Summaries

14th September 2009

Session 1.7

 

Marginal, Joint and Conditional Probabilities

 

Suppose that we have fair dice: d4 with face values {1,2,3,4}, d6 with face values {1,2,3,4,5,6} and d8 with face values {1,2,3,4,5,6,7,8}.  Our experiment consists of first randomly selecting one of the dice and then tossing that die and noting the face value.

 

The first stage probabilities:

 

Pr{Select d4} = 1/3( = P4)

Pr{Select d6} = 1/3( = P6)

Pr{Select d8} = 1/3( = P8)

 

The conditional probabilities:

 

Pr{1 shows | d4 selected} = 1/4

Pr{2 shows | d4 selected} = 1/4

Pr{3 shows | d4 selected} = 1/4

Pr{4 shows | d4 selected} = 1/4

 

Pr{1 shows | d6 selected} = 1/6

Pr{2 shows | d6 selected} = 1/6

Pr{3 shows | d6 selected} = 1/6

Pr{4 shows | d6 selected} = 1/6

Pr{5 shows | d6 selected} = 1/6

Pr{6 shows | d6 selected} = 1/6

 

Pr{1 shows | d8 selected} = 1/8

Pr{2 shows | d8 selected} = 1/8

Pr{3 shows | d8 selected} = 1/8

Pr{4 shows | d8 selected} = 1/8

Pr{5 shows | d8 selected} = 1/8

Pr{6 shows | d8 selected} = 1/8

Pr{7 shows | d8 selected} = 1/8

Pr{8 shows | d8 selected} = 1/8

 

The joint probabilities

 

Pr{1 shows} = Pr{1 shows | d4 selected}*Pr{d4 selected} + Pr{1 shows | d6 selected}*Pr{d6 selected} + Pr{1 shows | d8 selected}*Pr{d8 selected} = (1/3)*(1/4) + (1/3)*(1/6) + (1/3)*(1/8) = (1/3)*(13/24) = 13/72 ≈ 0.1806

 

Pr{2 shows} = Pr{2 shows | d4 selected}*Pr{d4 selected} + Pr{2 shows | d6 selected}*Pr{d6 selected} + Pr{2 shows | d8 selected}*Pr{d8 selected} = (1/3)*(1/4) + (1/3)*(1/6) + (1/3)*(1/8) = (1/3)*(13/24) = 13/72 ≈ 0.1806

 

Pr{3 shows} = Pr{3 shows | d4 selected}*Pr{d4 selected} + Pr{3 shows | d6 selected}*Pr{d6 selected} + Pr{3 shows | d8 selected}*Pr{d8 selected} = (1/3)*(1/4) + (1/3)*(1/6) + (1/3)*(1/8) = (1/3)*(13/24) = 13/72 ≈ 0.1806

 

Pr{4 shows} = Pr{4 shows | d4 selected}*Pr{d4 selected} + Pr{4 shows | d6 selected}*Pr{d6 selected} + Pr{4 shows | d8 selected}*Pr{d8 selected} = (1/3)*(1/4) + (1/3)*(1/6) + (1/3)*(1/8) = (1/3)*(13/24) = 13/72 ≈ 0.1806

 

Pr{5 shows} = Pr{5 shows | d6 selected}*Pr{d6 selected} + Pr{5 shows | d8 selected}*Pr{d8 selected} = (1/3)*(1/6) + (1/3)*(1/8) = (1/3)*(7/24) = 7/72 ≈ 0.0972

 

Pr{6 shows} = Pr{6 shows | d6 selected}*Pr{d6 selected} + Pr{6 shows | d8 selected}*Pr{d8 selected} = (1/3)*(1/6) + (1/3)*(1/8) = (1/3)*(7/24) = 7/72 ≈ 0.0972

 

Pr{7 shows} = Pr{7 shows | d8 selected}*Pr{d8 selected} = (1/3)*(1/8) = 3/72 ≈ 0.0417

 

Pr{8 shows} = Pr{8 shows | d8 selected}*Pr{d8 selected} = (1/3)*(1/8) = 3/72 ≈ 0.0417

 

Sample Tables from Summer 2009

 

Sample #1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d4

 

 

 

d6

 

 

 

d8

 

 

 

Joint

 

 

 

Face

n

p

P

Face

n

p

P

Face

n

p

P

Face

n

p

P

1

17

0.261538

0.25

1

9

0.1232877

0.1666667

1

5

0.0806452

0.125

1

31

0.155

0.1805556

2

19

0.292308

0.25

2

13

0.1780822

0.1666667

2

7

0.1129032

0.125

2

39

0.195

0.1805556

3

13

0.2

0.25

3

15

0.2054795

0.1666667

3

6

0.0967742

0.125

3

34

0.17

0.1805556

4

16

0.246154

0.25

4

8

0.109589

0.1666667

4

7

0.1129032

0.125

4

31

0.155

0.1805556

Total

65

1

1

5

15

0.2054795

0.1666667

5

11

0.1774194

0.125

5

26

0.13

0.0972222

p4

65

0.325

 

6

13

0.1780822

0.1666667

6

5

0.0806452

0.125

6

18

0.09

0.0972222

P4

 

0.333333

 

Total

73

1

1

7

10

0.1612903

0.125

7

10

0.05

0.0416667

 

 

 

 

p6

73

0.365

 

8

11

0.1774194

0.125

8

11

0.055

0.0416667

 

 

 

 

P6

 

0.3333333

 

Total

62

1

1

Total

200

1

1

 

 

 

 

 

 

 

 

p8

62

0.31

 

 

 

 

 

 

 

 

 

 

 

 

 

P8

 

0.333333

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Sample #2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d4

 

 

 

d6

 

 

 

d8

 

 

 

Joint

 

 

 

Face

n

p

P

Face

n

p

P

Face

n

p

P

Face

n

p

P

1

15

0.234375

0.25

1

8

0.1269841

0.1666667

1

8

0.109589

0.125

1

31

0.155

0.1805556

2

18

0.28125

0.25

2

9

0.1428571

0.1666667

2

8

0.109589

0.125

2

35

0.175

0.1805556

3

17

0.265625

0.25

3

12

0.1904762

0.1666667

3

15

0.2054795

0.125

3

44

0.22

0.1805556

4

14

0.21875

0.25

4

11

0.1746032

0.1666667

4

10

0.1369863

0.125

4

35

0.175

0.1805556

Total

64

1

1

5

13

0.2063492

0.1666667

5

4

0.0547945

0.125

5

17

0.085

0.0972222

p4

64

0.32

 

6

10

0.1587302

0.1666667

6

6

0.0821918

0.125

6

16

0.08

0.0972222

P4

 

0.333333

 

Total

63

1

1

7

6

0.0821918

0.125

7

6

0.03

0.0416667

 

 

 

 

p6

63

0.315

 

8

16

0.2191781

0.125

8

16

0.08

0.0416667

 

 

 

 

P6

 

0.3333333

 

Total

73

1

1

Total

200

1

1

 

 

 

 

 

 

 

 

p8

73

0.365

 

 

 

 

 

 

 

 

 

 

 

 

 

P8

 

0.333333

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Sample #3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d4

 

 

 

d6

 

 

 

d8

 

 

 

Joint

 

 

 

Face

n

p

P

Face

n

p

P

Face

n

p

P

Face

n

p

P

1

14

0.21875

0.25

1

8

0.1333333

0.1666667

1

10

0.1315789

0.125

1

32

0.16

0.1805556

2

13

0.203125

0.25

2

10

0.1666667

0.1666667

2

8

0.1052632

0.125

2

31

0.155

0.1805556

3

22

0.34375

0.25

3

12

0.2

0.1666667

3

12

0.1578947

0.125

3

46

0.23

0.1805556

4

15

0.234375

0.25

4

11

0.1833333

0.1666667

4

6

0.0789474

0.125

4

32

0.16

0.1805556

Total

64

1

1

5

10

0.1666667

0.1666667

5

8

0.1052632

0.125

5

18

0.09

0.0972222

p4

64

0.32

 

6

9

0.15

0.1666667

6

11

0.1447368

0.125

6

20

0.1

0.0972222

P4

 

0.333333

 

Total

60

1

1

7

6

0.0789474

0.125

7

6

0.03

0.0416667

 

 

 

 

p6

60

0.3

 

8

15

0.1973684

0.125

8

15

0.075

0.0416667

 

 

 

 

P6

 

0.3333333

 

Total

76

1

1

Total

200

1

1

 

 

 

 

 

 

 

 

p8

76

0.38

 

 

 

 

 

 

 

 

 

 

 

 

 

P8

 

0.333333

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Sample #4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d4

 

 

 

d6

 

 

 

d8

 

 

 

Joint

 

 

 

Face

n

p

P

Face

n

p

P

Face

n

p

P

Face

n

p

P

1

21

0.283784

0.25

1

12

0.2105263

0.1666667

1

10

0.1449275

0.125

1

43

0.215

0.1805556

2

24

0.324324

0.25

2

10

0.1754386

0.1666667

2

5

0.0724638

0.125

2

39

0.195

0.1805556

3

14

0.189189

0.25

3

8

0.1403509

0.1666667

3

12

0.173913

0.125

3

34

0.17

0.1805556

4

15

0.202703

0.25

4

8

0.1403509

0.1666667

4

7

0.1014493

0.125

4

30

0.15

0.1805556

Total

74

1

1

5

10

0.1754386

0.1666667

5

4

0.057971

0.125

5

14

0.07

0.0972222

p4

74

0.37

 

6

9

0.1578947

0.1666667

6

10

0.1449275

0.125

6

19

0.095

0.0972222

P4

 

0.333333

 

Total

57

1

1

7

13

0.1884058

0.125

7

13

0.065

0.0416667

 

 

 

 

p6

57

0.285

 

8

8

0.115942

0.125

8

8

0.04

0.0416667

 

 

 

 

P6

 

0.3333333

 

Total

69

1

1

Total

200

1

1

 

 

 

 

 

 

 

 

p8

69

0.345

 

 

 

 

 

 

 

 

 

 

 

 

 

P8

 

0.333333

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Pooled All

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d4

 

 

 

d6

 

 

 

d8

 

 

 

Joint

 

 

 

Face

n

p

P

Face

n

p

P

Face

n

p

P

Face

n

p

P

1

67

0.250936

0.25

1

37

0.1462451

0.1666667

1

33

0.1178571

0.125

1

137

0.17125

0.1805556

2

74

0.277154

0.25

2

42

0.1660079

0.1666667

2

28

0.1

0.125

2

144

0.18

0.1805556

3

66

0.247191

0.25

3

47

0.1857708

0.1666667

3

45

0.1607143

0.125

3

158

0.1975

0.1805556

4

60

0.224719

0.25

4

38

0.1501976

0.1666667

4

30

0.1071429

0.125

4

128

0.16

0.1805556

Total

267

1

1

5

48

0.1897233

0.1666667

5

27

0.0964286

0.125

5

75

0.09375

0.0972222

p4

267

0.33375

 

6

41

0.1620553

0.1666667

6

32

0.1142857

0.125

6

73

0.09125

0.0972222

P4

 

0.333333

 

Total

253

1

1

7

35

0.125

0.125

7

35

0.04375

0.0416667

 

 

 

 

p6

253

0.31625

 

8

50

0.1785714

0.125

8

50

0.0625

0.0416667

 

 

 

 

P6

 

0.3333333

 

Total

280

1

1

Total

800

1

1

 

 

 

 

 

 

 

 

p8

280

0.350

 

 

 

 

 

 

 

 

 

 

 

 

 

P8

 

0.333333

 

 

 

 

 

 

 

 

Conditional Probability

 

Conditional = Joint / Prior

 

Pr{A|B} = Pr{A∩B} / Pr{B}

 

How much of B is tied up in A ?

 

Case Study 1.11

Conditional Probability

Case Study Description: Compute conditional probabilities associated with the color sequence experiment.

Suppose that we have a special box - each time we press a button on the box, it prints out a sequence of colors, in order - it prints four colors at a time. Suppose the box follows the following Probabilities for each Color Sequence:

 

 

 

 

Color Sequence

Probability CS Prints Out

BBBB

.10 = 10%

BGGB

.25 = 25%

RGGR

.05 = 05%

YYYY

.30 = 30%

BYRG

.15 = 15%

RYYB

.15 = 15%

Total

1.00 = 100%

 

Let's define the experiment: We push the button, and then the box prints out exactly one (1) of the above listed color sequences. We then note the resulting (printed out) color sequence.

Compute Pr{ blue shows 1st | blue shows 4th };

Pr{ B 1st and B 4th } = Pr{ exactly one of BBBB, BGGB shows } = Pr{ BBBB} + Pr{BGGB} =.10 + .25 = .35

 

Pr{ B 4th } = Pr{ exactly one of BBBB, BGGB, RYYB shows } = Pr{BBBB} + Pr{BGGB} +

Pr{RYYB} = .10+.25+.15 = .50

 

So, Pr{ B 1st | B 4th } = .35/ .50= .70

Compute Pr{ green shows 2nd or 3rd | yellow shows };

Pr{ G 2nd or 3rd and Y shows } = 0, since no sequences meet this requirement

 

Pr{ Y shows } = Pr{ exactly one of YYYY, BYRG, RYYB shows } = Pr{YYYY}+ Pr{BYRG}+ Pr{RYYB} = .30+.15+.15 = .60

 

So, Pr{ G 2nd or 3rd | Y shows } = 0 / .60= 0

Compute Pr{ yellow shows | red shows }.

Pr{ Y and R show } = Pr{ exactly one of BYRG, RYYB shows } = Pr{BYRG}+ Pr{RYYB } = .15 + .15 = .30

 

Pr{ R shows } = Pr{ exactly one of RGGR, BYRG, RYYB shows } = Pr{RGGR}+ Pr{BYRG}+

Pr{RYYB } = .05+.15+.15 = .35

 

So, Pr{ Y shows | R shows } = .30/.35 = 6/7 = .8571

 

Case Study 1.12

Conditional Probability II: Pair of Dice

Case Description: Compute conditional probabilities.

Suppose we have a pair of fair dice: d4(faces 1,2,3,4), d6(faces 1,2,3,4,5,6). In our experiment, we toss this pair of dice, and note the face value from each die. For simplicity, we write the outcome as (d4 result, d6 result). Assume that the dice operate independently and separately.

Case Objectives:

Identify the simple (basic) events. Compute (and justify) a probability for each simple event. 

As before, Pr{ ( any d4 face, any d6 face) } = Pr{ any d4 face }*Pr{ any d6 face } = (1/4)*(1/6) = 1/24

We have 24 equally likely pairs.

 

1

2

3

4

1

(1,1)

(2,1)

(3,1)

(4,1)

2

(1,2)

(2,2)

(3,2)

(4,2)

3

(1,3)

(2,3)

(3,3)

(4,3)

4

(1,4)

(2,4)

(3,4)

(4,4)

5

(1,5)

(2,5)

(3,5)

(4,5)

6

(1,6)

(2,6)

(3,6)

(4,6)

Suppose we observe the sum of the faces in the pair of dice. Identify the possible values of this sum, and compute (and justify) a probability for each value.

Now for the sums:

 

1

2

3

4

1

(1,1) @ 2

(2,1) @ 3

(3,1) @ 4

(4,1) @ 5

2

(1,2) @ 3

(2,2) @ 4

(3,2) @ 5

(4,2) @ 6

3

(1,3) @ 4

(2,3) @ 5

(3,3) @ 6

(4,3) @ 7

4

(1,4) @ 5

(2,4) @ 6

(3,4) @ 7

(4,4) @ 8

5

(1,5) @ 6

(2,5) @ 7

(3,5) @ 8

(4,5) @ 9

6

(1,6) @ 7

(2,6) @ 8

(3,6) @ 9

(4,6) @ 10

 

Compute the conditional probability Pr{Sum is Even|d4 shows Even}.

Pr{ Sum is Even and d4 shows Even } = Pr{ exactly one of (2,2), (2,4), (2,6), (4,2), (4,4), (4,6) shows } = 6/24 = 1/4 = .25

Pr{ d4 shows Even } =

Pr{ exactly one of (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6) shows } = 12/24 = 1/2 = .50

So, Pr{ Sum is Even | d4 shows Even } = .25 / .50 = .50

Continuing,…

 

1

2

3

4

1

(1,1) @ 2

(2,1) @ 3

(3,1) @ 4

(4,1) @ 5

2

(1,2) @ 3

(2,2) @ 4

(3,2) @ 5

(4,2) @ 6

3

(1,3) @ 4

(2,3) @ 5

(3,3) @ 6

(4,3) @ 7

4

(1,4) @ 5

(2,4) @ 6

(3,4) @ 7

(4,4) @ 8

5

(1,5) @ 6

(2,5) @ 7

(3,5) @ 8

(4,5) @ 9

6

(1,6) @ 7

(2,6) @ 8

(3,6) @ 9

(4,6) @ 10

 

Compute the conditional probability Pr{Sum is Odd|d6 shows Odd}.

Pr{ Sum is Odd and d6 shows Odd } = Pr{ exactly one of (2,1), (2,3), (2,5), (4,1), (4,3), (4,5) shows } = 6/24 = 1/4 = .25

Pr{ d6 shows Odd } =

Pr{ exactly one of (1,1), (1,3), (1,5), (2,1), (2,3), (2,5), (3,1), (3,3), (3,5), (4,1), (4,3), (4,5) shows } = 12/24 = 1/2 = .50

So, Pr{ Sum is Odd | d6 shows Odd } = .25/.50 = 1/2 = .50

HR1 – Summer Version A, Case Three

 

Case Three | Color Slot Machine | Conditional Probabilities

 

Here is our slot machine – on each trial, it produces a 10-color sequence, using the table below:

 

Sequence*

Probability

RRBBRRYRRR

.10

RRGGRGBRRB

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

BGYGYRYGYY

.25

RRYYGRRBBY

.10

YYGBYYBGRR

.20

Total

1.00

*B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered from left to right: (1st 2nd 3rd 4th 5th6th7th 8th 9th 10th )

Compute the following conditional probabilities:

 

Pr{ Yellow Shows Exactly Twice | Blue Shows}

 

 

Sequence*

Probability

RRBBRRYRRR

.10

RRGGRGBRRB

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

BGYGYRYGYY

.25

RRYYGRRBBY

.10

YYGBYYBGRR

.20

Total

1.00

 

Pr{Blue Shows} = Pr{One of RRBBRRYRRR, RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG, BGYGYRYGYY, RRYYGRRBBY, YYGBYYBGRR Shows} =

Pr{RRBBRRYRRR} + Pr{RRGGRGBRRB} + Pr{BBYYGGYGBR} + Pr{GRRGGYBRGG} + Pr{BGYGYRYGYY} + Pr{RRYYGRRBBY} + Pr{YYGBYYBGRR} = .1+.1+.15+.1+.25+.1+.2 = 1.00

 

Sequence*

Probability

 

 

Total

0

 

 

Pr{ Yellow Shows Exactly Twice and  Blue Shows} = 0

 

 

Pr{ Yellow Shows Exactly Twice | Blue Shows} = Pr{ Yellow Shows Exactly Twice and Blue Shows}/Pr{Blue Shows} = 0/1 =0

 

Pr{ Green Shows | “BR” Shows }

 

Sequence*

Probability

RRBBRRYRRR

.10

RRGGRGBRRB

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

Total

.45

 

Pr{ “BR” Shows } = Pr{One of RRBBRRYRRR, RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG Shows} = Pr{RRBBRRYRRR}+ Pr{ RRGGRGBRRB}+

Pr{BBYYGGYGBR}+ Pr{GRRGGYBRGG} = .1+.1+.15+.1 = .45

 

 

Pr{Green Shows and “BR” Shows}

 

Sequence*

Probability

RRGGRGBRRB

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

Total

.35

 

Pr{ Green Shows and “BR” Shows } = Pr{One of RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG Shows} = Pr{ RRGGRGBRRB}+

Pr{BBYYGGYGBR}+ Pr{GRRGGYBRGG} =.1+.15+.1 = .35

 

Pr{ Green Shows | “BR” Shows } = Pr{ Green Shows and “BR” Shows }/Pr{ “BR” Shows } = .35/.45 = 7/9

 

 

Pr{ Red Shows | Green Shows}

 

Sequence*

Probability

RRGGRGBRRB

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

BGYGYRYGYY

.25

RRYYGRRBBY

.10

YYGBYYBGRR

.20

Total

.90

 

Pr{Green Shows} = Pr{One of RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG, BGYGYRYGYY, RRYYGRRBBY, YYGBYYBGRR Shows} =

Pr{RRBBRRYRRR} + Pr{RRGGRGBRRB} + Pr{BBYYGGYGBR} + Pr{GRRGGYBRGG} + Pr{BGYGYRYGYY} + Pr{RRYYGRRBBY} + Pr{YYGBYYBGRR} = .1+.15+.1+.25+.1+.2 = .90

 

Pr{ Red Shows and Green Shows}

 

Sequence*

Probability

RRGGRGBRRB

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

BGYGYRYGYY

.25

RRYYGRRBBY

.10

YYGBYYBGRR

.20

Total

.90

 

Pr{ Red and Green Show } = Pr{One of RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG, BGYGYRYGYY, RRYYGRRBBY, YYGBYYBGRR Shows} =

Pr{RRBBRRYRRR} + Pr{RRGGRGBRRB} + Pr{BBYYGGYGBR} + Pr{GRRGGYBRGG} + Pr{BGYGYRYGYY} + Pr{RRYYGRRBBY} + Pr{YYGBYYBGRR} = .1+.15+.1+.25+.1+.2 = .90

 

Pr{ Red Shows | Green Shows} = Pr{ Red Shows and Green Shows}/Pr{ Green Shows} = .90/.90 = 1

 

HR1 – Spring 2008, Case Four

 

Case Four: Color Slot Machine, Computation of Conditional Probabilities

 

Here is our slot machine – on each trial, it produces a 10-color sequence, using the table below:

 

Sequence*

Probability

RRBBR RYRRB

.10

RRGGRGBRRB

.10

BBYYRGYGBR

.15

GRRGRGBRGB

.10

BGYGYRYGYY

.25

RRGYGRRBBB

.10

YYGBYYBGRR

.20

Total

1.00

*B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered as 1st to 6th , from left to right: (1st 2nd 3rd 4th 5th6th7th 8th 9th 10th )

Compute the following conditional probabilities:

 

1. Pr{Red Shows Somewhere in the 1st ─ 4th slots | Yellow Shows Somewhere in the 7th ─ 10th slots}

 

Pr{Red Shows in the 1st – 4th slots|Yellow Shows in the 7th – 10th slots} =

 

Pr{Red Shows in the 1st – 4th slots and Yellow Shows in the 7th – 10th slots}/ Pr{ Yellow Shows in the 7th – 10th slots}

 

Sequence*

Probability

RRBBR RYRRB

.10

BBYYRGYGBR

.15

BGYGYRYGYY

.25

Total

0.50

 

Pr{ Yellow Shows in the 7th – 10th slots} = Pr{One of RRBBRRYRRB, BBYYRGYGBR, BGYGYRYGYY shows} =

Pr{RRBBRRYRRB}+ Pr{BBYYRGYGBR}+ Pr{BGYGYRYGYY} = .10+.15+.25 = .50

 

Sequence*

Probability

RRBBR RYRRB

.10

Total

0.10

 

Pr{ Red Shows in the 1st – 4th slots and Yellow Shows in the 7th – 10th slots } = Pr{One of RRBBRRYRRB shows} = .10

 

Pr{Red Shows in the 1st – 4th slots|Yellow Shows in the 7th – 10th slots} = .10/.50 = .20

 

2. Pr{Green Shows Anywhere  | “RB” Shows Anywhere}

 

Pr{Green Shows Anywhere|”RB” Shows Anywhere} =

 

Pr{ Green Shows Anywhere and ”RB” Shows Anywhere }/ Pr{”RB” Shows Anywhere}

 

Sequence*

Probability

RRBBR RYRRB

.10

RRGGRGBRRB

.10

RRGYGRRBBB

.10

Total

0.30

Pr{”RB” Shows Anywhere} = Pr{One of  RRBBRRYRRB, RRGGRGBRRB, RRGYGRRBBB Shows} = Pr{RRBBRRYRRB}+Pr{RRGGRGBRRB}+Pr{RRGYGRRBBB} =.10+.10+.10 = .30

 

Sequence*

Probability

RRGGRGBRRB

.10

RRGYGRRBBB

.10

Total

0.20

 

Pr{ Green Shows Anywhere and ”RB” Shows Anywhere } = Pr{One of  RRGGRGBRRB, RRGYGRRBBB Shows} = Pr{RRGGRGBRRB}+Pr{RRGYGRRBBB} =.10+.10 = .20

 

Pr{Green Shows Anywhere|”RB” Shows Anywhere} = .20/.30

 

3. Pr{Yellow Shows Anywhere | Blue Shows Anywhere}

 

Pr{Yellow Shows Anywhere | Blue Shows Anywhere} =

Pr{Yellow Shows Anywhere and Blue Shows Anywhere}/Pr{ Blue Shows Anywhere}

 

Sequence*

Probability

RRBBR RYRRB

.10

RRGGRGBRRB

.10

BBYYRGYGBR

.15

GRRGRGBRGB

.10

BGYGYRYGYY

.25

RRGYGRRBBB

.10

YYGBYYBGRR

.20

Total

1.00

Pr{ Blue Shows Anywhere} = Pr{one of RRBBRRYRRB, RRGGRGBRRB, BBYYRGYGBR, GRRGRGBRGB, BGYGYRYGYY, RRGYGRRBBB, YYGBYYBGRR Shows} =Pr{RRBBRRYRRB}+Pr{RRGGRGBRRB}+Pr{ BBYYRGYGBR}+Pr{GRRGRGBRGB}+Pr{BGYGYRYGYY}+Pr{RRGYGRRBBB}+Pr{YYGBYYBGRR} = .10+.10+.15+.10+.25+.10+.20 = 1.00

 

Sequence*

Probability

RRBBR RYRRB

.10

BBYYRGYGBR

.15

BGYGYRYGYY

.25

RRGYGRRBBB

.10

YYGBYYBGRR

.20

Total

0.80

 

 

Pr{Yellow Shows Anywhere and Blue Shows Anywhere} = Pr{one of RRBBRRYRRB, BBYYRGYGBR, BGYGYRYGYY, RRGYGRRBBB, YYGBYYBGRR Shows} =Pr{RRBBRRYRRB}+ Pr{BBYYRGYGBR}+Pr{BGYGYRYGYY}+Pr{RRGYGRRBBB}+Pr{YYGBYYBGRR} = .10+.15+.25+.10+.20 = .80

 

Pr{Yellow Shows Anywhere | Blue Shows Anywhere} = .80/1.00 = .80

 

 

Case Study 1.13

Conditional Probability

Case Description: Compute conditional probabilities for pairs of draws (without replacement).

Here is our bowl, in tabular form:

Color

# in Bowl

Proportion of Bowl

Blue

5

5/9

Green

3

3/9

Red

1

1/9

Total

9

1

 

Suppose that on each trial of this experiment that we make two (2) draws without replacement from the bowl.

Compute Pr{ green shows 2nd | red shows 1st };

Here is our bowl, after "red shows 1st", in tabular form:

 

Color

# in Bowl – Before 1st Draw

# in Bowl – After 1st Draw

Blue

5

5 – 0 = 5

Green

3

3 – 0 = 3

Red

1

1 – 1 = 0

Total

9

8

With the red chip out of the bowl, 3 of the 8 surviving chips are green. So, Pr{G 2nd | R 1st} = (3-0) / (9-1) = 3/8

Compute Pr{ red shows 2nd | red shows 1st };

Pr{R 2nd | R 1st} = (1-1) / (9-1) = 0/8 = 0. There are 1-1=0 surviving red chips after the first draw.

Compute Pr{ blue shows 2nd | blue shows 1st }.

Here is our bowl, after "blue shows 1st", in tabular form:

Color

# in Bowl – Before 1st Draw

# in Bowl – After 1st Draw

Blue

5

5 – 1 = 4

Green

3

3 – 0 = 3

Red

1

1 – 0 = 1

Total

9

8

Pr{B 2nd | B 1st} = (5-1)/(9-1) = 4/8. After the first draw, 4 of 8 surviving chips are blue.

HR1 – Fall 2004, Case Three

Case Three

Conditional Probability

Color Bowl/Draws without Replacement

 

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

 

Blue @ 5, Green @ 1, Red @ 2, Yellow @ 3

 

Each trial of our experiment consists of three (3) draws without replacement from the bowl.

 

Compute these directly.

 

Color

Count

B

5

G

1

R

2

Y

3

Total

11

 

Pr{ green shows 2nd | green shows 1st}

 

Color

Count

B

5

G

1

R

2

Y

3

Total

11

ß green shows 1st

Color

Count

B

5

G

0

R

2

Y

3

Total

10

 

Pr{ green shows 2nd  | green shows 1st} = 0/10

  

Pr{ yellow shows 3rd | green shows 1st, blue shows 2nd}

 

Color

Count

B

5

G

1

R

2

Y

3

Total

11

ß green shows 1st

Color

Count

B

5

G

0

R

2

Y

3

Total

10

ß blue shows 2nd

Color

Count

B

4

G

0

R

2

Y

3

Total

9

 

Pr{ yellow shows 3rd | green shows 1st, blue shows 2nd} = 3/9

 

Pr{ red shows 3rd | green shows 1st, red shows 2nd }

 

Color

Count

B

5

G

1

R

2

Y

3

Total

11

ß green shows 1st

Color

Count

B

5

G

0

R

2

Y

3

Total

10

ß red shows 2nd

Color

Count

B

5

G

0

R

1

Y

3

Total

9

 

Pr{ red shows 3rd | green shows 1st, red shows 2nd } = 1/9