Probability Theory: Foundation for Data Science Coursera Quiz Answers 2022 | All Weeks Assessment Answers [💯Correct Answer]

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About Probability Theory: Foundation for Data Science Course

Learn the principles underlying probability and how it relates to statistics and data analytics. We’ll discover the definition of probability calculation, independent and dependent outcomes, and conditional occurrences. We’ll investigate discrete and continuous random variables to discover how data collecting fits in. The Central Limit Theorem and Gaussian (normal) random variables will be the course’s final topics, and we will comprehend its critical significance for all of statistics and data science.

As a component of the Master of Science in Data Science (MS-DS) degree offered by CU Boulder on the Coursera platform, this course can be taken for academic credit. The MS-DS is a multidisciplinary degree that brings together academics from Applied Mathematics, Computer Science, Information Science, and other departments at CU Boulder. The MS-DS is best suited for people with a wide range of undergraduate education and/or professional experience in computer science, information science, mathematics, and statistics because it has performance-based admissions and no application process. Visit https://www.coursera.org/degrees/master-of-science-data-science-boulder to learn more about the MS-DS program.

Christopher Burns’ Unsplash shot was used as inspiration for the logo.

WHAT YOU WILL LEARN

  • Explain why probability is important to statistics and data science.
  • See the relationship between conditional and independent events in a statistical experiment.
  • Calculate the expectation and variance of several random variables and develop some intuition.

SKILLS YOU WILL GAIN

  • Probability
  • central limit theorem
  • continuous random variables
  • Bayes’ Theorem
  • discrete random variables

Course Apply Link – Probability Theory: Foundation for Data Science

Probability Theory: Foundation for Data Science Quiz Answers

Week 1 Quiz Answers

Quiz 1: Module 1 Quiz

Q1. Prompt 1: For each of the following scenarios, find the requested probability. Assume the sets AA, BB, and CC are events from the same sample space SS. (Hint: Venn diagrams may help with the visualization, although they are not required to answer the questions.)

If P(A) = .4P(A)=.4, P(B^c) = .7P(B
c
)=.7, and P(A ∩ B^c) = .2P(A∩B
c
)=.2, find P(A ∩ B)P(A∩B)

Enter answer here

Q2. Prompt 1: For each of the following scenarios, find the requested probability. Assume the sets AA, BB, and CC are events from the same sample space SS. (Hint: Venn diagrams may help with the visualization, although they are not required to answer the questions.)

If P(A) = 0.9P(A)=0.9 and P(B) = 0.9P(B)=0.9, what is the lower bound for P(A \cup B)P(A∪B).

Enter answer here

Q3. Prompt 2: Three popular options on a certain type of car are AA leather seats, BB a sunroof, and CC heated seats. In the past, P(A) = 0.55P(A)=0.55 (i.e. 55% of the customers have requested option A), P(B) = 0.45P(B)=0.45, P(C) = 0.4P(C)=0.4. Furthermore, P(A ∩ B) = 0.25P(A∩B)=0.25, P(A ∩ C) = 0.2P(A∩C)=0.2, P(B ∩ C) = 0.15P(B∩C)=0.15 and P(A ∩ B ∩ C) = 0.1P(A∩B∩C)=0.1.

Find the probability that a customer will ask for at least one of the three options.

Enter answer here

Q4. Prompt 2: Three popular options on a certain type of car are AA leather seats, BB a sunroof, and CC heated seats. In the past, P(A) = 0.55P(A)=0.55 (i.e. 55% of the customers have requested option A), P(B) = 0.45P(B)=0.45, P(C) = 0.4P(C)=0.4. Furthermore, P(A ∩ B) = 0.25P(A∩B)=0.25, P(A ∩ C) = 0.2P(A∩C)=0.2, P(B ∩ C) = 0.15P(B∩C)=0.15 and P(A ∩ B ∩ C) = 0.1P(A∩B∩C)=0.1.

Find the probability that a customer will not ask for any of these three options.

Enter answer here

Q5. Prompt 2: Three popular options on a certain type of car are AA leather seats, BB a sunroof, and CC heated seats. In the past, P(A) = 0.55P(A)=0.55 (i.e. 55% of the customers have requested option A), P(B) = 0.45P(B)=0.45, P(C) = 0.4P(C)=0.4. Furthermore, P(A ∩ B) = 0.25P(A∩B)=0.25, P(A ∩ C) = 0.2P(A∩C)=0.2, P(B ∩ C) = 0.15P(B∩C)=0.15 and P(A ∩ B ∩ C) = 0.1P(A∩B∩C)=0.1.

Find the probability that a customer will ask for heated leather seats but not a sunroof.

Enter answer here

Q6. Prompt 2: Three popular options on a certain type of car are AA leather seats, BB a sunroof, and CC heated seats. In the past, P(A) = 0.55P(A)=0.55 (i.e. 55% of the customers have requested option A), P(B) = 0.45P(B)=0.45, P(C) = 0.4P(C)=0.4. Furthermore, P(A ∩ B) = 0.25P(A∩B)=0.25, P(A ∩ C) = 0.2P(A∩C)=0.2, P(B ∩ C) = 0.15P(B∩C)=0.15 and P(A ∩ B ∩ C) = 0.1P(A∩B∩C)=0.1.

Find the probability that a customer will ask for at most two of the options.

Enter answer here

Q7. Prompt 2: Three popular options on a certain type of car are AA leather seats, BB a sunroof, and CC heated seats. In the past, P(A) = 0.55P(A)=0.55 (i.e. 55% of the customers have requested option A), P(B) = 0.45P(B)=0.45, P(C) = 0.4P(C)=0.4. Furthermore, P(A ∩ B) = 0.25P(A∩B)=0.25, P(A ∩ C) = 0.2P(A∩C)=0.2, P(B ∩ C) = 0.15P(B∩C)=0.15 and P(A ∩ B ∩ C) = 0.1P(A∩B∩C)=0.1.

Find the probability that a customer will ask for exactly two of the options.

Enter answer here

Q8. Prompt 3: A message of length 5 digits is to be sent. Each digit can be a 0, 1, or 2.

What is the cardinality of the sample space?

Enter answer here

Q9. Prompt 3: A message of length 5 digits is to be sent. Each digit can be a 0, 1, or 2.

If every message is equally likely, what is the probability that the message consists of 2 zeros, 2 ones, and 1 two? Round your answer to have four decimal places.

Enter answer here

Q10. Prompt 3: A message of length 5 digits is to be sent. Each digit can be a 0, 1, or 2.

What is the probability that the message contains at least one zero? Round your answer to have three decimal places.

Enter answer here

Week 2 Quiz Answers

Quiz 1: Module 2 Quiz

Q1. Prompt 1: For each of the following scenarios, answer the question with a numerical answer or a “Yes” or “No”. If you do not have enough information, write “Can’t tell”. For all situations, assume 0 < P(A) < 10<P(A)<1 and 0 < P(B) < 10<P(B)<1.

If AA and BB are two events such that P(A) = 0.7P(A)=0.7, P(B) = 0.5P(B)=0.5, and P(B|A) = 0.6P(B∣A)=0.6, find P(A ∪ B)P(A∪B). If possible, round your answer to two decimal places.

Enter answer here

Q2. Prompt 1: For each of the following scenarios, answer the question with a numerical answer or a “Yes” or “No”. If you do not have enough information, write “Can’t tell”. For all situations, assume 0 < P(A) < 10<P(A)<1 and 0 < P(B) < 10<P(B)<1.

If AA is a subset of BB, can AA and BB be independent? (Answer Yes, No, or Can’t tell.)

Enter answer here

Q3. Prompt 1: For each of the following scenarios, answer the question with a numerical answer or a “Yes” or “No”. If you do not have enough information, write “Can’t tell”. For all situations, assume 0 < P(A) < 10<P(A)<1 and 0 < P(B) < 10<P(B)<1.

If A is a subset of B, calculate P(B|A), if you can. (If you can’t, answer Can’t tell.)

Enter answer here

Q4. Prompt 1: For each of the following scenarios, answer the question with a numerical answer or a “Yes” or “No”. If you do not have enough information, write “Can’t tell”. For all situations, assume 0 < P(A) < 10<P(A)<1 and 0 < P(B) < 10<P(B)<1.

If A is a subset of B, calculate P(A|B), if you can. (If you can’t, answer Can’t tell.)

Enter answer here

Q5. Prompt 1: For each of the following scenarios, answer the question with a numerical answer or a “Yes” or “No”. If you do not have enough information, write “Can’t tell”. For all situations, assume 0 < P(A) < 10<P(A)<1 and 0 < P(B) < 10<P(B)<1.

If P(A) = 0.8, P(B|A) = 0.5 and A and B are dependent (not independent), then is P(B) ≤ P(A)? (Answer Yes, No, or Can’t tell)

Enter answer here

Q6. Prompt 2: An unfair coin is flipped four times. The probability of a heads is 0.6 and the probability of a tails is 0.4. Every flip is independent of every other flip.

Find the probability of getting exactly 2 heads and 2 tails. Round your answer to four decimal places.

Enter answer here

Q7. Prompt 2: An unfair coin is flipped four times. The probability of a heads is 0.6 and the probability of a tails is 0.4. Every flip is independent of every other flip.

Find the probability of getting exactly 2 heads and 2 tails given that the first coin flip was a head. Round your answer to three decimal places.

Enter answer here

Q8. Prompt 3: 70\%70% of the light aircraft that disappear while in flight in a certain country are subsequently found. Of the aircraft that are found, 60\%60% have an emergency locator, whereas only 10\%10% of the aircraft not found have such a locator. Suppose a light aircraft has disappeared.

If the aircraft has an emergency locator, what is the probability that it will not be found? Round your answer to three decimal places.

Enter answer here

Q9. Prompt 3: 70\%70% of the light aircraft that disappear while in flight in a certain country are subsequently found. Of the aircraft that are found, 60\%60% have an emergency locator, whereas only 10\%10% of the aircraft not found have such a locator. Suppose a light aircraft has disappeared.

Are the two events L=L=the plane has a locator and F=F=the plane is found independent?

  • Yes
  • No

Week 3 Quiz Answers

Quiz 1: Module 3 Quiz

Q1. Prompt 1: A critical system has 55 different components, each of which works with probability 0.900.90. Assume each component works independently of the others. Let XX be the number of components that are working.

Find P(X = 2)P(X=2). Round your answer to four decimal places.

Q2. Prompt 1: A critical system has 55 different components, each of which works with probability 0.900.90. Assume each component works independently of the others. Let XX be the number of components that are working.

Find E(X)E(X).

Q3. Prompt 1: A critical system has 55 different components, each of which works with probability 0.900.90. Assume each component works independently of the others. Let XX be the number of components that are working.

Find V(X)V(X). Round your answer to two decimal places.

Q4. Prompt 2: A certain type of item produced by a factory has a 6\%6% chance of being defective. Draw a random sample until you get the first defective. Let XX be the number of items that are drawn.

Find P(X = 2)P(X=2). Round your answer to four decimal places.

Q5. Prompt 2: A certain type of item produced by a factory has a 6\%6% chance of being defective. Draw a random sample until you get the first defective. Let XX be the number of items that are drawn.

Find E(X)E(X). Round answer to two decimal places.

Q6. Prompt 2: A certain type of item produced by a factory has a 6\%6% chance of being defective. Draw a random sample until you get the first defective. Let XX be the number of items that are drawn.

Find V(X)V(X). Round answer to two decimal places.

Q7. Prompt 3: A certain system can experience three different types of defects. Let A_iAi​, i = 1, 2, 3 be the event that the system has a defect of type ii. Suppose that P(A_1) = 0.17P(A1​)=0.17, P(A_2) = 0.07P(A2​)=0.07, P(A_3) = 0.13P(A3​)=0.13, P(A_1 ∪ A_2) =0.18P(A1​∪A2​)=0.18, P(A_2 ∪ A_3) = 0.18P(A2​∪A3​)=0.18, P(A_1 ∪ A_3) = 0.19P(A1​∪A3​)=0.19, and P(A_1 ∩ A_2 ∩ A_3) = 0.01P(A1​∩A2​∩A3​)=0.01. Let the random variable XX be the number of defects that are present. Calculate P(X=0)P(X=0).

Q8. Prompt 3: A certain system can experience three different types of defects. Let A_iAi​, i = 1, 2, 3i=1,2,3 be the event that the system has a defect of type ii. Suppose that P(A_1) = 0.17P(A1​)=0.17, P(A_2) = 0.07P(A2​)=0.07, P(A_3) = 0.13P(A3​)=0.13, P(A_1 ∪ A_2) =0.18P(A1​∪A2​)=0.18, P(A_2 ∪ A_3) = 0.18P(A2​∪A3​)=0.18, P(A_1 ∪ A_3) = 0.19P(A1​∪A3​)=0.19, and P(A_1 ∩ A_2 ∩ A_3) = 0.01P(A1​∩A2​∩A3​)=0.01. Let the random variable XX be the number of defects that are present.

Calculate P(X=1)P(X=1).

Q9. Prompt 3: A certain system can experience three different types of defects. Let A_iAi​, i = 1, 2, 3 be the event that the system has a defect of type i. Suppose that P(A_1) = .17P(A1​)=.17, P(A_2) = 0.07P(A2​)=0.07, P(A_3) = 0.13P(A3​)=0.13, P(A_1 ∪ A_2) =0.18P(A1​∪A2​)=0.18, P(A_2 ∪ A_3) = 0.18P(A2​∪A3​)=0.18, P(A_1 ∪ A_3) = 0.19P(A1​∪A3​)=0.19, and P(A_1 ∩ A_2 ∩ A_3) = .01P(A1​∩A2​∩A3​)=.01. Let the random variable XX be the number of defects that are present.

Calculate P(X=2)P(X=2)

Q10. Prompt 3: A certain system can experience three different types of defects. Let A_iAi​, i = 1, 2, 3i=1,2,3 be the event that the system has a defect of type ii. Suppose that P(A_1) = 0.17P(A1​)=0.17, P(A_2) = 0.07P(A2​)=0.07, P(A_3) = 0.13P(A3​)=0.13, P(A_1 ∪ A_2) =0.18P(A1​∪A2​)=0.18, P(A_2 ∪ A_3) = 0.18P(A2​∪A3​)=0.18, P(A_1 ∪ A_3) = 0.19P(A1​∪A3​)=0.19, and P(A_1 ∩ A_2 ∩ A_3) = 0.01P(A1​∩A2​∩A3​)=0.01. Let the random variable XX be the number of defects that are present.

Calculate P(X=3)P(X=3)

Q11. Prompt 3: A certain system can experience three different types of defects. Let A_iAi​, i = 1, 2, 3 be the event that the system has a defect of type i. Suppose that P(A_1) = .17P(A1​)=.17, P(A_2) = 0.07P(A2​)=0.07, P(A_3) = 0.13P(A3​)=0.13, P(A_1 ∪ A_2) =0.18P(A1​∪A2​)=0.18, P(A_2 ∪ A_3) = 0.18P(A2​∪A3​)=0.18, P(A_1 ∪ A_3) = 0.19P(A1​∪A3​)=0.19, and P(A_1 ∩ A_2 ∩ A_3) = .01P(A1​∩A2​∩A3​)=.01. Let the random variable XX be the number of defects that are present.

Find E(X)E(X)

Q12. Prompt 3: A certain system can experience three different types of defects. Let A_iAi​, i = 1, 2, 3 be the event that the system has a defect of type i. Suppose that P(A_1) = .17P(A1​)=.17, P(A_2) = 0.07P(A2​)=0.07, P(A_3) = 0.13P(A3​)=0.13, P(A_1 ∪ A_2) =0.18P(A1​∪A2​)=0.18, P(A_2 ∪ A_3) = 0.18P(A2​∪A3​)=0.18, P(A_1 ∪ A_3) = 0.19P(A1​∪A3​)=0.19, and P(A_1 ∩ A_2 ∩ A_3) = .01P(A1​∩A2​∩A3​)=.01. Let the random variable XX be the number of defects that are present.

Find V (X)V(X). Give four decimal places for your answer.

Q13. Prompt 3: A certain system can experience three different types of defects. Let A_iAi​, i = 1, 2, 3 be the event that the system has a defect of type i. Suppose that P(A_1) = .17P(A1​)=.17, P(A_2) = 0.07P(A2​)=0.07, P(A_3) = 0.13P(A3​)=0.13, P(A_1 ∪ A_2) =0.18P(A1​∪A2​)=0.18, P(A_2 ∪ A_3) = 0.18P(A2​∪A3​)=0.18, P(A_1 ∪ A_3) = 0.19P(A1​∪A3​)=0.19, and P(A_1 ∩ A_2 ∩ A_3) = .01P(A1​∩A2​∩A3​)=.01. Let the random variable XX be the number of defects that are present.

Find σσ, the standard deviation of XX. Round your answer to three decimal places.

Week 4 Quiz Answers

Quiz 1: Module 4 Quiz

Q1. Prompt 1: A skilled worked requires at least 1010 minutes, and no more than 2020 minutes, to complete a certain task. The completion time XX is a continuous random variable with density function f(x) = c/x^2f(x)=c/x2 for 10 ≤ x ≤ 2010≤x≤20 and f(x) = 0 f(x)=0 for all other values of xx.

What is the value of cc?

Q2. Prompt 1: A skilled worked requires at least 1010 minutes, and no more than 2020 minutes, to complete a certain task. The completion time XX is a continuous random variable with density function f(x) = c/x^2f(x)=c/x2 for 10 ≤ x ≤ 2010≤x≤20 and f(x) = 0 f(x)=0 for all other values of xx.

Find the probability that the worker completes the task in 15 minutes or less. Round your answer to three decimal places.

Q3. Prompt 1: A skilled worked requires at least 1010 minutes, and no more than 2020 minutes, to complete a certain task. The completion time XX is a continuous random variable with density function f(x) = c/x^2f(x)=c/x2 for 10 ≤ x ≤ 2010≤x≤20 and f(x) = 0 f(x)=0 for all other values of xx.

Find the expected time expected time for the worker to complete the task. Round your answer to three decimal places.

Q4. Prompt 2: The number of eggs laid on a tree leaf by an insect of a certain type is a Poisson random variable, XX, with parameter cc, where E(X) = cE(X)=c. However, such a random variable can only be observed if it is positive, since if it is 0, then we cannot know that such an insect was on the leaf. If we let YY denote the observed number of eggs, then P(Y = i) = P(X = i|X > 0)P(Y=i)=P(X=iX>0), where XX is Poisson with parameter cc.

Find E(Y )E(Y).

(Note 1: your answer will be a mathematical expression with cc. When you type in your answer, the preview box will display what the mathematical expression looks like.

Note 2: Coursera uses the capital letter “E” to represent the mathematical constant “e” equivalent to approximately 2.72.)

Preview will appear here…

Q5. Prompt 3: Suppose XX is a random variable X \sim N(12, 4)XN(12,4).

Find the probability that XX is within 1.51.5 standard deviations of the mean. Round your answer to four decimal places.

Q6. Prompt 3: Suppose XX is a random variable X \sim N(12, 4)XN(12,4).

Find kk such that P(X > k) = 0.10P(X>k)=0.10. Round your answer to two decimal places.

Week 6 Quiz Answers

Quiz 1: Module 5 Quiz

Q1. Prompt 1: A store orders 1010 units of a certain item. For each unit sold, it makes \$10$10. For each item that remains unsold, it loses \$5$5. Let XX be the number of items sold in a week. If the store knows that XX is a discrete uniform random variable, with XX taking values \{0, 1, \dots , 10\}{0,1,…,10}, find the expected profit.

Q2. Prompt 2: Suppose that the service time at a call center is exponentially distributed with the average call lasting 10 minutes. If 33 workers are each on a call, what is the probability that they have each completed their service in 1212 minutes? Assume that the length of each call is independent of every other call. Round answer to four decimal places.

y=1y = 4y = 16
x=10.200.250.05
x=50.100.150.25

Q3. Prompt 3: Suppose that XX and YY are random variables with the joint probability mass function above.

Find P(X = 1).P(X=1).

Q4. Prompt 3: Suppose that XX and YY are random variables with the joint probability mass function below.

y=1y = 4y = 16
x=10.200.250.05
x=50.100.150.25

Find P(X > Y)P(X>Y).

Q5. Prompt 3: Suppose that XX and YY are random variables with the joint probability mass function below.

y=1y = 4y = 16
x=10.200.250.05
x=50.100.150.25

Determine Cov(X, Y)Cov(X,Y).

Q6. Prompt 3: Suppose that XX and YY are random variables with the joint probability mass function below.

y=1y = 4y = 16
x=10.200.250.05
x=50.100.150.25

Find \rhoρ, the correlation coefficient of XX and YY. Round answer to four decimal places.

Q7. Prompt 3: Suppose that X and Y are random variables with the joint probability mass function above.

y=1y = 4y = 16
x=10.200.250.05
x=50.100.150.25

Are X and Y independent? (Answer “Yes”, “No”, or “Can’t determine”)

Week 6 Quiz Answers

Quiz 1: Module 6 Quiz

Q1. Prompt 1: Suppose 3030 numbers are selected at random from the interval [0, 1][0,1]. That is, X_i \sim U[0, 1]Xi​∼U[0,1] for i =1, 2, . . . 30i=1,2,…30. Given \bar{X} = (1/30) \sum_{i=1}^{30} X_i Xˉ=(1/30)∑i=130​Xi​ estimate P(0.5 \leq \bar{X} \leq 0.6)P(0.5≤Xˉ≤0.6). (Round your answer to three decimal places.)

Q2. Prompt 2: A company wants to compare the efficiency of two types of fuel for a car. 55 identical cars will be driven 500500 kilometers each, two with the first type of fuel and three with the second type. Let X_1X1​ and X_2X2​ be the observed fuel efficiency for the first type and Y_1Y1​, Y_2Y2​, and Y_3Y3​ be the efficiency for the second type. Suppose these variables are independent and X_i \sim N(20, 4)Xi​∼N(20,4) for i = 1, 2i=1,2 and Y_j \sim N(18, 9)Yj​∼N(18,9) for j=1, 2, 3j=1,2,3. Define a new random variable: W = (X_1 + X_2)/2 – (Y_1 + Y_2 + Y_3)/3W=(X1​+X2​)/2−(Y1​+Y2​+Y3​)/3.

Find E(W)E(W).

Q3. Prompt 2: A company wants to compare the efficiency of two types of fuel for a car. 55 identical cars will be driven 500500 kilometers each, two with the first type of fuel and three with the second type. Let X_1X1​ and X_2X2​ be the observed fuel efficiency for the first type and Y_1Y1​, Y_2Y2​, and Y_3Y3​ be the efficiency for the second type. Suppose these variables are independent and X_i \sim N(20, 4)Xi​∼N(20,4) for i = 1, 2i=1,2 and Y_j \sim N(18, 9)Yj​∼N(18,9) for j=1, 2, 3j=1,2,3. Define a new random variable: W = (X_1 + X_2)/2 – (Y_1 + Y_2 + Y_3)/3W=(X1​+X2​)/2−(Y1​+Y2​+Y3​)/3.

Find Var(W)Var(W).

Q4. Prompt 2: A company wants to compare the efficiency of two types of fuel for a car. 55 identical cars will be driven 500500 kilometers each, two with the first type of fuel and three with the second type. Let X_1X1​ and X_2X2​ be the observed fuel efficiency for the first type and Y_1Y1​, Y_2Y2​, and Y_3Y3​ be the efficiency for the second type. Suppose these variables are independent and X_i \sim N(20, 4)Xi​∼N(20,4) for i = 1, 2i=1,2 and Y_j \sim N(18, 9)Yj​∼N(18,9) for j=1, 2, 3j=1,2,3. Define a new random variable: W = (X_1 + X_2)/2 – (Y_1 + Y_2 + Y_3)/3W=(X1​+X2​)/2−(Y1​+Y2​+Y3​)/3.

Find P(W ≥ 0). P(W≥0). Round answer to four decimal places.

Conclusion

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500 thoughts on “Probability Theory: Foundation for Data Science Coursera Quiz Answers 2022 | All Weeks Assessment Answers [💯Correct Answer]”

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