Key

1st Hourly

Math 1107

Spring Semester 2004

 

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);

           This copy of the hourly.

 

Do not share these resources with anyone else. Do not collaborate or share information with anyone else during the test session.

 

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

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

Conditional Probability

Pair of Dice

Random Variable

Suppose we have a pair of fair dice: d2(faces 1,2), d5(faces 1,2,3,4,5). In our experiment, we toss this pair of dice, and note the face value from each die. For simplicity, we write the outcome as (d5 result, d2 result). Assume that the dice operate independently and separately.

Write the pairs as (d2,d5). Then:

Pr{(1,1)} = Pr{1 from d5}*Pr{1 from d2} = (1/5)*(1/2) = 1/10 = .10 or 10%;                                                                                                                                                           

Pr{(1,2)} = Pr{1 from d5}*Pr{2 from d2} = (1/5)*(1/2) = 1/10 = .10 or 10%;                                                                                                                                                               

Pr{(1,3)} = Pr{1 from d5}*Pr{3 from d2} = (1/5)*(1/2) = 1/10 = .10 or 10%;                                                                                                                                                                

Pr{(1,4)} = Pr{1 from d5}*Pr{4 from d2} = (1/5)*(1/2) = 1/10 = .10 or 10%;                                                                                                                                                           

Pr{(1,5)} = Pr{1 from d5}*Pr{5 from d2} = (1/5)*(1/2) = 1/10 = .10 or 10%;                                                                                                                                                                  

Pr{(2,1)} = Pr{2 from d5}*Pr{1 from d2} = (1/5)*(1/2) = 1/10 = .10 or 10%;                                                                                                                                                            

Pr{(2,2)} = Pr{2 from d5}*Pr{2 from d2} = (1/5)*(1/2) = 1/10 = .10 or 10%;                                                                                                                                                                

Pr{(2,3)} = Pr{2 from d5}*Pr{3 from d2} = (1/5)*(1/2) = 1/10 = .10 or 10%;                                                                                                                                                                 

Pr{(2,4)} = Pr{2 from d5}*Pr{4 from d2} = (1/5)*(1/2) = 1/10 = .10 or 10%;                                                                                                                                                           

Pr{(2,5)} = Pr{2 from d5}*Pr{5 from d2} = (1/5)*(1/2) = 1/10 = .10 or 10%;                                                                                                                                                                

Compute the conditional probability Pr{SUM ³ 4 | d5 shows Even}.

Pr{d5 shows even}=Pr{2 or 4 shows} = (1/5)+(1/5) = 2/5 = .40 or 40%                                                                                                                                                                                  

Pr{ SUM ³ 4 and d5 shows Even} = Pr{ one of (2,2), (1,4), (2,4) shows } = (1/10)+(1/10)+(1/10) = .30 or 30%                                                                                                       

Pr{SUM ³ 4 | d5 shows Even} = Pr{ SUM ³ 4 and d5 shows Even} / Pr{d5 shows even} = .30/.40 = .75 or 75%

Compute the conditional probability Pr{SUM < 6 | d2 shows Odd}.

Pr{d2 shows odd} =Pr{1 shows} = (1/2) = .50 or 50%                                                                                                                                                                                                           

Pr{SUM < 6 and d2 shows Odd}. = Pr{ one of (1,4),(1,3),(1,2),(1,1) shows } = (1/10)+ (1/10)+(1/10)+(1/10) = .40 or 40%                                                                                   

Pr{SUM < 6 | d2 shows Odd} = Pr{SUM < 6 and d2 shows Odd} / Pr{d2 shows odd} = .40/.50 = .80 or 80%

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

GBRB

.15

BBGG

.05

GGGG

.10

GBBR

.25

BBBB

.05

RGYB

.06

YYYY

.04

BBYY

.18

RRYY

.03

RRRR

.025

YBGR

.075

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.

             

2.1 Pr{Green Shows}

 

Pr{Green Shows} =

Pr{one of GBRB, BBGG, GGGG, GBBR, RGYB, YBGR shows} =

Pr{GBRB} + Pr{BBGG} + Pr{GGGG} + Pr{GBBR} + Pr{RGYB} + Pr{YBGR}=

.15 + .05 + .10 + .25 + .06 + .075 = .685 or 68.5%

 

In long runs of box-plays, approximately 68.5% of plays will show some green in the color sequence.

 

2.2 Pr{Yellow Shows and Red Does Not Show}

 

Pr{Yellow Shows and Red Does Not Show} =

Pr{one of YYYY,BBYY shows} =

Pr{YYYY} + Pr{BBYY} = .04 + .18 = .22 or 22%

 

In long runs of box-plays, approximately 22% of plays will show some yellow but not red in the color sequence.

 

2.3 Pr{Yellow Shows 3rd} - Use the Complementary Rule.   

 

Pr{Yellow Does Not Shows 3rd} =

Pr{one of GBRB, BBGG, GGGG, GBBR, BBBB, RRRR, YBGR shows} =
Pr{GBRB}+Pr{BBGG}+Pr{GGGG}+Pr{GBBR}+Pr{BBBB}+Pr{RRRR}+Pr{YBGR} =

.15+.05+.10+.25+.05+.025+.075 = .70 or 70%

Pr{Yellow Shows 3rd} = 1 - Pr{Yellow Does Not Shows 3rd} = 1 - .70 = .30 or 30%

 

In long runs of box-plays, approximately 30% of plays will not show red in the 3rd slot of the color sequence.

Case Three

Random Variables

Pair of Dice

 

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

 

D4

 

 

d3

 

Face

Probability

 

Face

Probability

0

0.40

 

1

.10

2

0.30

 

3

.30

4

0.20

 

5

.60

6

0.10

 

 

 

 

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 pair of faces.

 

(d4,d3)

Pr{Pair}

SUM

LOWTIE

HIGHTIE

 0

@

.4

2

@

.3 

4

@

.2

6

@

.1

1

@

.1

(0,1)

.4*.1=.04

1

0

1

(2,1)

.3*.1=.03

3

1

2 

(4,1)

.2*.1=.02

5

1

4

(6,1)

.1*.1=.01

7

1

6

3

@

.3

(0,3)

.4*.3=.12

3

0

3

(2,3)

.3*.3=.09

5

2

3 

(4,3)

.2*.3=.06

7

3

4

(6,3)

.1*.3=.03

9

3

6

5

@

.6

(0,5)

.4*.6=.24

5

0

5

(2,5)

.3*.6=.18

7

2

5 

(4,5)

.2*.6=.12

9

4

5

(6,5)

.1*.6=.06

11

5

6

 

 

3.1        List the possible pairs, and compute a probability for each.

 

Write the pair as (d4,d3).

 

Pr{(0,1)} = Pr{0 from d4}*Pr{1 from d3} = .4*.1 = .04 or 4%;

Pr{(0,3)} = Pr{0 from d4}*Pr{3 from d3} = .4*.3 = .12 or 12%;

Pr{(0,5)} = Pr{0 from d4}*Pr{5 from d3} = .4*.6 = .24 or 24%;

Pr{(2,1)} = Pr{2 from d4}*Pr{1 from d3} = .3*.1 = .03 or 3%;

Pr{(2,3)} = Pr{2 from d4}*Pr{3 from d3} = .3*.3 = .09 or 9%;

Pr{(2,5)} = Pr{2 from d4}*Pr{5 from d3} = .3*.6 = .18 or 18%;

Pr{(4,1)} = Pr{4 from d4}*Pr{1 from d3} = .2*.1 = .02 or 2%;

Pr{(4,3)} = Pr{4 from d4}*Pr{3 from d3} = .2*.3 = .06 or 6%;

Pr{(4,5)} = Pr{4 from d4}*Pr{5 from d3} = .2*.6 = .12 or 12%;

Pr{(6,1)} = Pr{6 from d4}*Pr{1 from d3} = .1*.1 = .01 or 1%;

Pr{(6,3)} = Pr{6 from d4}*Pr{3 from d3} = .1*.3 = .03 or 3%;

Pr{(6,5)} = Pr{6 from d4}*Pr{5 from d3} = .1*.6 = .06 or 6%.

 

3.2        Define LOWTIE as either the lesser of the two faces when unequal, or the common face value when equal. List the possible values for LOWTIE, and compute a probability for each value of LOWTIE.

 

 

Pr{LOWTIE=0} = Pr{(0,1)} + Pr{(0,3)} + Pr{(0,5)} = .04 + .12 +.24 = .40 or 40%

Pr{LOWTIE=1} = Pr{(2,1)} + Pr{(4,1)} + Pr{(6,1)} = .03+.02+.01 = .06 or 6%

Pr{LOWTIE=2} = Pr{(2,3)} + Pr{(2,5)} = .09 + .18 = .27 or 27%

Pr{LOWTIE=3} = Pr{(4,3)} + Pr{(6,3)} = .06+.03 = .09 or 9%

Pr{LOWTIE=4} = Pr{(4,5)} = .12 or 12%;

Pr{LOWTIE=5} = Pr{(6,5)} = .06 or 6%.

 

3.3        Define HIGHTIE as either the greater of the two faces when unequal, or the common face value when equal. List the possible values for HIGHTIE, and compute a probability for each value of HIGHTIE.

 

Pr{HIGHTIE=1} = Pr{(0,1)} = .04 or 4%

Pr{HIGHTIE=2} = Pr{(2,1)} = .03 or 3%

Pr{HIGHTIE=3} = Pr{(0,3)} + Pr{(2,3)} = .12 + .09 = .21 or 21%

Pr{HIGHTIE=4} = Pr{(4,1)} + Pr{(4,3)} = .02 +.06 = .08 or 8%

Pr{HIGHTIE=5} = Pr{(0,5)} + Pr{(2,5)} + Pr{(4,5)} =.24 +.18 +.12 = .54 or 54%;

Pr{HIGHTIE=6} = Pr{(6,1)} + Pr{(6,3)} + Pr{(6,5)} =.01 + .03 + .06 = .10 or 10%.

 

 

Show full work and detail for full credit. Be sure that you have worked each part of all three cases.

Performance Summary and Notes

 

Performance Summary

 

There were 108 tests written (n=108).

 

The maximum score was 100%, and there were five (5) scores at this level.

Approximately 10% of the scores were at or above 97.

Approximately 25% of the scores were at or above 86.

Approximately 50% of the scores were at or above 73.

Approximately 25% of the scores were at or below 63.

Approximately 10% of the scores were at or below 49.

The minimum score was 40, and there were two (2) scores at this level.

 

Approximately 15.7% of the scores were in the interval [90,100].

Approximately 35.2% of the scores were in the interval [80,100].

Approximately 53.7% of the scores were in the interval [70,100].

Approximately 26.9%f the scores were in the interval [60,70).

Approximately 19.5%f the scores were in the interval [40,60).

 

I really can't complain about the upper portion of the scores, but the first test typically has a heavy lower score group. Bear in mind the scoring rule for this course:

 

If your score falls in the lower portion, recall the scoring rules for the course:

We will have three (3) in-class examinations (called hourlies). Your best performance among these three will comprise 40% of your course grade. Your second best performance among these three will comprise 20% of your course grade. Your worst performance among these three will comprise 0% of your course grade. We will have a comprehensive final examination, which will comprise 40% of your course grade.

Performance Notes

 

Please follow the protocol and directions. In particular, many students did not place the work correctly. Use one side only of the blank work sheets provided. Moreover, apace out your work, and show full detail.

 

In case one, the primary failures involved severe abuses of the definition of conditional probability. This case was quite similar to case study 1.12 in the part one sessions.

 

In case two, many students failed to discuss or interpret their probabilities - be more careful about this in the future. A number of students also did not employ the complementary rule in part 2.3 of the case. When I say to use a particular method, use that method.

 

In case three, there was some confusion regarding the pair of dice - I repeatedly clarified the issue during the test session. Additionally, many students did not supply full detail in their work.