AP Statistics Objectives Ch17
Know the three parts of a
Bernoulli trial
Use Geometric model when
interested in the number of
Bernoulli trials until the next
success
AP Statistics Objectives Ch17
Use Binomial model when
interested in the number of
successes in a certain number of
Bernoulli trials
Use Normal model to
approximate a Binomial model
when expecting at least 10
successes and 10 failures
Vocabulary
Bernoulli trial
Geometric probability model
Binomial probability model
Success/Failure condition
Expected Value
Chapter 17 Notes
Additional
Examples
Chapter 17
Assignment
Chapter 16
Part I
Chapter 16
Part II
Chapter 17
Answers
Chapter 17 Assignment
Part I: Page 398 #2, 6
Part II: Pages 398 – 400 #14&22, 24
Part III: Page 400 #30
Chapter 17
1. Bernoulli Trial –
Events that meet the following 3 conditions:
i. There are only two outcomes.
ii. The probability of success is constant.
iii. The trials are independent.
(Must meet all 3 conditions above.)
EXAMPLES: flipping heads on a coin, finding a defect,
finding a prize, making a basket, getting a hit.
Chapter 17
2. 10% Condition –
Recall that for the events A & B to be independent,
P(A) = P(A|B). This isn’t true when we sample
without replacement. However, the change is
insignificant if the sample is smaller than 10% of the
population.
Chapter 17
Geometric Probability Model – Counts the
number Bernoulli Trials until the first Success.
Binomial Probability Model – Counts the number
of successes within a fixed number of Bernoulli
Trials.
Chapter 17
3. Geometric Probability Model – Geom(p)
Counts the number Bernoulli Trials until the first
Success.
P(X=x) = 𝒒𝒙−𝟏 𝒑
Geom(p)
p = probability of success <------ ONLY PARAMETER
q = probability of failure (q=1-p)
X = number of trials until the first success occurs
E(X) = 𝝁 =
𝟏
𝒑
Memorize this.
SD(X) = 𝝈 =
𝒒
𝒑𝟐
EXAMPLE 1
A new sales gimmick has 30% of the M&M’s covered with
speckles. These “groovy” candies are mixed randomly with
the normal candies as they are put into the bags for
distribution and sale. You buy a bag and remove candies one
at a time looking for the speckles.
This is a Bernoulli Trial because
i. There are only two outcomes which are…
1) Find speckled 2) Don’t find speckled
ii. The probability of success is constant.
p = .30
Success: Speckled
iii. The trials are not independent, because…
I sample without replacement, but the sample is less
than 10% of all M&M’s.
A new sales gimmick has 30% of the M&M’s covered with speckles.
These “groovy” candies are mixed randomly with the normal candies as
they are put into the bags for distribution and sale. You buy a bag and
remove candies one at a time looking for the speckles.
a) How many M&M’s do you expect to check before finding
a speckled M&M?
• Note that you keep checking until you find a speckled
M&M (success). This is how we identify that we need the
Geometric Probability Model.
• Also note that we are asked for an expected #. This is how
we identify that we need the Expected Value for
Geom(0.3).
E(X) = 𝝁 =
𝟏
𝒑
=
𝟏
.𝟑
≈ 𝟑. 𝟑
A new sales gimmick has 30% of the M&M’s covered with speckles.
These “groovy” candies are mixed randomly with the normal candies as
they are put into the bags for distribution and sale. You buy a bag and
remove candies one at a time looking for the speckles.
b) What is the probability that the first speckled M&M will
be the third one that you check? Geom(0.3)
• Note that if the third is the first success, then you started
with two failures.
• We are still using the Geometric Probability Model
so P(X=x) = 𝒒𝒙−𝟏 𝒑
P(X=3) = 𝟎. 𝟕𝟐 . 𝟑 = . 𝟏𝟒𝟕
c) What is the probability that the first speckled M&M will
Geom(0.3)
be the fourth one that you check?
P(X=4) = 𝟎. 𝟕𝟑 . 𝟑 =. 𝟏𝟎𝟐𝟗
A new sales gimmick has 30% of the M&M’s covered with speckles.
These “groovy” candies are mixed randomly with the normal candies as
they are put into the bags for distribution and sale. You buy a bag and
remove candies one at a time looking for the speckles.
d) What is the probability that the first speckled M&M will
be one of the first three that you check? Geom(0.3)
• Note that includes the third is the first success, second is
the first success, OR a success the first time you check.
P(X≤3) = P(X=3)
+ P(X=2)
+ P(X=1)
= (.7)(.7)(.3) + (.7)(.3) + .3
= 𝟎. 𝟕𝟐 . 𝟑 + 𝟎. 𝟕𝟏 . 𝟑 + 𝟎. 𝟕𝟎 . 𝟑
= .657
A new sales gimmick has 30% of the M&M’s covered with speckles.
These “groovy” candies are mixed randomly with the normal candies as
they are put into the bags for distribution and sale. You buy a bag and
remove candies one at a time looking for the speckles.
f) What is the probability that the first speckled M&M will
be one of the first four that you check? Geom(0.3)
P(X<4) = 𝟎. 𝟕𝟑 . 𝟑 + 𝟎. 𝟕𝟐 . 𝟑 + 𝟎. 𝟕𝟏 . 𝟑 + 𝟎. 𝟕𝟎 (. 𝟑)
= .7599
4. TI-84 steps to calculate the P(X=x) or
P(X<x) for Geom(p):
2nd VARS
For P(X = x) --- a specific value use…
Don’t use CALCULATOR
E: geometpdf(
SPEAK in your work.
For P(X < x) --- a specific value OR FEWER use…
F: geometcdf(
Don’t use CALCULATOR
SPEAK in your work.
TI-84 and Geom(p)
P(X=x)
Must add x
P(X=3) for Geom(.1)
4. TI-84 steps to calculate the P(X=x) or
P(X<x) for Geom(p):
2nd VARS
For P(X = x) --- a specific value use…
E: geometpdf(
For P(X < x) --- a specific value OR FEWER use…
F: geometcdf(
Practice using Geom(.3) for the values of X that we just calculated.
P(X = 3) = . 𝟏𝟒𝟕
P(X = 4) = . 𝟏𝟎𝟐𝟗
TI-84 and Geom(p)
P(X<x)
Must add x
P(X<4) for Geom(.1)
4. TI-84 steps to calculate the P(X=x) or
P(X<x) for Geom(p):
2nd VARS
For P(X = x) --- a specific value use…
E: geometpdf(
For P(X < x) --- a specific value OR FEWER use…
F: geometcdf(
Practice using Geom(.3) for the values of X that we just calculated.
P(X = 3) = . 𝟏𝟒𝟕
P(X = 4) = . 𝟏𝟎𝟐𝟗
P(X < 3) =
.657
P(X < 4) =
.7599
Factorial
5. 3! is read “3 factorial” is 3*2*1 = 6
6. What is 5! ?
5*4*3*2*1 = 120
Combinations
7. What is the formula for calculating a
combination of n things taken k at a time?
(Don’t worry. This is background information
only.)
“ A combination of n things taken k at a time”
𝑛
=
𝑘
𝑛𝐶𝑘
=
𝑛!
𝑘! 𝑛−𝑘 !
Combinations
8. How many combinations of 3 can you get
from 4 items?
“ A combination of 4 things taken 3 at a time”
4
=
3
4 𝐶3 =
4!
3! 4−3 !
=
4∗3∗2∗1
3∗2∗1(1)
= 4
Combinations
8. How many combinations of 3 can you get
from 4 items?
“ A combination of n things taken k at a time”
𝑛
=
𝑘
4
=
3
𝑛𝐶𝑘
4 𝐶3 =
=
𝑛!
𝑘! 𝑛−𝑘 !
4!
3! 4−3 !
=
4∗3∗2∗1
3∗2∗1(1)
= 4
Combinations
9. How many combinations of 5 can you get
from 8 items?
8∗7∗6∗5∗4∗3∗2∗1
8!
8
= 8 𝐶5 =
=
5! 8−5 ! 5∗4∗3∗2∗1(3∗2∗1)
5
8∗7∗6
=
6
= 8*7 = 56
10. TI-84 and Combinations
i. Type n first for
𝒏𝑪𝒌
ii.
iii. Under PRB choose 𝒏𝑪𝒌
iv. Then type k
Geometric Probability Model – Counts the number
Bernoulli Trials until the first Success.
11. Binomial Probability Model Binom(n,p)
Counts the number of successes within a fixed
number of Bernoulli Trials.
𝒏 𝒙 𝒏−𝒙
P(X=x) =
𝒑 𝒒
Binom(n,p)
𝒙
n = number of trials <------ 1 of 2 parameters
p = probability of success <------ 2 of 2 parameters
q = probability of failure (q=1-p)
X = number of successes in n trials
E(X) = 𝝁 = np
SD(X) = 𝝈 = 𝒏𝒑𝒒
𝒏
Remember:
=
𝒙!
𝒙
𝒏!
𝒏−𝒙 !
12. TI-84 steps to calculate the P(X=x) or
P(X<x) for Binom(n,p):
2nd VARS
For P(X = x) --- a specific value use…
A: binompdf(
For P(X < x) --- a specific value OR FEWER use…
B: Binomcdf(
13.
a) n = 12, p = 0.2, find P(2 successes)
Binom(12, 0.2)
P(X=2) = . 𝟐𝟖𝟑
b) n = 10, p = 0.4, find P(1 successes)
Binom(10, 0.4)
P(X=1) = . 𝟎𝟒𝟎
USE TI-84’s binompdf ,
but don’t write that
down.
13.
c) n = 20, p = 0.5, find P(10 successes)
Binom(20, 0.5)
P(X=10) = . 𝟏𝟕𝟔
d) n = 15, p = 0.9, find P(11 successes)
Binom(15, 0.9)
P(X=11) = . 𝟎𝟒𝟑
13.
𝟏
𝟑
e) n = 7, p = , find P(4 successes)
Binom(7, 1/3 )
P(X=4) = . 𝟏𝟐𝟖
13.
f) n = 11, p = 0.05, find P(3 failures)
Binom(11, .05 )
3 failures is the same as 8 successes
P(X=8) = . 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟓
OR q = 0.95 so
Binom(11, .95 )
P(X=3) = . 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟓
13.
g) n = 15, p = 0.99, find P(1 failure)
Binom(15, .99 )
1 failures is the same as 14 successes
P(X=14) =. 𝟏𝟑
OR q = 0.01
Binom(15, .01 )
P(X=1) = . 𝟏𝟑
14.
a) n = 6, p = 0.35, find P(at least 3 successes)
Binom(6, .35)
P(X>3) = P(X=3) OR P(X=4) OR P(X=5) OR P(X=6)
USE TI-84’s binomcdf ,
but don’t write that
down.
= 1 – P(X<2)
= 1 - . 𝟔𝟒𝟕
= . 𝟑𝟓𝟑
Careful! This is a discrete model…
limited values for X.
X can only be equal to …
0,1, 2, 3, 4, 5, or 6
P(X<2)
P(X>3)
14.
b) n = 100, p = 0.01, find P(no more than 3 successes)
Binom(100, .01)
P(X<3) = P(X=3) OR P(X=2) OR P(X=1) OR P(X=0)
= . 𝟗𝟖𝟐
Chapter 17 – Binomial
15. In history class, Colin takes a multiple choice quiz.
This is a Bernoulli Trial because
i. There are only two outcomes which are
1) Answer Right 2) Answer Wrong
ii. The probability of success is constant.
p = .20
Success: Answering correctly
iii. The trials are independent, if
he chooses randomly.
The model is Binom(10,0.2)
P(X > 7) = 1 - P(X < 6) = 0.00086
Chapter 17 – Binomial
16.
This is a Bernoulli Trial because
i. There are only two outcomes which are
1) Defective 2) Not Defective
ii. The probability of success is constant.
p = .01
Success: Defective
iii. The trials are independent, if
the components are randomly selected
The model is Binom(50,.01)
P(X = 0) = 0.605
NOTE
a) None defective
b) At least 1 defective
Binom(50,.99)
P(X = 50) = 0.605
Chapter 17 – Binomial
16.
Binom(50,.01)
P(X > 1) = 1 – P(X = 0)
= 1 – 0.605
= 0.395
Chapter 17 – Binomial
17.
This is a Bernoulli Trial because
i. There are only two outcomes which are
1) Cracked 2) Not Cracked
ii. The probability of success is constant.
p = .03
Success: Cracked egg
iii. The trials are independent, if
the eggs are randomly selected
The model is Binom(24,.03)
P(X = 0) = 0.481
Chapter 17 – Binomial WS II
17.
Binom(24,.03)
P(X > 1) = 1 – P(X = 0)
= 1 – 0.481
= 0.519
Chapter 17 – Binomial WS II
17.
Binom(24,.03)
P(X = 2) = 0.127
Chapter 17 – Binomial
18.
This is a Bernoulli Trial because
i. There are only two outcomes which are
1) Sum is 5
2) Sum is not 5
ii. The probability of success is constant.
𝟒
p=
Success: Sum of 5
𝟑𝟔
iii. The trials are independent, because
the dice rolls are independent of each other
The model is
𝟒
Binom(20, )
𝟑𝟔
P(X = 6) = 0.014
Chapter 17 – Binomial WS
18.
𝟒
Binom(20, )
𝟑𝟔
P(X > 4) = 1 – P(X < 3) = 0.175
Careful! This is a discrete model…
limited values for X.
X can only be equal to …
0,1, 2, 3, 4, 5, 6, 7, 8, 9, …, 19, 20
P(X<3)
P(X>4)
Chapter 17 – Binomial WS II
18.
𝟒
Binom(20, )
𝟑𝟔
P(X < 5) = 0.982
Chapter 17 – Binomial
19.
This is a Bernoulli Trial because
i. There are only two outcomes which are
1) Sinks shot
2) Doesn’t sink shot
ii. The probability of success is constant.
p = .𝟕
Success: sinks shot
iii. The trials are independent?
May not be true. Since Tim may get tired or “warmed up”
the more he shots.
The model is Binom(30, .7)
P(X = 21) = 0.157
If the shots are independent of each other.
Chapter 17 – Binomial WS II
19.
Binom(30, .7)
P(X > 21) = 1 – P(X < 20) = 0.589
If the shots are independent of each other.
Careful! This is a discrete model…
limited values for X.
X can only be equal to …
0,1, 2, 3,…20, 21, 22, 23, …, 29, 30
P(X<20)
P(X>21)
Chapter 17 – Binomial
19.
Binom(30, .7)
P(X < 21) = 0.568
If the shots are independent of each other.
Chapter 17 – Binomial
19.
P(X<17)
0,1, 2,
2, 3,…17,
3,…17, 18,
18, 19,
19, 20,
20, 21,
21, 22,
22, …
… 29,
29, 30
30
0,1,
Binom(30, .7)
P(X<20)
P(18 < X < 20)
= P(X < 20) – P(X < 17)
= 0.411 – 0.084 = 0.327
If the shots are independent of each other.
Chapter 17 – Binomial
20.
This is a Bernoulli Trial because
i. There are only two outcomes which are
1) Red drawn 2) Red not drawn
ii. The probability of success is constant.
p = .𝟒
Success: Red drawn
iii.The trials are independent, because
the marbles are replaced after each draw.
The model is Binom(50, .4)
P(X = 15) = 0.042
Chapter 17 – Binomial WS II
20.
Binom(50, .4)
P(X > 15) = 1 – P(X < 14) = 0.946
Chapter 17 – Binomial WS II
20.
Binom(50, .4)
P(X < 20) = 0.561
Chapter 17 – Binomial WS II
20.
P(X<16)
1,
1, 2,
2, 3,…16,
3,…16, 17,
17, 18,
18, 19,
19, 20,
20, 21,
21, 22,
22, 23,
23, 24,
24, 25,
25, …
… 49,
49, 50
50
Binom(50, .4)
P(17 < X < 25)
P(X<25)
= P(X < 25) – P(X < 16) =
= 0.943 – 0.156 = 0.787
Chapter 17
21. Binomial Model can be approximated
using the Normal Model if the number of trials
is large enough. HOW LARGE?
•
The SUCCESS/FAILURE CONDITION: A
binomial model is approximately Normal if
we expect at least 10 successes and 10
failures:
np > 10 and nq > 10
Chapter 17
22.Given the number of trials and the probability of success,
determine the probability indicated using the Normal Model
as an estimate.
a) n = 600, p = 0.35, find P(at least 220 successes)
SUCCESS/FAILURE CONDITION:
np = (600)(0.35) = 210 > 10 and
nq = (600)(0.65) = 390 > 10
𝝁 = (600)(0.35) = 210
𝝈=
z=
𝟔𝟎𝟎 . 𝟑𝟓 (. 𝟔𝟓) = 11.68
𝟐𝟐𝟎 −𝟐𝟏𝟎
𝟏𝟏.𝟔𝟖
= 0.856
P(z > 0.856 ) = 0.196
Chapter 17
23. Given the number of trials and the probability of success,
determine the probability indicated using the Normal Model
as an estimate.
b) n = 1000, p = 0.48, find P(no more than 450 successes)
SUCCESS/FAILURE CONDITION:
np = (1000)(0.48) = 480 > 10 and
nq = (1000)(0.52) = 520 > 10
𝝁 = (1000)(0.48) = 480
𝝈=
z=
𝟏𝟎𝟎𝟎 . 𝟒𝟖 (. 𝟓𝟐) = 15.80
𝟒𝟓𝟎 − 𝟒𝟖𝟎
𝟏𝟓.𝟖𝟎
= -1.90
P(z < -1.90 ) = 0.029
Chapter 17
23. Given the number of trials and the probability of success,
determine the probability indicated using the Normal Model
as an estimate.
b) n = 1000, p = 0.48, find P(no more than 450 successes)
SUCCESS/FAILURE CONDITION:
np = (1000)(0.48) = 480 > 10 and
nq = (1000)(0.52) = 520 > 10
𝝁 = (1000)(0.48) = 480
𝝈=
z=
𝟏𝟎𝟎𝟎 . 𝟒𝟖 (. 𝟓𝟐) = 15.80
𝟒𝟓𝟎 − 𝟒𝟖𝟎
𝟏𝟓.𝟖𝟎
= -1.90
P(z < -1.90 ) = 0.029
Bell Work
Suppose a used car dealer runs autos through a two-stage process to
get them ready to sell. The mechanical checkup costs $50 per hour
and takes an average of 90 minutes, with a standard deviation of 15
minutes. The appearance prep (wash, polish, etc.) costs $6 per hour
and takes an average of 60 minutes, with a standard deviation of 5
minutes.
1. What are the mean and standard deviation of the total time spent
preparing a car? (Note that we cannot find the standard deviation if
we do not believe that the two phases of the process are
independent, an important assumption to check.)
Chapter 16 Part I
2&10. Find the expected value and standard deviation of each
random variable. (Show work)
a)
x 0
P(X=x) 0.2
1
0.4
2
0.4
Chapter 16 Part I
2&10. Find the expected value and standard deviation of each
random variable. (Show work)
b)
x
P(X=x)
100
0.1
200
0.2
300
0.5
400
0.2
Chapter 16 Part I
18. An insurance policy costs $100 and will pay
policyholders $10,000 if they suffer a major injury
(resulting in hospitalization) or $3000 if they suffer a
minor injury (resulting in lost time from work). The
company estimates that each year 1 in every 2000
policyholders may have a major injury, and 1 in 500 a
minor injury.
a) Create a probability model for the profit on a policy.
Chapter 16 Part I
18. An insurance policy costs $100 and will pay
policyholders $10,000 if they suffer a major injury
(resulting in hospitalization) or $3000 if they suffer a
minor injury (resulting in lost time from work). The
company estimates that each year 1 in every 2000
policyholders may have a major injury, and 1 in 500 a
minor injury.
b) What’s the company’s expected profit on this policy?
Chapter 16 Part I
18. An insurance policy costs $100 and will pay
policyholders $10,000 if they suffer a major injury
(resulting in hospitalization) or $3000 if they suffer a
minor injury (resulting in lost time from work). The
company estimates that each year 1 in every 2000
policyholders may have a major injury, and 1 in 500 a
minor injury.
c) What’s the standard deviation for profit?
Chapter 16 Part I
20. Your company bids for two contracts. You believe
the probability you get contract #1 is 0.8. If you get
contract #1, the probability you also get contract #2 will
be 0.2, and if you do not get #1, the probability you get
#2 will be 0.3.
a) Are the two contracts independent? Explain.
Chapter 16 Part I
20. Your company bids for two contracts. You believe
the probability you get contract #1 is 0.8. If you get
contract #1, the probability you also get contract #2 will
be 0.2, and if you do not get #1, the probability you get
#2 will be 0.3.
b) Find the probability you get both contracts.
Chapter 16 Part I
20. Your company bids for two contracts. You believe
the probability you get contract #1 is 0.8. If you get
contract #1, the probability you also get contract #2 will
be 0.2, and if you do not get #1, the probability you get
#2 will be 0.3.
c) Find the probability you get no contract.
Chapter 16 Part I
20. Your company bids for two contracts. You believe
the probability you get contract #1 is 0.8. If you get
contract #1, the probability you also get contract #2 will
be 0.2, and if you do not get #1, the probability you get
#2 will be 0.3.
d) Let X be the number of contacts you get. Find the
probability model for X.
Chapter 16 Part I
20. Your company bids for two contracts. You believe
the probability you get contract #1 is 0.8. If you get
contract #1, the probability you also get contract #2 will
be 0.2, and if you do not get #1, the probability you get
#2 will be 0.3.
e) Find the expected value and standard deviation of X.
Chapter 16 Part I
24. Given independent random variables with means and
standard deviations as shown, find the mean and
standard deviation of each of these variables. (SHOW WORK)
Mean
X
80
Y
12
a) X – 20
SD
12
3
Chapter 16 Part I
24. Given independent random variables with means and
standard deviations as shown, find the mean and
standard deviation of each of these variables. (SHOW WORK)
Mean
X
80
Y
12
b) 0.5Y
SD
12
3
Chapter 16 Part I
24. Given independent random variables with means and
standard deviations as shown, find the mean and
standard deviation of each of these variables. (SHOW WORK)
Mean
X
80
Y
12
c) X + Y
SD
12
3
Chapter 16 Part I
24. Given independent random variables with means and
standard deviations as shown, find the mean and
standard deviation of each of these variables. (SHOW WORK)
Mean
X
80
Y
12
d) X – Y
SD
12
3
Chapter 16 Part I
24. Given independent random variables with means and
standard deviations as shown, find the mean and
standard deviation of each of these variables. (SHOW WORK)
Mean
X
80
Y
12
e) 𝒀𝟏 + 𝒀𝟐
SD
12
3
Chapter 16 Part II
27. A grocery supplier believes that in a dozen eggs, the
mean number of broken ones is 0.6 with a standard
deviation of 0.5 eggs. You buy 3 dozen eggs without
checking them.
a) How many broken eggs do you expect to get?
Chapter 16 Part II
27. A grocery supplier believes that in a dozen eggs, the
mean number of broken ones is 0.6 with a standard
deviation of 0.5 eggs. You buy 3 dozen eggs without
checking them.
b) What’s the standard deviation?
𝜎 = 𝑆𝐷 𝐵𝑟𝑜𝑘𝑒𝑛 𝑒𝑔𝑔𝑠 𝑖𝑛 3 𝑑𝑜𝑧𝑒𝑛
= .52 +.52 +.52
≈ 0.87 𝑒𝑔𝑔𝑠
Chapter 16 Part II
27. A grocery supplier believes that in a dozen eggs, the
mean number of broken ones is 0.6 with a standard
deviation of 0.5 eggs. You buy 3 dozen eggs without
checking them.
c) What assumptions did you have to make about
the eggs in order to answer this question?
The cartons of eggs must be independent of
each other.
Chapter 16 Part II
32. A casino knows that people play the slot machines in
hopes of hitting the jackpot, but that most of them lose
their dollar. Suppose a certain machine pays out an
average of $0.92, with a standard deviation of $120.
a) Why is the standard deviation so large?
Chapter 16 Part II
32. A casino knows that people play the slot machines in
hopes of hitting the jackpot, but that most of them lose
their dollar. Suppose a certain machine pays out an
average of $0.92, with a standard deviation of $120.
b) If you play 5 times, what are the mean and
standard deviation of the casino’s profit?
Chapter 16 Part II
32. A casino knows that people play the slot machines in
hopes of hitting the jackpot, but that most of them lose
their dollar. Suppose a certain machine pays out an
average of $0.92, with a standard deviation of $120.
c) If gamblers play this machine 1000 times in a day,
what are the mean and standard deviation of the
casino’s profit?
Chapter 16 Part II
32. A casino knows that people play the slot machines in
hopes of hitting the jackpot, but that most of them lose
their dollar. Suppose a certain machine pays out an
average of $0.92, with a standard deviation of $120.
d) Do you think the casino is likely to be profitable?
Explain.
Chapter 16 Part II
34&36. The American Veterinary Association claims that
the annual cost of medical care for dogs averages $100,
with a standard deviation of $30, and for cats averages
$120, with a standard deviation of $35.
a) Answer the following questions:
i. What’s the expected difference in the cost of medical care for dogs
and cats?
Chapter 16 Part II
34&36. The American Veterinary Association claims that
the annual cost of medical care for dogs averages $100,
with a standard deviation of $30, and for cats averages
$120, with a standard deviation of $35.
a) Answer the following questions:
ii. What’s the standard deviation of that difference?
Chapter 16 Part II
34&36. The American Veterinary Association claims that
the annual cost of medical care for dogs averages $100,
with a standard deviation of $30, and for cats averages
$120, with a standard deviation of $35.
a) Answer the following questions:
iii. If the difference in costs can be described by a Normal model,
what’s the probability that medical expenses are higher for someone’s
dog than for her cat?
P(z>0.4338) = 0.332
Chapter 16 Part II
34&36. The American Veterinary Association claims that
the annual cost of medical care for dogs averages $100,
with a standard deviation of $30, and for cats averages
$120, with a standard deviation of $35.
b) You’re thinking about getting two dogs and a cat.
Assume the annual veterinary expenses are
independent. Answer the following questions:
i. Define appropriate variables and express the total annual
veterinary costs you may have.
Chapter 16 Part II
34&36. The American Veterinary Association claims that
the annual cost of medical care for dogs averages $100,
with a standard deviation of $30, and for cats averages
$120, with a standard deviation of $35.
b) You’re thinking about getting two dogs and a cat.
Assume the annual veterinary expenses are
independent. Answer the following questions:
ii. Describe the model for this total cost. Be sure to specify
its name, expected value, and standard deviation.
Chapter 16 Part II
34&36. The American Veterinary Association claims that
the annual cost of medical care for dogs averages $100,
with a standard deviation of $30, and for cats averages
$120, with a standard deviation of $35.
b) You’re thinking about getting two dogs and a cat.
Assume the annual veterinary expenses are
independent. Answer the following questions:
iii. What’s the probability that your total expenses will
exceed $400?
P(z>1.45) = 0.073
Chapter 16 Part II
38. Bicycles arrive at a bike shop in boxes. Before they
can be sold, they must be unpacked, assembled, and
turned (lubricated, adjusted, etc.). Based on past
experience, the shop manager makes the following
assumptions about how long this may take:
*The times for each setup phase are independent.
*The times for each phase follow a Normal model.
*The means and standard deviations of the times (in minutes) are as shown:
Phase
Unpacking
Assembly
Tuning
Mean
3.5
21.8
12.3
SD
0.7
2.4
2.7
a) What are the mean and standard deviation for the
total bicycle setup time?
Chapter 16 Part II
38. Bicycles arrive at a bike shop in boxes. Before they can be sold, they must be unpacked,
assembled, and turned (lubricated, adjusted, etc.). Based on past experience, the shop manager
makes the following assumptions about how long this may take:
*The times for each setup phase are independent.
*The times for each phase follow a Normal model.
*The means and standard deviations of the times (in minutes) are as shown:
Phase
Mean
SD
Unpacking
3.5
0.7
Assembly
21.8
2.4
Tuning
12.3
2.7
a) What are the mean and standard deviation for the
total bicycle setup time?
Chapter 16 Part II
38. Bicycles arrive at a bike shop in boxes. Before they can be sold, they must be unpacked,
assembled, and turned (lubricated, adjusted, etc.). Based on past experience, the shop manager
makes the following assumptions about how long this may take:
*The times for each setup phase are independent.
*The times for each phase follow a Normal model.
*The means and standard deviations of the times (in minutes) are as shown:
Phase
Unpacking
Assembly
Tuning
Mean
3.5
21.8
12.3
SD
0.7
2.4
2.7
b) A customer decides to buy a bike like one of the display
models but wants a different color. The shop has one, still in
the box. The manager says they can have it read in half an
hour. Do you think the bike will be set up and ready to go as
promised? Explain.
Chapter 16 Part II
40. The bicycle shop will be offering 2 specially priced
children’s models at a sidewalk sale. The basic model
will sell for $120 and the deluxe model for $150. Past
experience indicates that sales of the basic model will
have a mean of 5.4 bikes with a standard deviation of
1.2, and sales of the deluxe model will have a mean of
3.2 bikes with a standard deviation of 0.8 bikes. The cost
of setting up for the sidewalk sale is $200.
a) Define random variables and use them to express
the bicycle shop’s net income.
Chapter 16 Part II
40. The bicycle shop will be offering 2 specially priced children’s models at a
sidewalk sale. The basic model will sell for $120 and the deluxe model for $150.
Past experience indicates that sales of the basic model will have a mean of 5.4
bikes with a standard deviation of 1.2, and sales of the deluxe model will have a
mean of 3.2 bikes with a standard deviation of 0.8 bikes. The cost of setting up
for the sidewalk sale is $200.
a) Define random variables and use them to express the
bicycle shop’s net income.
Chapter 16 Part II
40. The bicycle shop will be offering 2 specially priced children’s models at a
sidewalk sale. The basic model will sell for $120 and the deluxe model for $150.
Past experience indicates that sales of the basic model will have a mean of 5.4
bikes with a standard deviation of 1.2, and sales of the deluxe model will have a
mean of 3.2 bikes with a standard deviation of 0.8 bikes. The cost of setting up
for the sidewalk sale is $200.
b) What’s the mean of the net income?
Chapter 16 Part II
40. The bicycle shop will be offering 2 specially priced children’s models at a
sidewalk sale. The basic model will sell for $120 and the deluxe model for $150.
Past experience indicates that sales of the basic model will have a mean of 5.4
bikes with a standard deviation of 1.2, and sales of the deluxe model will have a
mean of 3.2 bikes with a standard deviation of 0.8 bikes. The cost of setting up
for the sidewalk sale is $200.
c) What’s the standard deviation of the net income?
Chapter 16 Part II
40. The bicycle shop will be offering 2 specially priced children’s models at a
sidewalk sale. The basic model will sell for $120 and the deluxe model for $150.
Past experience indicates that sales of the basic model will have a mean of 5.4
bikes with a standard deviation of 1.2, and sales of the deluxe model will have a
mean of 3.2 bikes with a standard deviation of 0.8 bikes. The cost of setting up
for the sidewalk sale is $200.
d) Do you need to make any assumptions in calculating the
mean? How about the standard deviation?
Chapter 16 Part II
40. The bicycle shop will be offering 2 specially priced children’s models at a
sidewalk sale. The basic model will sell for $120 and the deluxe model for $150.
Past experience indicates that sales of the basic model will have a mean of 5.4
bikes with a standard deviation of 1.2, and sales of the deluxe model will have a
mean of 3.2 bikes with a standard deviation of 0.8 bikes. The cost of setting up
for the sidewalk sale is $200.
d) Do you need to make any assumptions in calculating the
mean? How about the standard deviation?
Chapter 17
2. Can we use probability models based on Bernoulli trials to
investigate the following situations? Explain by listing all
conditions met and those that are not met.
a) You are rolling 5 dice and need to get at least two
6’s to win the game.
This can be a Bernoulli trial, because…
1) There are only 2 outcomes
- Roll a 6
- Don’t roll a 6
2) The probability of success is constant
p = 1/6
Success: Roll a 6
3) The rolls are independent of each other.
Chapter 17
2. Can we use probability models based on Bernoulli trials to
investigate the following situations? Explain by listing all
conditions met and those that are not met.
b) We record the eye colors found in a group of 500
people.
This cannot be a Bernoulli trial, because…
1) There are more than two possible outcomes for
eye color.
Chapter 17
2. Can we use probability models based on Bernoulli trials to
investigate the following situations? Explain by listing all
conditions met and those that are not met.
c) A manufacturer recalls a doll because about 3% have
buttons that are not properly attached. Customers return 37
of these dolls to the local toy store. Is the manufacturer likely
to find any dangerous buttons?
This can be a Bernoulli trial, because…
1) There are only 2 outcomes
- Dangerous button - Not dangerous button
2) The probability of success is constant
p = .03
Success: Find dangerous button
3) The trials are not independent, since the total
number of dolls is finite, but 37 is probably less than
10% of all dolls.
Chapter 17
2. Can we use probability models based on Bernoulli trials to
investigate the following situations? Explain by listing all
conditions met and those that are not met.
d) A city council of 11 Republicans and 8 Democrats
picks a committee of 4 at random. What’s the
probability they choose all Democrats?
This cannot be a Bernoulli trial, because…
the probability of choosing a Democrat changes depending
on who has already been chosen. The 10% condition is not
met, because 4 of 19 people is more than 10%.
Chapter 17
2. Can we use probability models based on Bernoulli trials to investigate the
following situations? Explain by listing all conditions met and those that are not
met.
e) A 2002 Rutgers University study found that 74% of highschool students have cheated on a test at least once. Your
local high-school principal conducts a survey in homerooms
and gets responses that admit to cheating from 322 of the
481 students.
This can be a Bernoulli trial, because…
1) There are only 2 outcomes
- Admit cheating
- Don’t admit cheating
2) The probability of success is constant
p = .74
Success: Admit cheating
3) The trials are not independent, since the total
number of students is finite, but 481 is less than
10% of all students.
Chapter 17
6. Suppose 75% of all drivers always wear their seatbelts. Let’s
investigate how many of the drivers might be belted among five
cars waiting at a traffic light.
a) Describe how you would simulate the number of
seatbelt-wearing drivers among the five cars.
Component: Check one driver
Outcomes: 00 – 74 Driver wearing seatbelt
75 – 99 Driver is not wearing seatbelt
Trail: Check five drivers
Response variable: Number of drivers wearing seatbelts
out of 5 drivers
Chapter
17
6. Suppose 75% of all drivers always wear their seatbelts. Let’s investigate
how many of the drivers might be belted among five cars waiting at a traffic
light.
b) Run 30 trials
5
5
3
4
4
2
4
4
4
4
3
4
3
3
3
3
# Wearing Seatbelt
Wearing Seatbelt
3
4
4
2
4
5
4
5
4
4
5
3
5
5
4
Chapter 17
6. Suppose 75% of all drivers always wear their
seatbelts. Let’s investigate how many of the
drivers might be belted among five cars waiting
at a traffic light.
c) Based on you simulation, estimate the
probabilities there are no belted drivers, exactly 1,
exactly 2, exactly 3, exactly 4, and exactly 5.
x
P(X=x)
0
𝟎
=0
𝟑𝟎
1
𝟎
=0
𝟑𝟎
2
3
4
𝟕
𝟏𝟒
𝟐
𝟑𝟎
𝟑𝟎
𝟑𝟎
≈. 𝟐𝟑𝟑 ≈. 𝟒𝟔𝟕
≈. 𝟎𝟔𝟕
5
𝟕
≈
𝟑𝟎
. 𝟐𝟑𝟑
Chapter 17
6. Suppose 75% of all drivers always wear their
seatbelts. Let’s investigate how many of the
drivers might be belted among five cars waiting
at a traffic light.
d) Calculate the actual probability model.
Binom(5, .75)
P(X=0)
P(X=1) P(X=2)
P(X=3)
P(X=4)
P(X=5)
Chapter 17
6. Suppose 75% of all drivers always wear their
seatbelts. Let’s investigate how many of the
drivers might be belted among five cars waiting
at a traffic light.
e) Compare the distribution of outcomes in your
simulation to the probability model.
x
0
P(X=x) 0
1
0
2
3
4
5
. 𝟎𝟔𝟕 . 𝟐𝟑𝟑 . 𝟒𝟔𝟕 . 𝟐𝟑𝟑
x
P(X=x)
0
0
1
2
.001 . 𝟎𝟗
3
4
5
. 𝟐𝟔
. 𝟒𝟎
. 𝟐𝟒
The differences were small (the largest being 6.7% for 4
cars) and by the Law of Large Numbers they would get
even smaller as the number of trials increased.
Chapter 17
14 & 22. An Olympic archer is able to hit the
bull’s-eye 80% of the time. Assume each shot is
independent of the others.
This can be a Bernoulli trial, because…
1) There are only 2 outcomes
- Hit bull’s-eye - Don’t hit bull’s-eye
2) The probability of success is constant
p = .80
Success: Hit bull’s-eye
3) We are told to assume each shot is independent.
Chapter 17
14. An Olympic archer is able to hit the bull’seye 80% of the time. Assume each shot is
independent of the others.
i. If she shoots 6 arrows, what’s the probability of each result
described below. Be sure to name the model and use proper
notation each time. (First two are set-up for you.)
a) Her first bull’s-eye comes on the third arrow.
Geom(______)
0.80
P(X = 3) = 0.032
Chapter 17
14. An Olympic archer is able to hit the bull’seye 80% of the time. Assume each shot is
independent of the others.
i. If she shoots 6 arrows, what’s the probability of each result
described below. Be sure to name the model and use proper
notation each time. (First two are set-up for you.)
b) She misses the bull’s-eye at least once.
6 0.20
Binom(_____,_____)
P(Y > 1) = 1 – P(Y = 0) = 0.738
Chapter 17
14. An Olympic archer is able to hit the bull’seye 80% of the time. Assume each shot is
independent of the others.
i. If she shoots 6 arrows, what’s the probability of each result
described below. Be sure to name the model and use proper
notation each time. (First two are set-up for you.)
c) Her first bull’s-eye comes on the fourth or fifth arrow.
Geom(______)
0.80
P(X = 4 OR X =5)
= P(X = 4) + P(X = 5)
= 0.006 + 0.001
= 0.007
Chapter 17
14. An Olympic archer is able to hit the bull’seye 80% of the time. Assume each shot is
independent of the others.
i. If she shoots 6 arrows, what’s the probability of each result
described below. Be sure to name the model and use proper
notation each time. (First two are set-up for you.)
d) She gets exactly 4 bull’s-eyes.
6 0.80
Binom(_____,_____)
P(Y = 4) = 0.246
Chapter 17
14. An Olympic archer is able to hit the bull’seye 80% of the time. Assume each shot is
independent of the others.
i. If she shoots 6 arrows, what’s the probability of each result
described below. Be sure to name the model and use proper
notation each time. (First two are set-up for you.)
e) She gets at least 4 bull’s-eyes.
6 0.80
Binom(_____,_____)
P(Y > 4) = 1 – P(X < 3) = 0.901
Chapter 17
14. An Olympic archer is able to hit the bull’seye 80% of the time. Assume each shot is
independent of the others.
i. If she shoots 6 arrows, what’s the probability of each result
described below. Be sure to name the model and use proper
notation each time. (First two are set-up for you.)
f) She gets at most 4 bull’s-eyes.
6 0.80
Binom(_____,_____)
P(Y < 4) = 0.345
Chapter 17
22. An Olympic archer is able to hit the bull’seye 80% of the time. Assume each shot is
independent of the others.
ii. If she shoots 200 arrows, what’s the probability of each result
described below.
a) What are the mean and standard deviation of the number of
bull’s-eyes she might get? (Make sure that you name the model
first.)
200 0.80
Binom(_____,_____)
E(Y) = np = 200(0.8) = 160 bull’s eyes
SD(Y) = 𝒏𝒑𝒒 = 𝟐𝟎𝟎 . 𝟖 (. 𝟐) ≈ 5.66 bull’s eyes
Chapter 17
22. An Olympic archer is able to hit the bull’seye 80% of the time. Assume each shot is
independent of the others.
ii. If she shoots 200 arrows, what’s the probability of each result
described below.
b) Verify that you can use a Normal model to approximate the
distribution of the number of good first serves.
Since np = 160 and nq = 40 are both greater than 10,
Binom(200, 0.8) may be approximated by N(160, 5.66).
Chapter 17
22. An Olympic archer is able to hit the bull’seye 80% of the time. Assume each shot is
independent of the others.
ii. If she shoots 200 arrows, what’s the probability of each result
described below.
c) Use the 68-95-99.7 Rule to describe this distribution.
Chapter 17
22. An Olympic archer is able to hit the bull’seye 80% of the time. Assume each shot is
independent of the others.
ii. If she shoots 200 arrows, what’s the probability of each result
described below.
d) Would you be surprised if she made only 140 bull’s-eyes? Explain.
I would be surprised if she only made 140 bull’s-eyes out of 200,
because that should happen only about 0.02% of the time.
Chapter 17
24. An orchard owner knows that he’ll have to use about 6% of
the apples he harvests for cider because they will have bruises
or blemishes. He expects a tree to produce about 300 apples.
a) Describe an appropriate model for the number of cider apples that
may come from that tree. Justify your model. (You justify the model
by showing its conditions are met.)
This is a Bernoulli Trial because
i. There are only two outcomes which are
1) Apple for cider 2) Apple not for cider
ii. The probability of success is constant.
p = . 𝟎𝟔
Success: Apple for cider
iii.The trials are not independent, because
there are a finite number of apples, but 300 apples is probably
less than 10% of all the farmer’s apples.
The model is Binom(300, .06)
Chapter 17
24. An orchard owner knows that he’ll have to use about 6% of
the apples he harvests for cider because they will have bruises
or blemishes. He expects a tree to produce about 300 apples.
a) Describe an appropriate model for the number of cider apples that
may come from that tree. Justify your model. (You justify the model
by showing its conditions are met.)
For Binom(300, .06) …
E(X) = np = 300(0.𝟎𝟔) = 𝟏𝟖 cider apples
SD(X) = 𝒏𝒑𝒒 = 𝟑𝟎𝟎 . 𝟎𝟔 (. 𝟗𝟒) ≈ 4.11 cider apples
Since np = 18 and nq = 282 are both greater than 10,
Binom(300, .06) can be approximated by N(18, 4.11).
Chapter 17
24. An orchard owner knows that he’ll have to use about 6% of
the apples he harvests for cider because they will have bruises
or blemishes. He expects a tree to produce about 300 apples.
b) Find the probability there will be no more than a dozen cider
apples.
Binom(300, .06)
P(X<12) ≈ 0.085
Chapter 17
24. An orchard owner knows that he’ll have to use about 6% of
the apples he harvests for cider because they will have bruises
or blemishes. He expects a tree to produce about 300 apples.
b) Find the probability there will be no more than a dozen cider
apples.
Chapter 17
24. An orchard owner knows that he’ll have to
use about 6% of the apples he harvests for cider
because they will have bruises or blemishes. He
expects a tree to produce about 300 apples.
c) Is it likely there will be more than 50 cider apples? Explain.
Binom(300, 0.06) P(X > 50) = 1 - P(X < 50) = 2.2 x 𝟏𝟎−𝟏𝟏
N(18, 4.11) z =
𝟓𝟎 −𝟏𝟖
𝟒.𝟏𝟏
= 7.79
P(z > 50) = 3.4 x 𝟏𝟎−𝟏𝟓
It is nearly impossible for there to be more than 50 cider apples,
because for both models the probability is nearly zero.
Chapter 17
30. Vitamin D is essential for strong, healthy bones. Our bodies
produce vitamin D naturally when sunlight falls upon the skin, or
it can be taken as a dietary supplement. Although the bone
disease rickets was largely eliminated in England during the
1950s, some people there are concern that this generation of
children is at increased risk because they are more likely to
watch TV or play computer games than spend time outdoors.
Recent research indicated that about 20% of British children are
deficient in vitamin D. Suppose doctors test a group of
elementary school children.
Chapter 17
30. Vitamin D is essential for strong, healthy bones. Our bodies produce
vitamin D naturally when sunlight falls upon the skin, or it can be taken as a
dietary supplement. Although the bone disease rickets was largely eliminated
in England during the 1950s, some people there are concern that this
generation of children is at increased risk because they are more likely to
watch TV or play computer games than spend time outdoors. Recent research
indicated that about 20% of British children are deficient in vitamin D. Suppose
doctors test a group of elementary school children.
This is a Bernoulli Trial because
i. There are only two outcomes which are
1) Vitamin D deficient 2) Not vitamin D deficient
ii. The probability of success is constant.
p = . 𝟐𝟎
Success: Being Vitamin D deficient
iii.The trials are not independent, because
there are a finite number of British children, but children at this
school represent fewer than 10% of all British children.
Chapter 17
30. Vitamin D is essential for strong, healthy bones. Our bodies produce
vitamin D naturally when sunlight falls upon the skin, or it can be taken as a
dietary supplement. Although the bone disease rickets was largely eliminated
in England during the 1950s, some people there are concern that this
generation of children is at increased risk because they are more likely to
watch TV or play computer games than spend time outdoors. Recent research
indicated that about 20% of British children are deficient in vitamin D. Suppose
doctors test a group of elementary school children.
a) What’s the probability that the first vitamin D-deficient child is the
8th one tested?
Geom(0.2)
P(X = 8) = 0.042
Chapter 17
30. Vitamin D is essential for strong, healthy bones. Our bodies produce
vitamin D naturally when sunlight falls upon the skin, or it can be taken as a
dietary supplement. Although the bone disease rickets was largely eliminated
in England during the 1950s, some people there are concern that this
generation of children is at increased risk because they are more likely to
watch TV or play computer games than spend time outdoors. Recent research
indicated that about 20% of British children are deficient in vitamin D. Suppose
doctors test a group of elementary school children.
b) What’s the probability that the first 10 children tested are all okay?
Binom(10, 0.8)
P(X = 10) = 0.107
Chapter 17
30. Vitamin D is essential for strong, healthy bones. Our bodies produce
vitamin D naturally when sunlight falls upon the skin, or it can be taken as a
dietary supplement. Although the bone disease rickets was largely eliminated
in England during the 1950s, some people there are concern that this
generation of children is at increased risk because they are more likely to
watch TV or play computer games than spend time outdoors. Recent research
indicated that about 20% of British children are deficient in vitamin D. Suppose
doctors test a group of elementary school children.
c) How many kids do they expect to test before finding one who has
this vitamin deficiency?
Geom(0.2)
𝟏
𝒑
E(X) = =
𝟏
𝟎.𝟐
= 5 children
Chapter 17
30. Vitamin D is essential for strong, healthy bones. Our bodies produce
vitamin D naturally when sunlight falls upon the skin, or it can be taken as a
dietary supplement. Although the bone disease rickets was largely eliminated
in England during the 1950s, some people there are concern that this
generation of children is at increased risk because they are more likely to
watch TV or play computer games than spend time outdoors. Recent research
indicated that about 20% of British children are deficient in vitamin D. Suppose
doctors test a group of elementary school children.
d) They will test 50 students at the third grade level. Find the mean
and standard deviation of the number who may be deficient in
vitamin D.
Binom(50, 0.2)
E(X) = np = 50(0.2) = 10 children
SD(X) = 𝒏𝒑𝒒 =
𝟓𝟎 𝟎. 𝟐 (𝟎. 𝟖) ≈2.83 children
Chapter 17
30. Vitamin D is essential for strong, healthy bones. Our bodies produce
vitamin D naturally when sunlight falls upon the skin, or it can be taken as a
dietary supplement. Although the bone disease rickets was largely eliminated
in England during the 1950s, some people there are concern that this
generation of children is at increased risk because they are more likely to
watch TV or play computer games than spend time outdoors. Recent research
indicated that about 20% of British children are deficient in vitamin D. Suppose
doctors test a group of elementary school children.
e) If they test 320 children at this school, what’s the probability that
no more than 50 of them have the vitamin deficiency?
Binom(320, 0.2)
P(X < 50) = 0.027
Chapter 17
30. Vitamin D is essential for strong, healthy bones. Our bodies produce
vitamin D naturally when sunlight falls upon the skin, or it can be taken as a
dietary supplement. Although the bone disease rickets was largely eliminated
in England during the 1950s, some people there are concern that this
generation of children is at increased risk because they are more likely to
watch TV or play computer games than spend time outdoors. Recent research
indicated that about 20% of British children are deficient in vitamin D. Suppose
doctors test a group of elementary school children.
e) If they test 320 children at this school, what’s the probability that
no more than 50 of them have the vitamin deficiency?
Binom(320, 0.2)
P(X < 50) = 0.027
Additional Examples
Which of the following are true statements?
I. The histogram of a binomial distribution with p = .5 is always
symmetric no matter what the value of n, the number of trials.
II. The histogram of a binomial distribution with p = .2 is skewed
to the left.
III. The histogram of a binomial distribution with p = .9 looks
more and more symmetric, the larger the value of n.
I and III
Additional Examples
Five managers and five employees are on a grievance
committee. A three-person subcommittee is formed by
a random selection from the tem committee members.
What is the probability that all three members of the
committee are managers?
Additional Examples
Among the 125 at a small college, 75 are registered
Democrats, 35 are registered Republicans, and the rest
are independents. If a 10-person committee is randomly
picked, what is the probability that at least two
independents are chosen.
Additional Examples
Among the 125 at a small college, 75 are registered
Democrats, 35 are registered Republicans, and the rest
are independents. If a 10-person committee is randomly
picked, what is the probability that at least two
independents are chosen.