The probability for selecting a single person.

Question

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Answer 1

Question

Chapter 4.3, Problem 55E

(a)

To determine

To find: The probability for selecting a single person.

(a)

Expert Solution

Answer to Problem 55E

Solution: Theprobability for selecting a single twitter user or personis 0.19.

Explanation of Solution

Given: According to Question 4.54, a random sample in which 19% are Twitter users who use Twitter to update their post or see another person's post.

Calculation: Consider the sample space (S) for choosing one person is,

S=[Y,N]

Here, Y represents the Twitter user in the sample and N represents those users who do not use Twitter.

The probabilities for S=[Y,N] are,

P(Y)=P(selecting a single person)=0.19

(b)

To determine

To find: The probability for selecting three people.

(b)

Expert Solution

Answer to Problem 55E

Solution: The probabilitiesare:P(TTT)=0.0069,P(TTTc)=P(TTcT)=P(TcTT)=0.0292P(TcTTc)=P(TcTcT)=P(TTcTc)=0.1247,P(TcTcTc)=0.5314

Explanation of Solution

Calculation: Consider the sample space (S) for choosing three person is,

S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc]

P(T)=0.19P(Tc)=1−0.19=0.81

The probabilities for sample space are,

P(TTT)=0.19×0.19×0.19=0.0069

P(TTTc)=P(TTcT)=P(TcTT)=0.19×0.19×(1−0.19)=0.0292

P(TcTTc)=P(TcTcT)=P(TTcTc)=0.19×(1−0.19)×(1−0.19)=0.1247

P(TcTcTc)=(1−0.19)×(1−0.19)×(1−0.19)=0.5314

(c)

To determine

To find: The probability for selecting a single person.

(c)

Expert Solution

Answer to Problem 55E

Solution: The possible values are 0, 1, 2, 3 and their probabilities are 0.0069, 0.0876, 0.3741 and 0.5314 respectively.

Explanation of Solution

Calculation: Consider the sample space (S) for choosing three person is,

S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc]

Here, T represents the Twitter user in the sample and T^c represents those users who do not use Twitter. Now, the sample space for the random variable X, which expresses the number of Twitter users in a sample size 3, is

X={0,1,2,3}

The probabilities for S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc] are,

P(T)=0.19P(Tc)=1−0.19=0.81

Now, the probabilities are,

P(X=3)=P(TTT)=0.0069

P(X=2)=P(TTTc)+P(TTcT)+P(TcTT)=0.0292+0.0292+0.0292=0.0876

P(X=1)=P(TcTTc)+P(TcTcT)+P(TTcTc)=0.1247+0.1247+0.1247=0.3741

P(X=0)=P(TcTcTc)=0.5314

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Answer 2

Question

Chapter 4.3, Problem 55E

(a)

To determine

To find: The probability for selecting a single person.

(a)

Expert Solution

Answer to Problem 55E

Solution: Theprobability for selecting a single twitter user or personis 0.19.

Explanation of Solution

Given: According to Question 4.54, a random sample in which 19% are Twitter users who use Twitter to update their post or see another person's post.

Calculation: Consider the sample space (S) for choosing one person is,

S=[Y,N]

Here, Y represents the Twitter user in the sample and N represents those users who do not use Twitter.

The probabilities for S=[Y,N] are,

P(Y)=P(selecting a single person)=0.19

(b)

To determine

To find: The probability for selecting three people.

(b)

Expert Solution

Answer to Problem 55E

Solution: The probabilitiesare:P(TTT)=0.0069,P(TTTc)=P(TTcT)=P(TcTT)=0.0292P(TcTTc)=P(TcTcT)=P(TTcTc)=0.1247,P(TcTcTc)=0.5314

Explanation of Solution

Calculation: Consider the sample space (S) for choosing three person is,

S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc]

P(T)=0.19P(Tc)=1−0.19=0.81

The probabilities for sample space are,

P(TTT)=0.19×0.19×0.19=0.0069

P(TTTc)=P(TTcT)=P(TcTT)=0.19×0.19×(1−0.19)=0.0292

P(TcTTc)=P(TcTcT)=P(TTcTc)=0.19×(1−0.19)×(1−0.19)=0.1247

P(TcTcTc)=(1−0.19)×(1−0.19)×(1−0.19)=0.5314

(c)

To determine

To find: The probability for selecting a single person.

(c)

Expert Solution

Answer to Problem 55E

Solution: The possible values are 0, 1, 2, 3 and their probabilities are 0.0069, 0.0876, 0.3741 and 0.5314 respectively.

Explanation of Solution

Calculation: Consider the sample space (S) for choosing three person is,

S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc]

Here, T represents the Twitter user in the sample and T^c represents those users who do not use Twitter. Now, the sample space for the random variable X, which expresses the number of Twitter users in a sample size 3, is

X={0,1,2,3}

The probabilities for S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc] are,

P(T)=0.19P(Tc)=1−0.19=0.81

Now, the probabilities are,

P(X=3)=P(TTT)=0.0069

P(X=2)=P(TTTc)+P(TTcT)+P(TcTT)=0.0292+0.0292+0.0292=0.0876

P(X=1)=P(TcTTc)+P(TcTcT)+P(TTcTc)=0.1247+0.1247+0.1247=0.3741

P(X=0)=P(TcTcTc)=0.5314

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Answer 3

Question

Chapter 4.3, Problem 55E

(a)

To determine

To find: The probability for selecting a single person.

(a)

Expert Solution

Answer to Problem 55E

Solution: Theprobability for selecting a single twitter user or personis 0.19.

Explanation of Solution

Given: According to Question 4.54, a random sample in which 19% are Twitter users who use Twitter to update their post or see another person's post.

Calculation: Consider the sample space (S) for choosing one person is,

S=[Y,N]

Here, Y represents the Twitter user in the sample and N represents those users who do not use Twitter.

The probabilities for S=[Y,N] are,

P(Y)=P(selecting a single person)=0.19

(b)

To determine

To find: The probability for selecting three people.

(b)

Expert Solution

Answer to Problem 55E

Solution: The probabilitiesare:P(TTT)=0.0069,P(TTTc)=P(TTcT)=P(TcTT)=0.0292P(TcTTc)=P(TcTcT)=P(TTcTc)=0.1247,P(TcTcTc)=0.5314

Explanation of Solution

Calculation: Consider the sample space (S) for choosing three person is,

S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc]

P(T)=0.19P(Tc)=1−0.19=0.81

The probabilities for sample space are,

P(TTT)=0.19×0.19×0.19=0.0069

P(TTTc)=P(TTcT)=P(TcTT)=0.19×0.19×(1−0.19)=0.0292

P(TcTTc)=P(TcTcT)=P(TTcTc)=0.19×(1−0.19)×(1−0.19)=0.1247

P(TcTcTc)=(1−0.19)×(1−0.19)×(1−0.19)=0.5314

(c)

To determine

To find: The probability for selecting a single person.

(c)

Expert Solution

Answer to Problem 55E

Solution: The possible values are 0, 1, 2, 3 and their probabilities are 0.0069, 0.0876, 0.3741 and 0.5314 respectively.

Explanation of Solution

Calculation: Consider the sample space (S) for choosing three person is,

S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc]

Here, T represents the Twitter user in the sample and T^c represents those users who do not use Twitter. Now, the sample space for the random variable X, which expresses the number of Twitter users in a sample size 3, is

X={0,1,2,3}

The probabilities for S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc] are,

P(T)=0.19P(Tc)=1−0.19=0.81

Now, the probabilities are,

P(X=3)=P(TTT)=0.0069

P(X=2)=P(TTTc)+P(TTcT)+P(TcTT)=0.0292+0.0292+0.0292=0.0876

P(X=1)=P(TcTTc)+P(TcTcT)+P(TTcTc)=0.1247+0.1247+0.1247=0.3741

P(X=0)=P(TcTcTc)=0.5314

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Answer 4

Question

Chapter 4.3, Problem 55E

(a)

To determine

To find: The probability for selecting a single person.

(a)

Expert Solution

Answer to Problem 55E

Solution: Theprobability for selecting a single twitter user or personis 0.19.

Explanation of Solution

Given: According to Question 4.54, a random sample in which 19% are Twitter users who use Twitter to update their post or see another person's post.

Calculation: Consider the sample space (S) for choosing one person is,

S=[Y,N]

Here, Y represents the Twitter user in the sample and N represents those users who do not use Twitter.

The probabilities for S=[Y,N] are,

P(Y)=P(selecting a single person)=0.19

(b)

To determine

To find: The probability for selecting three people.

(b)

Expert Solution

Answer to Problem 55E

Solution: The probabilitiesare:P(TTT)=0.0069,P(TTTc)=P(TTcT)=P(TcTT)=0.0292P(TcTTc)=P(TcTcT)=P(TTcTc)=0.1247,P(TcTcTc)=0.5314

Explanation of Solution

Calculation: Consider the sample space (S) for choosing three person is,

S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc]

P(T)=0.19P(Tc)=1−0.19=0.81

The probabilities for sample space are,

P(TTT)=0.19×0.19×0.19=0.0069

P(TTTc)=P(TTcT)=P(TcTT)=0.19×0.19×(1−0.19)=0.0292

P(TcTTc)=P(TcTcT)=P(TTcTc)=0.19×(1−0.19)×(1−0.19)=0.1247

P(TcTcTc)=(1−0.19)×(1−0.19)×(1−0.19)=0.5314

(c)

To determine

To find: The probability for selecting a single person.

(c)

Expert Solution

Answer to Problem 55E

Solution: The possible values are 0, 1, 2, 3 and their probabilities are 0.0069, 0.0876, 0.3741 and 0.5314 respectively.

Explanation of Solution

Calculation: Consider the sample space (S) for choosing three person is,

S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc]

Here, T represents the Twitter user in the sample and T^c represents those users who do not use Twitter. Now, the sample space for the random variable X, which expresses the number of Twitter users in a sample size 3, is

X={0,1,2,3}

The probabilities for S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc] are,

P(T)=0.19P(Tc)=1−0.19=0.81

Now, the probabilities are,

P(X=3)=P(TTT)=0.0069

P(X=2)=P(TTTc)+P(TTcT)+P(TcTT)=0.0292+0.0292+0.0292=0.0876

P(X=1)=P(TcTTc)+P(TcTcT)+P(TTcTc)=0.1247+0.1247+0.1247=0.3741

P(X=0)=P(TcTcTc)=0.5314

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Answer 5

Chapter 4.3, Problem 55E

(a)

To determine

To find: The probability for selecting a single person.

(a)

Expert Solution

Answer to Problem 55E

Solution: Theprobability for selecting a single twitter user or personis 0.19.

Explanation of Solution

Given: According to Question 4.54, a random sample in which 19% are Twitter users who use Twitter to update their post or see another person's post.

Calculation: Consider the sample space (S) for choosing one person is,

S=[Y,N]

Here, Y represents the Twitter user in the sample and N represents those users who do not use Twitter.

The probabilities for S=[Y,N] are,

P(Y)=P(selecting a single person)=0.19

(b)

To determine

To find: The probability for selecting three people.

(b)

Expert Solution

Answer to Problem 55E

Solution: The probabilitiesare:P(TTT)=0.0069,P(TTTc)=P(TTcT)=P(TcTT)=0.0292P(TcTTc)=P(TcTcT)=P(TTcTc)=0.1247,P(TcTcTc)=0.5314

Explanation of Solution

Calculation: Consider the sample space (S) for choosing three person is,

S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc]

P(T)=0.19P(Tc)=1−0.19=0.81

The probabilities for sample space are,

P(TTT)=0.19×0.19×0.19=0.0069

P(TTTc)=P(TTcT)=P(TcTT)=0.19×0.19×(1−0.19)=0.0292

P(TcTTc)=P(TcTcT)=P(TTcTc)=0.19×(1−0.19)×(1−0.19)=0.1247

P(TcTcTc)=(1−0.19)×(1−0.19)×(1−0.19)=0.5314

(c)

To determine

To find: The probability for selecting a single person.

(c)

Expert Solution

Answer to Problem 55E

Solution: The possible values are 0, 1, 2, 3 and their probabilities are 0.0069, 0.0876, 0.3741 and 0.5314 respectively.

Explanation of Solution

Calculation: Consider the sample space (S) for choosing three person is,

S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc]

Here, T represents the Twitter user in the sample and T^c represents those users who do not use Twitter. Now, the sample space for the random variable X, which expresses the number of Twitter users in a sample size 3, is

X={0,1,2,3}

The probabilities for S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc] are,

P(T)=0.19P(Tc)=1−0.19=0.81

Now, the probabilities are,

P(X=3)=P(TTT)=0.0069

P(X=2)=P(TTTc)+P(TTcT)+P(TcTT)=0.0292+0.0292+0.0292=0.0876

P(X=1)=P(TcTTc)+P(TcTcT)+P(TTcTc)=0.1247+0.1247+0.1247=0.3741

P(X=0)=P(TcTcTc)=0.5314

Answer 6

Chapter 4.3, Problem 55E

(a)

To determine

To find: The probability for selecting a single person.

(a)

Expert Solution

Answer to Problem 55E

Solution: Theprobability for selecting a single twitter user or personis 0.19.

Explanation of Solution

Given: According to Question 4.54, a random sample in which 19% are Twitter users who use Twitter to update their post or see another person's post.

Calculation: Consider the sample space (S) for choosing one person is,

S=[Y,N]

Here, Y represents the Twitter user in the sample and N represents those users who do not use Twitter.

The probabilities for S=[Y,N] are,

P(Y)=P(selecting a single person)=0.19

Answer 7

Chapter 4.3, Problem 55E

(a)

To determine

To find: The probability for selecting a single person.

Answer 8

(a)

Expert Solution

Answer to Problem 55E

Solution: Theprobability for selecting a single twitter user or personis 0.19.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given: According to Question 4.54, a random sample in which 19% are Twitter users who use Twitter to update their post or see another person's post.

Calculation: Consider the sample space (S) for choosing one person is,

S=[Y,N]

Here, Y represents the Twitter user in the sample and N represents those users who do not use Twitter.

The probabilities for S=[Y,N] are,

P(Y)=P(selecting a single person)=0.19

Answer 13

(b)

To determine

To find: The probability for selecting three people.

(b)

Expert Solution

Answer to Problem 55E

Solution: The probabilitiesare:P(TTT)=0.0069,P(TTTc)=P(TTcT)=P(TcTT)=0.0292P(TcTTc)=P(TcTcT)=P(TTcTc)=0.1247,P(TcTcTc)=0.5314

Explanation of Solution

Calculation: Consider the sample space (S) for choosing three person is,

S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc]

P(T)=0.19P(Tc)=1−0.19=0.81

The probabilities for sample space are,

P(TTT)=0.19×0.19×0.19=0.0069

P(TTTc)=P(TTcT)=P(TcTT)=0.19×0.19×(1−0.19)=0.0292

P(TcTTc)=P(TcTcT)=P(TTcTc)=0.19×(1−0.19)×(1−0.19)=0.1247

P(TcTcTc)=(1−0.19)×(1−0.19)×(1−0.19)=0.5314

Answer 14

(b)

To determine

To find: The probability for selecting three people.

Answer 15

(b)

Expert Solution

Answer to Problem 55E

Solution: The probabilitiesare:P(TTT)=0.0069,P(TTTc)=P(TTcT)=P(TcTT)=0.0292P(TcTTc)=P(TcTcT)=P(TTcTc)=0.1247,P(TcTcTc)=0.5314

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation: Consider the sample space (S) for choosing three person is,

S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc]

P(T)=0.19P(Tc)=1−0.19=0.81

The probabilities for sample space are,

P(TTT)=0.19×0.19×0.19=0.0069

P(TTTc)=P(TTcT)=P(TcTT)=0.19×0.19×(1−0.19)=0.0292

P(TcTTc)=P(TcTcT)=P(TTcTc)=0.19×(1−0.19)×(1−0.19)=0.1247

P(TcTcTc)=(1−0.19)×(1−0.19)×(1−0.19)=0.5314

Answer 20

(c)

To determine

To find: The probability for selecting a single person.

(c)

Expert Solution

Answer to Problem 55E

Solution: The possible values are 0, 1, 2, 3 and their probabilities are 0.0069, 0.0876, 0.3741 and 0.5314 respectively.

Explanation of Solution

Calculation: Consider the sample space (S) for choosing three person is,

S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc]

Here, T represents the Twitter user in the sample and T^c represents those users who do not use Twitter. Now, the sample space for the random variable X, which expresses the number of Twitter users in a sample size 3, is

X={0,1,2,3}

The probabilities for S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc] are,

P(T)=0.19P(Tc)=1−0.19=0.81

Now, the probabilities are,

P(X=3)=P(TTT)=0.0069

P(X=2)=P(TTTc)+P(TTcT)+P(TcTT)=0.0292+0.0292+0.0292=0.0876

P(X=1)=P(TcTTc)+P(TcTcT)+P(TTcTc)=0.1247+0.1247+0.1247=0.3741

P(X=0)=P(TcTcTc)=0.5314

Answer 21

(c)

To determine

To find: The probability for selecting a single person.

Answer 22

(c)

Expert Solution

Answer to Problem 55E

Solution: The possible values are 0, 1, 2, 3 and their probabilities are 0.0069, 0.0876, 0.3741 and 0.5314 respectively.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation: Consider the sample space (S) for choosing three person is,

S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc]

Here, T represents the Twitter user in the sample and T^c represents those users who do not use Twitter. Now, the sample space for the random variable X, which expresses the number of Twitter users in a sample size 3, is

X={0,1,2,3}

The probabilities for S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc] are,