Transcript Chapter 5 - Sample Surveys and Experiments

Section 5.2 - Using Simulation to Estimate Probabilities
P12. A catastrophic accident is one that involves severe skull
or spinal damage. The National Center for Catastrophic Sports
Injury Research reports that over the last 21 years, there have
been 101 catastrophic accidents among female high school
and college athletes. Fifty-five of these resulted from
cheerleading.
Suppose you want to study catastrophic accidents in more
detail, and you take a random sample, without replacement,
of 8 of these 101 accidents. Estimate the probability that at
least half of your eight sampled accidents resulted from
cheerleading.
Start at the beginning of row 17 of Table D on page 828, and
add ten runs to the frequency table.
Section 5.2 - Using Simulation to Estimate Probabilities
P12. Of the 101 catastrophic accidents among female high
school and college athletes, 55 of resulted from cheerleading.
Take a random sample, without replacement, of 8 of these 101
accidents. Estimate the probability that at least half of your
eight sampled accidents resulted from cheerleading.
Assumptions:
You have a random sample (without replacement) of size n = 8
from a population of 101 accidents, of which 55 resulted from
cheerleading. (Each sample of size n = 8 is equally likely.)
Section 5.2 - Using Simulation to Estimate Probabilities
P12. Model:
There are 1000 three-digit triples and you wish to represent
101 accidents, so it is convenient to assign 9 three-digit
numbers to each accident, as follows. (If you assign the
numbers 001 - 101, you will have to reject 90% of the numbers
selected. This method rejects only 91 or 9.1%)
Triple-Digit Numbers
Accident Number
011-019
1
021-029
2
…
991-999
99
001-009
100
010, 020, 030, 040, 050,
060, 070, 080, 090
101
Section 5.2 - Using Simulation to Estimate Probabilities
P12. Model:
Assign 9 three-digit numbers to each accident, as follows.
Ignore all triples not on the table. Accidents numbered from 1
through 55 will be considered as resulting from cheerleading.
Stop a run when you have a sample of 8 accidents (no
repeats). Repeat until you have 10 runs.
Triple-Digit Numbers
Accident Number
011-019
1
021-029
2
…
…
991-999
99
001-009
100
010, 020, 030, 040, 050,
060, 070, 080, 090
101
Section 5.2 - Using Simulation to Estimate Probabilities
P12. Repetition:
Starting on line 17 of Table D, divide the digits into triples.
Triples to be ignored are crossed out and the remaining ones
are grouped into sequences representing samples of 8
accidents, making sure there are no repeated accidents in any
one run.
Run 1 = 4 from cheerleading
Triple:
801 243 563 517 727 080 154 531
Accident:
80
56
51
72
101 15
Result:
no yes no
yes
no
no yes yes
24
53
Run 2 = 6 from cheerleading
Triple:
822 374 211 157 825 314 385 537 637
Accident:
82
Result:
no yes yes yes
37
21
15
82
x
31
38
53
63
yes yes yes
no
Section 5.2 - Using Simulation to Estimate Probabilities
P12. Repetition:
Run 3 = 5 from cheerleading
Triple:
435 099 817 774 027 721 443 236
Accident:
43
81
77
Result:
yes yes no
no
9
2
72
44
23
yes no yes yes
Run 4 = 6 from cheerleading
Triple:
002 104 552 164 237 962 860 265 569
Accident: 100 10
Result:
55
16
23
96
x
26
56
no yes yes yes yes no
x
yes
no
Run 5: 6 from cheerleading
Triple:
916 268 036 625 229 148 369 368 720
376
Accident:
91
Result:
no yes yes
26
3
62
22
14
36
36
x
37
no
yes yes yes
x
x
yes
Section 5.2 - Using Simulation to Estimate Probabilities
P12. Repetition:
Run 6 = 4 from cheerleading
Triple:
621 139 909 440 056 418 098 932 050
Accident:
62
90
x
Result:
no yes no
x
13
5
41
9
yes yes yes
93
101
no
no
Run 7 = 3 from cheerleading
Triple:
514 225 685 144 642 756 788 962
Accident:
51
68
14
64
75
78
96
Result:
yes yes no
yes
no
no
no
no
22
Run 8: 4 from cheerleading
Triple:
977 882 254 382 145 989 149 914 523
Accident:
97
88
98
14
91
52
Result:
no
no yes yes yes no
x
no
yes
25
38
14
Section 5.2 - Using Simulation to Estimate Probabilities
P12. Repetition:
Run 9 = 3 from cheerleading
Triple:
684 792 768 646 162 835 549 475
Accident:
68
79
76
64
16
Result:
no
no
no
no
yes no yes yes
83
54
47
Run 10 = 2 from cheerleading
Triple:
Accident:
Result:
089 923 370 892 004 880 336 945 982
8
694
92
x
89
100
x
33
94
98
69
yes no
x
no
no
x
yes
no
no
no
Summary: 4,6,5,5,6,4,3,4,3,2
Section 5.2 - Using Simulation to Estimate Probabilities
P12. Repetition: Update Display 5.22
Accidents from Cheerleading
Frequency
0
0
1
10
2
81
3
187
4
262
5
252
6
162
7
43
8
3
Total
1000
Section 5.2 - Using Simulation to Estimate Probabilities
P12. Conclusion:
Out of the 1000 runs, 722 samples had at least half (four or
more) of the accidents from cheerleading.
The estimated probability that a random sample of eight
catastrophic accidents would have at least half of the
accidents from cheerleading is about 0.722.
This is close to the theoretical probability of 0.7374 computed
using the hypergeometric distribution.
See Fathom file SIA Ch 5 P12.ftm
Section 5.2 - Using Simulation to Estimate Probabilities
P13. The winner of the World Series is the first team to win
four games. That means the series can be over in four games
or can go as many as seven games. Suppose the two teams
playing are evenly matched.
Estimate the probability that the World Series will go seven
games before there is a winner.
Start at the beginning of row 9 of Table D on page 828, and
add your ten runs to the frequency table.
Section 5.2 - Using Simulation to Estimate Probabilities
P13. The winner of the World Series is the first team to win
four games. Suppose the two teams playing are evenly
matched. Estimate the probability that the World Series will go
seven games before there is a winner.
Assumptions:
The probability of a given team winning a particular game is
50% and that the results of the games are independent of
each other. (This is probably not a realistic assumption!)
Section 5.2 - Using Simulation to Estimate Probabilities
P13. Model:
Assign digits 1-5 to team A winning and 6-9, 0 to team B
winning. Select digits from the table, allowing repeats, until one
of the teams has four wins.
Record the total number of digits needed to get four wins for
one of the teams.
Digits
Winning Team
1-5
A
6-9,0
B
Section 5.2 - Using Simulation to Estimate Probabilities
P13. Repetition:
Starting on line 09 of Table D:
Run
1
2
3
4
5
Digits
635733
2135
05325
470489
05535
Team
BAABAA
AAAA
BAAAA
ABBAB
B
BAAAA
Games
6
4
5
6
5
Run
6
7
8
9
10
Digits
7548284
68287
098349
12562
473796
Team
BAABABA
BBABB
BBBAAB AAABA
ABABBB
Games
7
5
6
6
5
Section 5.2 - Using Simulation to Estimate Probabilities
P13. Repetition: Update Display 5.23
Games in Series
Frequency
4
611
5
1307
6
1545
7
1537
Total
5000
Section 5.2 - Using Simulation to Estimate Probabilities
P13. Conclusion:
Out of the 5000 runs, 1537 series went the full seven games,
so the estimated probability of a World Series of two evenly
matched teams going seven games is 1537 / 5000, or 0.3074.
The theoretical probability is 0.3125. When the teams are not
evenly matched, the probability is less.
In the 66 World Series between 1940 and 2006, the World
Series went the full seven games 27 times out of 66, or 0.409.
This is not statistically significant, but close.
One reasonable explanation is that the probability of winning
changes from game to game.
Section 5.2 - Using Simulation to Estimate Probabilities
E15. About 10% of high school girls report that they rarely or
never wear a seat belt while riding in motor vehicles. Suppose
you randomly sample four high school girls.
Estimate the probability that no more than one of the girls says
that she rarely or never wears a seat belt.
Start at the beginning of row 36 of Table D on page 828, and
add your ten results to the frequency table.
Section 5.2 - Using Simulation to Estimate Probabilities
E15. About 10% of high school girls report that they rarely or
never wear a seat belt while riding in motor vehicles. Suppose
you randomly sample four high school girls.
Estimate the probability that no more than one of the girls says
that she rarely or never wears a seat belt.
Assumptions:
The probability that a randomly selected girl reports that she
rarely or never wears a seat belt is 10%, and that the girls are
selected independently of each other.
Section 5.2 - Using Simulation to Estimate Probabilities
E15. Model:
Assign the digits 0 to 9 as follows:
Reports that she rarely or never wears a seat belt: 0
Does not report that she rarely or never wears a seat belt: 1-9
Section 5.2 - Using Simulation to Estimate Probabilities
E15. Repetition:
Begin with row 36 of table D, separated into groups of 4:
Run:
Digits:
Result:
1
2
3
4
5
6
7
8
9
10
3217 9005 9787 3792 5241 0556 7070 0786 7431 7157
0
2
0
0
0
1
2
1
0
0
Section 5.2 - Using Simulation to Estimate Probabilities
E15. Repetition:
Add the results to the table in Display 5.24:
Number of Girls Who
Rarely or Never Wear a
Seat Belt
Frequency
0
6,641
1
2,863
2
464
3
32
4
0
Total
10,000
Section 5.2 - Using Simulation to Estimate Probabilities
E15. Conclusion:
Out of the 10,000 runs, 6641 + 2863 = 9504 had no more than
one 0.
The estimated probability that a random sample of four girls
contains no more than one that says she rarely or never wears
a seat belt is 0.9504
The theoretical probability is 0.9477.
Section 5.2 - Using Simulation to Estimate Probabilities
E18. A Harris poll estimated that 25% of U.S. residents believe
in astrology. Suppose you would like to interview a person who
believes in astrology.
Estimate the probability that you will have to ask four or more
U.S. residents to find one who believes in astrology.
Start at the beginning of row 15 of Table D on page 828, and
add your ten results to the frequency table.
Section 5.2 - Using Simulation to Estimate Probabilities
E18. A Harris poll estimated that 25% of U.S. residents believe
in astrology. Suppose you would like to interview a person who
believes in astrology.
Estimate the probability that you will have to ask four or more
U.S. residents to find one who believes in astrology.
Assumptions:
The probability that a randomly selected person believes in
astrology is 25%, and that each person was selected randomly
and independently from the population of U.S. residents.
Section 5.2 - Using Simulation to Estimate Probabilities
E18. Model:
Assign two digit-numbers as follows:
Believes in astrology: 01 - 25
Does not believe in astrology: 26 - 99, 00
Section 5.2 - Using Simulation to Estimate Probabilities
E18. Repetition:
Begin with row 15 of table D:
Run
1
2
3
Digits
99 59 46 73 48 87 51 76 49 69
91 82 60 89 28 93 78 56 13
68 23
47 83 41 13
Belief:
NNNNNNNNNN
NNNNNNNNY
NY
NNNY
People
19
2
4
Section 5.2 - Using Simulation to Estimate Probabilities
E18. Repetition:
Begin with row 15 of table D:
Run
4
5
6
7
8
Digits
65 48 11 76 74 17 46 80 09 50 58 04 77 69 74 73 03
Belief:
NNY
NNY
NNY
NNY
NNNNY
People
3
3
3
3
5
Run
9
10
Digits
95 71 86 40 21
81 65 44 80 12
Belief:
NNNNY
NNNNY
People
5
5
Section 5.2 - Using Simulation to Estimate Probabilities
E18. Repetition:
Add the results to the table in Display 5.24:
Number of
Frequency
People Asked
1
437
2
412 + 1
3
278 + 4
4
210 +1
5
173 +3
…
…
19
4+1
…
…
Total
2,000
Section 5.2 - Using Simulation to Estimate Probabilities
E18. Conclusion:
Out of the 2,000 runs, 2000 - (437 + 413 + 282) = 868 had four
or more.
The estimated probability that you would have to interview four
or more U.S. residents before getting a person who believes in
astrology is 868 / 2000 or 0.434
The theoretical probability is 0.422.
(You will learn how to compute this in Section 6.3)
Section 5.2 - Using Simulation to Estimate Probabilities
E19. The probability that a baby is a girl is about 0.49.
Suppose a large number of couples each plan to have babies
until they have a girl.
Estimate the average number of babies per couple.
Start at the beginning of row 9 of Table D on page 828, and
add your ten results to the frequency table in Display 5.28,
which gives the results of 1990 runs.
Section 5.2 - Using Simulation to Estimate Probabilities
E19. The probability that a baby is a girl is about 0.49.
Suppose a large number of couples each plan to have babies
until they have a girl.
Estimate the average number of babies per couple.
Assumptions:
Each baby has a 0.49 chance of being a girl.
The gender of each child is independent of the other children’s
gender in that family.
Section 5.2 - Using Simulation to Estimate Probabilities
E19. Model:
Assign pairs of digits as follows:
01 - 49 = girl baby
50 - 99, 00 = boy baby
A single run consists of selecting pairs of digits, allowing
repeats, until a pair in the range 01 - 49 is selected.
Record the number of pairs needed.
Section 5.2 - Using Simulation to Estimate Probabilities
E19. Repetition:
Start with Row 9 of Table D. Perform 10 repetitions.
63 / 57 / 33 // 21 // 35 // 05 // 32 // 54 / 70 / 48 //
90 / 55 / 35 // 75 / 48 // 28 // 46 //
The numbers needed are:
3, 1, 1, 1, 1, 3, 3, 2, 1, 1
Section 5.2 - Using Simulation to Estimate Probabilities
E19. Repetition:
Add the results of our ten runs to the frequency table:
Babies Frequency
Babies
Frequency
1
941 + 6 = 947
8
11
2
480 + 1 = 481
9
4
3
265 + 3 = 268
10
1
4
156
11
2
5
60
12
0
6
51
13
2
7
17
Total
1990 + 10 = 2000
Section 5.2 - Using Simulation to Estimate Probabilities
E19. Conclusion:
Apply the method of calculating a mean from a frequency
table: Enter the data into L1, L2; Run 1-Var Stats.
Babies Frequency
Babies
Frequency
1
941 + 6 = 947
8
11
2
480 + 1 = 481
9
4
3
265 + 3 = 268
10
1
4
156
11
2
5
60
12
0
6
51
13
2
7
17
Total
1990 + 10 = 2000
Section 5.2 - Using Simulation to Estimate Probabilities
E19. Conclusion:
Apply the method of calculating a mean from a frequency
table: Enter the data into L1, L2; Run 1-Var Stats.
The estimated average number of babies for a family that
keeps having babies until they have a girl is about 2.122
children.
The theoretical mean is about 2.04.
Section 5.2 - Using Simulation to Estimate Probabilities
E20. Boxes of cereal often have small prizes in them. Suppose
each box of one type of cereal contains one of four different
small cars.
Estimate the average number of boxes a parent will have to
buy until his or her child gets all four cars.
Start at the beginning of row 49 of Table D on page 828, and
add your ten results to the frequency table.
Section 5.2 - Using Simulation to Estimate Probabilities
E20. Boxes of cereal often have small prizes in them. Suppose
each box of one type of cereal contains one of four different
small cars.
Estimate the average number of boxes a parent will have to
buy until his or her child gets all four cars.
Assumptions:
The different types of cars are equally and randomly
distributed in the boxes of cereal so the probability if getting a
particular type of car is always 0.25. Each box of cereal is
selected independently.
Section 5.2 - Using Simulation to Estimate Probabilities
E20. Model:
Assign two digit-numbers as follows:
Car 1: 01 - 25
Car 2: 26 - 50
Car 3: 51 - 75
Car 4: 76 - 99, 00
Section 5.2 - Using Simulation to Estimate Probabilities
E20. Repetition:
Begin with row 49 of table D:
Run
1
Digits
48 32 47 79 28 31 24 96 47 10 02 29 53
Cars
2224221421123
Boxes
13
Run
2
Digits
68 70 32 30 75 75 46 15 02 09 99
Cars
33223321114
Boxes
11
Section 5.2 - Using Simulation to Estimate Probabilities
E20. Conclusion:
Boxes Frequency
Boxes Frequency
Boxes Frequency
4
887
14
263
24
7
5
1395
15
167
25
12
6
1488
16
137
26
5
7
1347
17
103
27+
25
8
1136
18
81
Total
10,000
9
903
19
67
10
679
20
44
11
491
21
29
12
369
22
20
13
331
23
14
Section 5.2 - Using Simulation to Estimate Probabilities
E20. Conclusion:
We were asked to estimate a mean, not a probability.
Count 27+ as 27.
x f

x
n
4  887  5 1395  6 1488  ...  27  25

10000
83402

10000
 8.3402
Section 5.2 - Using Simulation to Estimate Probabilities
E20. Conclusion:
We were asked to estimate a mean, not a probability.
The estimated mean number of boxes a parent must buy to
get all four cars is 8.3402.
Section 5.2 - Using Simulation to Estimate Probabilities
E21. A group of five friends have backpacks that all look alike.
They toss their backpacks on the ground and later pick up a
backpack at random.
Estimate the probability that everyone gets his of her own
backpack.
Start at the beginning of row 31 of Table D on page 828.
Section 5.2 - Using Simulation to Estimate Probabilities
E21. A group of five friends have backpacks that all look alike.
They toss their backpacks on the ground and later pick up a
backpack at random.
Estimate the probability that everyone gets his of her own
backpack.
Assumptions:
Each backpack is equally likely to be picked up by each
person. The backpacks are selected independently of each
other.
Section 5.2 - Using Simulation to Estimate Probabilities
E21. Model:
Backpack 1 belongs to person 1, backpack 2 belongs to
person 2, etc.
Assign digits to represent backpacks: digits 1 and 2 =
backpack 1; digits 3 and 4 = backpack 2; digits 5 and 6 =
backpack 3; digits 7 and 8 = backpack 4; digits 9 and 0 =
backpack 5.
Select 4 digits and ignore repeats.
Record the number of correct backpacks.
Section 5.2 - Using Simulation to Estimate Probabilities
E21. Repetition:
Begin with row 31 of table D. Perform 20 repetitions.
0449352 / 49475 / 246338 / 24458 / 6251025619627 /
933565337 / 124720 / 054997 / 65464051 / 88159 /
9611963 / 89654 / 6928 / 23912328 / 7295 / 2935 /
9631 / 5307 / 2689 / 80935
Section 5.2 - Using Simulation to Estimate Probabilities
E21. Repetition:
Run
Digits
Bags
Correct
1
0449352
52xxx31(4)
2
2
49475
25x43(1)
0
3
246338
123xx4(5)
5
4
24458
12x34(5)
5
5
6251025619627 31xx5xxxxxxx4(2)
1
6
933565337
52x3xxxx4(1)
3
7
124720
1x24x5(3)
2
8
054997
532xx4(1)
1
9
65464051
3x2xx5x1(4)
1
10
88159
4x135(2)
1
Section 5.2 - Using Simulation to Estimate Probabilities
E21. Repetition:
Run
Digits
Bags
Correct
11
9611963
531xxx2(4)
0
12
89654
453x2(1)
1
13
6928
3514(2)
1
14
23912328
125xxxx4(3)
3
15
7295
4153(2)
0
16
2935
1523(4)
1
17
9631
5321(4)
0
18
5307
3254(1)
2
19
2689
1345(2)
1
20
80935
45x23(1)
0
Section 5.2 - Using Simulation to Estimate Probabilities
E21. Repetition:
Correct
Frequency
Probability
0
5
5 / 20 = 0.25
1
8
8 / 20 = 0.40
2
3
3 / 20 = 0.15
3
2
2 / 20 = 0.10
4
0
0 / 20 = 0.00
5
2
2 / 20 = 0.10
Total
20
1.00
Section 5.2 - Using Simulation to Estimate Probabilities
E21. Conclusion:
The estimated probability that each person gets his or her own
backpack is 0.10.
The theoretical probability is 0.0083.
Section 5.2 - Using Simulation to Estimate Probabilities
E23. There are six different car keys in a drawer, including
yours. Suppose you grab one key at a time until you get your
car key.
Estimate the probability that you get your car key on the
second try.
Start at the beginning of row 13 of Table D on page 828.
Section 5.2 - Using Simulation to Estimate Probabilities
E23. There are six different car keys in a drawer, including
yours. Suppose you grab one key at a time until you get your
car key.
Estimate the probability that you get your car key on the
second try.
Assumptions:
You select a key at random and each key is equally likely to be
picked on any one grab
Section 5.2 - Using Simulation to Estimate Probabilities
E23. Model:
Assign the digit 1 to your key and the digits 2 - 6 to the other
keys. Ignore the digits 7, 8, 9, 0.
A single run consists of selecting two digits from 1 - 6 (ignore
repeats, ignore 7, 8, 9, 0).
There is no need to continue selecting after the second valid
digit.
Section 5.2 - Using Simulation to Estimate Probabilities
E23. Repetition:
Begin with row 13 of table D:
83452 99634 06288 98083 13746 70078 18475 40610 68711
77817 88685 40200 86507 58401 36766 67951 90364 76493
834 / 52 / 9963 / 406 / 28898083 / 13 / 746 / 70078184 /
754 / 061 / 06871 / 177817886 / 854 / 020086 / 507584 /
013 / 666795 / 1903 / 64 / 764 /
Section 5.2 - Using Simulation to Estimate Probabilities
E23. Repetition:
834 / 52 / 9963 / 406 / 28898083 / 13 / 746 / 70078184 /
754 / 061 / 06871 / 177817886 / 854 / 020086 / 507584 /
013 / 666795 / 1903 / 64 / 764 /
Second Key Yours? Frequency
Yes
2
No
18
Total
20
Section 5.2 - Using Simulation to Estimate Probabilities
E23. Conclusion:
This is a small simulation of only 20 runs.
The estimated probability that if you pull one key at a time out
of a drawer, you will get your key on the second draw is 2 / 20,
or 0.10. The theoretical probability is 1/6. Why?
Section 5.2 - Using Simulation to Estimate Probabilities
E23. Conclusion:
The estimated probability that if you pull one key at a time out
of a drawer, you will get your key on the second draw is 2 / 20,
or 0.10. The theoretical probability is 1/6. Why?
Prob(key on second draw)  P(no on first)P(yes on second)
5 1
 
6 5
1

6
Section 5.2 - Using Simulation to Estimate Probabilities
E24. A deck of cards contains 13 hearts. Suppose you draw
cards one at a time, without replacement.
Estimate the probability that it takes you four cards or more to
draw the first heart.
Start at the beginning of row 28 of Table D on page 828.
Section 5.2 - Using Simulation to Estimate Probabilities
E24. A deck of cards contains 13 hearts. Suppose you draw
cards one at a time, without replacement.
Estimate the probability that it takes you four cards or more to
draw the first heart.
Assumptions:
Each card is equally likely to be drawn and are selected
independently.
Section 5.2 - Using Simulation to Estimate Probabilities
E24. Model:
Assign the digits 01 - 52 to the cards in the deck.
Let the digits 01 - 13 represent the hearts.
Ignore all other digits: 53 - 99, 00.
A single run consists of selecting pairs of digits without repeats
until you get a heart (01 - 13)
Record the number of pairs needed to get a heart.
Section 5.2 - Using Simulation to Estimate Probabilities
E23. Repetition:
Begin with row 28 of table D:
Run 1:
94557 28573 67897 54387 54622 44431 91190 42592
94 55 72 85 73 67 89 75 43 87 54 62 24 44 31 91 19 04
Section 5.2 - Using Simulation to Estimate Probabilities
E24. Repetition:
Run Draw
s
Run Draw
s
Draws
Frequency
1, 2, 3
10
1
6
11
2
4+
10
2
4
12
7
Total
20
3
1
13
4
4
5
14
3
5
1
15
4
6
1
16
1
7
12
17
1
8
4
18
3
9
7
19
1
10
6
20
2
Section 5.2 - Using Simulation to Estimate Probabilities
E24. Conclusion:
This is a small simulation of only 20 runs.
The estimated probability that it takes four or more draws from
a standard deck to get a heart is 10 / 20, or 0.50.
The theoretical probability is 0.4135. Why?
Section 5.2 - Using Simulation to Estimate Probabilities
E24. Conclusion:
This is a small simulation of only 20 runs.
The estimated probability that it takes four or more draws from
a standard deck to get a heart is 10 / 20, or 0.50.
The theoretical probability is 0.4135. Why?
P(4  draws)  1  P(1 or 2 or 3 draws)
13 39 13 39 38 13
 1

 
 
52 52 51 52 51 50
Section 5.2 - Using Simulation to Estimate Probabilities
E24. Conclusion:
This is a small simulation of only 20 runs.
The estimated probability that it takes four or more draws from
a standard deck to get a heart is 10 / 20, or 0.50.
The theoretical probability is 0.4135. Why?
P(4  draws)  P(not on first 3 draws)
39 38 37

 
52 51 50
Section 5.2 - Using Simulation to Estimate Probabilities
E26. This question appeared in the “Ask Marilyn” column:
I work at a waste-treatment plant, and we do assessments of
the time-to-failure and time-to-repair of the equipment, then
input those figures into a computer model to make plans. But
when I need to explain the process to people in other
departments, I find it difficult.
Say a component has two failure modes. One occurs every 5
years, and the other occurs every 10 years. People usually say
that the time to failure is 7.5 years, but this is incorrect. It’s
between 3 and 4 years. Do you know of a way to explain this
that people will accept?
Section 5.2 - Using Simulation to Estimate Probabilities
E26. Say a component has two failure modes. One occurs
every 5 years, and the other occurs every 10 years. People
usually say that the time to failure is 7.5 years, but this is
incorrect. It’s between 3 and 4 years. Do you know of a way to
explain this that people will accept?
Use simulation to estimate the expected time to failure for this
component. Start at the beginning of row 40 of Table D.
Section 5.2 - Using Simulation to Estimate Probabilities
E26. Say a component has two failure modes. One occurs
every 5 years, and the other occurs every 10 years.
Use simulation to estimate the expected time to failure for this
component.
Assumptions:
The probability of a mode 1 failure in a given year is 20%. The
probability of a mode 2 failure in a given year is 10%. These
failures occur independently of each other.
Section 5.2 - Using Simulation to Estimate Probabilities
E26. Model:
Consider pairs of digits.
The first digit represents mode 1 status, with 1 or 2
representing failure.
The second digit represents mode 2 status, with 1
representing failure.
Thus, a failure is any one of the 28 pairs:
10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,
29,01,31,41,51,61,71,81,91 (Why 28 pairs?)
A single run consists of selecting pairs of digits until you get a
failure. Perform a large number of runs, and compute the
mean number of years until failure
Section 5.2 - Using Simulation to Estimate Probabilities
E26. Repetition:
Begin with row 40 of table D:
Run 1: 4 years Run 2: 3 years Run 3: 1 year
Run 4: 2 years
94864 319
94 3616
81
0851
94 86 43 19
94 36 16
81
08 51
ok ok ok fail
ok ok fail
fail
ok fail
Section 5.2 - Using Simulation to Estimate Probabilities
E26. Repetition:
Run Years
Run Years
1
4
11
16
2
3
12
1
3
1
13
3
4
2
14
1
5
4
15
3
6
2
16
2
7
6
17
7
8
2
18
4
9
6
19
9
10
2
20
5
Section 5.2 - Using Simulation to Estimate Probabilities
E26. Conclusion:
This is a small simulation of only 20 runs.
The estimated mean number of years until failure is 83/20 =
4.15 years.
The theoretical expected time to failure is 3.57.
P(failure) = 1 - P(no failures) = 1 - (0.8)(0.9) = 0.28 (failures/yr)
Expected time = 1 / 0.28 = 3.57 (yrs/failure)