRRB } + Pr{ YYGBYYBR } + Pr{ BBYYYGBR }= .10 + .20 + .15= .45

 

 

2. Compute Pr{“RY” Shows}.

 

Sequence

“RY” Shows

Probability

RBRRYRGG

Yes

.10

RRGGBRRB

No

.10

BBYYYGBR

No

.15

GRRGGRGG

No

.10

BGYYYRYY

Yes

.25

RRYYGRRB

No

.10

YYGBYYBR

No

.20

Total

 

1.00

 

Pr{“RY”} = Pr{One of RBRRYRGG, BGYYYRYY or RRYYGRRB  Shows} =

Pr{ RBRRYRGG } +  Pr{ BGYYYRYY } + Pr{RRYYGRRB} = .10 + .25 + .10 = .45

 

3. Compute Pr{Yellow Shows}. Use the complementary rule.

 

Sequence

Yellow Shows

Probability

RBRRYRGG

Yes

.10

RRGGBRRB

No

.10

BBYYYGBR

Yes

.15

GRRGGRGG

No

.10

BGYYYRYY

Yes

.25

RRYYGRRB

Yes

.10

YYGBYYBR

Yes

.20

Total

 

1.00

Other Event = “Yellow Does Not Show”

Pr{OE} = Pr{“Yellow Does Not Show} = Pr{One of RRGGBRRB or } = Pr{ RRGGBRRB } + Pr{ GRRGGRGG } = .10 + .10 = .20

Pr{“Yellow Shows”} = 1 – Pr{“Yellow Does Not Show} = 1 – .20 = .80

Check:

Pr{Yellow Shows} = Pr{One of  RBRRYRGG, BBYYYGBR, BGYYYRYY, RRYYGRRB or YGBYYBR Shows} = Pr{RBRRYRGG}+ Pr{BBYYYGBR}+ Pr{BGYYYRYY}+ Pr{RRYYGRRB}+ Pr{YGBYYBR} =

.10+.15+.25+.10+.20 = .80

Case Two | Long Run Argument, Perfect Samples | Maternal Age

Suppose that the following probability model applies to year 2006 United States Resident Live Births:

Maternal Age

Probability

Under 15

0.0015

15-19

0.1021

20-29

0.5304

30-39

0.3397

40+

0.0263

Total

1

 

Interpret each probability using the Long Run Argument.

 

In long runs of random sampling year 2006 United States Resident Live Births, approximately .15% of sampled births are to mothers aged less than 15 years.

 

In long runs of random sampling year 2006 United States Resident Live Births, approximately 10.21% of sampled births are to mothers aged between 15 and 19 years.

 

In long runs of random sampling year 2006 United States Resident Live Births, approximately 53.04% of sampled births are to mothers aged between 20 and 29 years.

 

In long runs of random sampling year 2006 United States Resident Live Births, approximately 33.97% of sampled births are to mothers aged between 30 and 39 years.

 

In long runs of random sampling year 2006 United States Resident Live Births, approximately 2.63% of sampled births are to mothers aged 40 or more years.

 

Compute and discuss Perfect Samples for n=3500.

 

Maternal Age

Probability

E3500

Under 15

0.0015

5.25

15-19

0.1021

357.35

20-29

0.5304

1856.4

30-39

0.3397

1188.95

40+

0.0263

92.05

Total

1

3500

 

E3500<15 = 3500*P<15 = 3500*0.0015 = 5.25

E350015-19 = 3500*P15-19 = 3500*0.1021 = 357.35

E350020-29 = 3500*P20-29 = 3500*0.5304 = 1856.4

E350030-39= 3500*P30-39 = 3500*0.3397 = 1188.95

E350040+ = 3500*P40+ = 3500*0.0263 = 92.05

 

In random samples of 3500 year 2006 United States Resident Live Births, approximately 5.25 sampled births are to mothers aged less than 15 years.

 

In random samples of 3500 year 2006 United States Resident Live Births, approximately 357.35 sampled births are to mothers aged between 15 and 19 years.

 

In random samples of 3500 year 2006 United States Resident Live Births, approximately 1856.4 sampled births are to mothers aged between 20 and 29 years.

 

In random samples of 3500 year 2006 United States Resident Live Births, approximately 1188.95 sampled births are to mothers aged between 30 and 39 years.

 

In random samples of 3500 year 2006 United States Resident Live Births, approximately 92.05 sampled births are to mothers aged 40 or more years.

 

Case Three | Random Variables Pair of Dice | Random Variable

We have a pair of dice– note the probability models for the dice below. 

1st d5

2nd d3

Face

Probability

Face

Probability

2

1/5

1

15/30

3

1/5

7

5/30

4

1/5

8

10/30

5

1/5

Total

30/30=1

6

1/5

 

 

Total

5/5=1

 

 

We assume that the dice operate separately and independently of each other. Suppose that our experiment consists of tossing the dice, and noting the resulting face-value-pair.

HIGH = “Highest of the Two Face Values in the Pair” and

 

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

 

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

 

Pr{(2,1)}=Pr{2 from d5}*Pr{1 from d3} = (1/5)*(15/30) = 15/150

Pr{(2,7)}=Pr{2 from d5}*Pr{7 from d3} = (1/5)*(5/30) = 5/150

Pr{(2,8)}=Pr{2 from d5}*Pr{8 from d3} = (1/5)*(10/30) = 10/150

 

Pr{(3,1)}=Pr{3 from d5}*Pr{1 from d3} = (1/5)*(15/30) = 15/150

Pr{(3,7)}=Pr{3 from d5}*Pr{7 from d3} = (1/5)*(5/30) = 5/150

Pr{(3,8)}=Pr{3 from d5}*Pr{8 from d3} = (1/5)*(10/30) = 10/150

 

Pr{(4,1)}=Pr{4 from d5}*Pr{1 from d3} = (1/5)*(15/30) = 15/150

Pr{(4,7)}=Pr{4 from d5}*Pr{7 from d3} = (1/5)*(5/30) = 5/150

Pr{(4,8)}=Pr{4 from d5}*Pr{8 from d3} = (1/5)*(10/30) = 10/150

 

Pr{(5,1)}=Pr{5 from d5}*Pr{1 from d3} = (1/5)*(15/30) = 15/150

Pr{(5,7)}=Pr{5 from d5}*Pr{7 from d3} = (1/5)*(5/30) = 5/150

Pr{(5,8)}=Pr{5 from d5}*Pr{8 from d3} = (1/5)*(10/30) = 10/150

 

Pr{(6,1)}=Pr{6 from d5}*Pr{1 from d3} = (1/5)*(15/30) = 15/150

Pr{(6,7)}=Pr{6 from d5}*Pr{7 from d3} = (1/5)*(5/30) = 5/150

Pr{(6,8)}=Pr{6 from d5}*Pr{8 from d3} = (1/5)*(10/30) = 10/150

 

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

 

HIGH{(2,1)}=2

HIGH {(2,7)}=7

HIGH {(2,8)}=8

 

HIGH{(3,1)}=3

HIGH{(3,7)}=7

HIGH{(3,8)}=8

 

HIGH{(4,1)}=4

HIGH{(4,7)}=7

HIGH{(4,8)}=8

 

HIGH{(5,1)}=5

HIGH{(5,7)}=7

HIGH{(5,8)}=8

 

HIGH{(6,1)}=6

HIGH{(6,7)}=7

HIGH{(6,8)}=8

 

3. Compute the probability for each value of HIGH from step 2, showing full detail.

 

Pr{HIGH=1} = 0

Pr{HIGH=2} = Pr{(2,1)} = 15/150

Pr{HIGH=3} = Pr{(3,1)} = 15/150

Pr{HIGH=4} = Pr{(4,1)} = 15/150

Pr{HIGH=5} = Pr{(5,1)} = 15/150

Pr{HIGH=6} = Pr{(6,1)} = 15/150

 

Pr{HIGH=7} = Pr{One of (2,7),(3,7),(4,7),(5,7) or (6,7)  Shows} = Pr{(2,7)}+Pr{(3,7) }+Pr{ (4,7) }+

Pr{(5,7)}+Pr{(6,7)} = (5/150) + (5/150) + (5/150) + (5/150) + (5/150) = 25/150

 

Pr{HIGH=8} = Pr{One of (2,8),(3,8),(4,8),(5,8) or (6,8)  Shows} = Pr{(2,8)}+Pr{(3,8) }+Pr{ (4,8) }+

Pr{(5,8)}+Pr{(6,8)} = (10/150) + (10/150) + (10/150) + (10/150) + (10/150) = 50/150

 

Show all work and detail for full credit.

 

Case Four | Color Slot Machine | Conditional Probabilities

Using the slot machine from Case One,

Sequence

Probability

RBRRYRGG

.10

RRGGBRRB

.10

BBYYYGBR

.15

GRRGGRGG

.10

BGYYYRYY

.25

RRYYGRRB

.10

YYGBYYBR

.20

Total

1.00

 

Compute Pr{Red Shows | Green Shows}.

 

Sequence

Probability

RBRRYRGG

.10

RRGGBRRB

.10

BBYYYGBR

.15

GRRGGRGG

.10

BGYYYRYY

.25

RRYYGRRB

.10

YYGBYYBR

.20

Total

1.00

 

Pr{Green} = Pr{One of  RBRRYRGG, RRGGBRRB, BBYYYGBR, GRRGGRGG, BGYYYRYY,

RRYYGRRB or YYGBYYBR Shows} = Pr{RBRRYRGG}+Pr{RRGGBRRB}+Pr{BBYYYGBR}+Pr{GRRGGRGG}+

Pr{BGYYYRYY}+Pr{RRYYGRRB}+Pr{YYGBYYBR} = .10+.10+.15+.10+.25+.10+.20 = 1.0

 

Sequence

Probability

RBRRYRGG

.10

RRGGBRRB

.10

BBYYYGBR

.15

GRRGGRGG

.10

BGYYYRYY

.25

RRYYGRRB

.10

YYGBYYBR

.20

Total

1.00

 

Pr{Red and Green} = Pr{One of  RBRRYRGG, RRGGBRRB, BBYYYGBR, GRRGGRGG, BGYYYRYY,

RRYYGRRB or YYGBYYBR Shows} = Pr{RBRRYRGG}+Pr{RRGGBRRB}+Pr{BBYYYGBR}+Pr{GRRGGRGG}+

Pr{BGYYYRYY}+Pr{RRYYGRRB}+Pr{YYGBYYBR} = .10+.10+.15+.10+.25+.10+.20 = 1.0

 

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

 

Compute Pr{Yellow Shows | Blue Shows}.

 

Sequence

Probability

RBRRYRGG

.10

RRGGBRRB

.10

BBYYYGBR

.15

BGYYYRYY

.25

RRYYGRRB

.10

YYGBYYBR

.20

Total

.90

Pr{Blue} = Pr{One of  RBRRYRGG, RRGGBRRB, BBYYYGBR, BGYYYRYY,

RRYYGRRB or YYGBYYBR Shows} = Pr{RBRRYRGG}+Pr{RRGGBRRB}+Pr{BBYYYGBR}+

Pr{BGYYYRYY}+Pr{RRYYGRRB}+Pr{YYGBYYBR} = .10+.10+.15+.25+.10+.20 = .90

 

Sequence

Probability

RBRRYRGG

.10

BBYYYGBR

.15

BGYYYRYY

.25

RRYYGRRB

.10

YYGBYYBR

.20

Total

.80

Pr{Yellow and Blue} = Pr{One of  RBRRYRGG, BBYYYGBR, BGYYYRYY,

RRYYGRRB or YYGBYYBR Shows} = Pr{RBRRYRGG}+ Pr{BBYYYGBR}+

Pr{BGYYYRYY}+Pr{RRYYGRRB}+Pr{YYGBYYBR} = .10+.15+.25+.10+.20 = .80

 

Pr{Yellow Shows | Blue Shows}= Pr{Yellow and Blue Show}/ Pr{ Blue Shows} = .80/.90

 

 

Compute Pr{Green Shows | Red Shows}.

 

 

Sequence

Probability

RBRRYRGG

.10

RRGGBRRB

.10

BBYYYGBR

.15

GRRGGRGG

.10

BGYYYRYY

.25

RRYYGRRB

.10

YYGBYYBR

.20

Total

1.00

 

Pr{Red} = Pr{One of  RBRRYRGG, RRGGBRRB, BBYYYGBR, GRRGGRGG, BGYYYRYY,

RRYYGRRB or YYGBYYBR Shows} = Pr{RBRRYRGG}+Pr{RRGGBRRB}+Pr{BBYYYGBR}+Pr{GRRGGRGG}+

Pr{BGYYYRYY}+Pr{RRYYGRRB}+Pr{YYGBYYBR} = .10+.10+.15+.10+.25+.10+.20 = 1.0

 

Sequence

Probability

RBRRYRGG

.10

RRGGBRRB

.10

BBYYYGBR

.15

GRRGGRGG

.10

BGYYYRYY

.25

RRYYGRRB

.10

YYGBYYBR

.20

Total

1.00

 

Pr{Red and Green} = Pr{One of  RBRRYRGG, RRGGBRRB, BBYYYGBR, GRRGGRGG, BGYYYRYY,

RRYYGRRB or YYGBYYBR Shows} = Pr{RBRRYRGG}+Pr{RRGGBRRB}+Pr{BBYYYGBR}+Pr{GRRGGRGG}+

Pr{BGYYYRYY}+Pr{RRYYGRRB}+Pr{YYGBYYBR} = .10+.10+.15+.10+.25+.10+.20 = 1.0

 

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

 

 

Work all four (4) cases. Show complete work and detail for full credit.