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 GREENCOUNT as the number of times that green shows in the color sequence.

Compute Pr{GREENCOUNT = 2 or 3}.

 

Sequence

Green Count

Probability

RBRRYRGG

2

.10

RRGGBRRB

2

.10

BBYYYGBR

1

.15

GRRGGRGG

5

.10

BGYYYRYY

1

.25

RRYYGRRB

1

.10

YYGBYYBR

1

.20

Total

 

1.00

 

Here are the complete calculations for GREENCOUNT:

 

Pr{GREENCOUNT=1} = Pr{One of BBYYYGBR, BGYYYRYY, RRYYGRRB or YYGBYYBR Shows} = Pr{BBYYYGBR} + Pr{BGYYYRYY} + Pr{RRYYGRRB} + Pr{YYGBYYBR} = .15 + .25 + .10 + .20 = .70 

 

Pr{GREENCOUNT=2} = Pr{One of RBRRYRGG or RRGGBRRB Shows} = Pr{RBRRYRGG} +

Pr{RRGGBRRB} = .10 + .10 = .20

 

Pr{GREENCOUNT=3} = 0

 

Pr{GREENCOUNT=4} = 0

 

Pr{GREENCOUNT=5} = Pr{GRRGGRGG} = .10

 

Here is the required calculation:

 

Pr{GREENCOUNT = 2 or 3 } = Pr{One of RBRRYRGG or RRGGBRRB Shows} = Pr{RBRRYRGG} +

Pr{RRGGBRRB} = .10 + .10 = .20

 

2. Compute Pr{“BY” Shows}.

 

Sequence

“BY” Shows

Probability

RBRRYRGG

No

.10

RRGGBRRB

No

.10

BBYYYGBR

Yes

.15

GRRGGRGG

No

.10

BGYYYRYY

No

.25

RRYYGRRB

No

.10

YYGBYYBR

Yes

.20

Total

 

1.00

 

Pr{“BY”} = Pr{One of BBYYYGBR or YYGBYYBR Shows} = Pr{BBYYYGBR} + Pr{YYGBYYBR} = .15 + .20 = .35

 

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

Sequence

Blue Shows

Probability

RBRRYRGG

Yes

.10

RRGGBRRB

Yes

.10

BBYYYGBR

Yes

.15

GRRGGRGG

No

.10

BGYYYRYY

Yes

.25

RRYYGRRB

Yes

.10

YYGBYYBR

Yes

.20

Total

 

1.00

 

Other Event = “Blue Does Not Show”

Pr{Other Event} = Pr{Blue Does Not Show} = Pr{GRRGGRGG} = .10

Pr{“Blue Shows”} = 1 – Pr{“Blue Does Not Show} = 1 – .10 = .90

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

15-24

0.3554

25-34

0.4999

35-44

0.1416

45+

0.0016

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 35.54% of sampled births are to mothers aged between 15 and 24 years.

 

In long runs of random sampling year 2006 United States Resident Live Births, approximately 49.99% of sampled births are to mothers aged between 25 and 34 years.

 

In long runs of random sampling year 2006 United States Resident Live Births, approximately 14.16% of sampled births are to mothers aged between 35 and 44 years.

 

In long runs of random sampling year 2006 United States Resident Live Births, approximately 0.16% of sampled births are to mothers aged 45 or more years.

 

 

Compute and discuss Perfect Samples for n=2500.

 

Maternal Age

Probability

E2500

Under 15

0.0015

3.75

15-24

0.3554

888.5

25-34

0.4999

1249.75

35-44

0.1416

354

45+

0.0016

4

Total

1

2500

 

E3500<15 = 2500*P<15 = 2500*0.0015 = 3.75

E350015-24 = 2500*P15-24 = 2500*0.3554 = 888.5

E350025-34 = 2500*P25-34 = 2500*0.4999 = 1249.75

E350035-44= 2500*P35-44 = 2500*0.1416 = 354

E350045+ = 2500*P45+ = 2500*0.0016 = 4

 

In random samples of 2500 year 2006 United States Resident Live Births, approximately 3.75 sampled births are to mothers aged less than 15 years.

 

In random samples of 2500 year 2006 United States Resident Live Births, approximately 888.5 sampled births are to mothers aged between 15 and 24 years.

 

In random samples of 2500 year 2006 United States Resident Live Births, approximately 1249.5 sampled births are to mothers aged between 25 and 34 years.

 

In random samples of 2500 year 2006 United States Resident Live Births, approximately 354 sampled births are to mothers aged between 35 and 44 years.

 

In random samples of 2500 year 2006 United States Resident Live Births, approximately 4 sampled births are to mothers aged 45 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.

SUM = “Sum 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 SUM, showing in detail how these values are computed from each pair of face values.

 

SUM{(2,1)}=3

SUM {(2,7)}=9

SUM {(2,8)}=10

 

SUM{(3,1)}=4

SUM{(3,7)}=10

SUM{(3,8)}=11

 

SUM{(4,1)}=5

SUM{(4,7)}=11

SUM{(4,8)}=12

 

SUM{(5,1)}=6

SUM{(5,7)}=12

SUM{(5,8)}=13

 

SUM{(6,1)}=7

SUM{(6,7)}=13

SUM{(6,8)}=14

 

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

 

Pr{SUM=1} = 0

Pr{SUM=2} = 0

Pr{SUM=3} = Pr{(2,1)} = 15/150

Pr{SUM=4} = Pr{(3,1)} = 15/150

Pr{SUM=5} = Pr{(4,1)} = 15/150

Pr{SUM=6} = Pr{(5,1)} = 15/150

Pr{SUM=7} = Pr{(6,1)} = 15/150

Pr{SUM=9} = Pr{(2,7)} = 5/15

Pr{SUM=10} = Pr{One of (2,8) or (3,7) Shows} = Pr{(2,8)} + Pr{(3,7)} = (10/150) + (5/150) = 15/150

Pr{SUM=11} = Pr{One of (3,8) or (4,7) Shows} = Pr{(3,8)} + Pr{(4,7)} = (10/150) + (5/150) = 15/150

Pr{SUM=12} = Pr{One of (4,8) or (5,7) Shows} = Pr{(4,8)} + Pr{(5,7)} = (10/150) + (5/150) = 15/150

Pr{SUM=13} = Pr{One of (5,8) or (6,7) Shows} = Pr{(5,8)} + Pr{(6,7)} = (10/150) + (5/150) = 15/150

Pr{SUM=14} = Pr{(6,8)} = 10/150

 

Show all work and detail for full credit.

 

Case Four | Color Slot Machine | Conditional Probabilities

Using the slot machine from Case One,

Compute Pr{Yellow 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

BBYYYGBR

.15

BGYYYRYY

.25

RRYYGRRB

.10

YYGBYYBR

.20

Total

.80

 

Pr{Yellow and Green} = 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 | Green Shows} = Pr{Yellow and Green Show}/ Pr{Green Shows} = .80/1.00

 

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{ Blue 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

BGYYYRYY

.25

RRYYGRRB

.10

YYGBYYBR

.20

Total

.90

 

Pr{Blue and Red} = 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

 

Pr{Blue Shows | Red Shows}= Pr{Blue and Red Show}/ Pr{ Red Shows} = .90/1.00

 

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