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Probability Homework

Site: Saylor Academy
Course: BUS204: Business Statistics
Book: Probability Homework
Printed by: Guest user
Date: Tuesday, May 13, 2025, 8:11 PM

Description

Solve these problems, then check your answers against the given solutions.

Table of contents

Exercises

Exercise 1

Suppose that you have 8 cards. 5 are green and 3 are yellow. The 5 green cards are numbered 1, 2, 3, 4, and 5. The 3 yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.

  • G = card drawn is green
  • E = card drawn is even-numbered
  1. List the sample space.
  2. P(G) =
  3. P(G\E) =
  4. P(G AND E) =
  5. P(G OR E) =
  6. Are G and E mutually exclusive? Justify your answer numerically.

Exercise 2
An experiment consists of tossing a nickel, a dime and a quarter. Of interest is the side the coin lands on.

  1. List the sample space.
  2. Let A be the event that there are at least two tails. Find P(A).
  3. Let B be the event that the first and second tosses land on heads. Are the events A and B mutually exclusive? Explain your answer in 1 - 3 complete sentences, including justification.

Exercise 3
E and F mutually exclusive events. P (E) = 0.4; P (F) = 0.5. Find P (E \ F).


Exercise 4
U and V are mutually exclusive events. P (17) = 0.26; P (V) = 0.37. Find:

  1. P(U AND V) =
  2. P(U I V) =
  3. P(U OR V) =

Exercise 5
The following table identifies a group of children by one of four hair colors, and by type of hair.

Hair Type
 Brown Blond Black Red Totals
Wavy 20
15 3 43
Straight 80 15
12
Totals
20

215

Table 3.2

  1. Complete the table above.
  2. What is the probability that a randomly selected child will have wavy hair?
  3. What is the probability that a randomly selected child will have either brown or blond hair?
  4. What is the probability that a randomly selected child will have wavy brown hair?
  5. What is the probability that a randomly selected child will have red hair, given that he has straight hair?
  6. If B is the event of a child having brown hair, find the probability of the complement of B.
  7. In words, what does the complement of B represent?

Exercise 6
Approximately 249,000,000 people live in the United States. Of these people, 31,800,000 speak a language other than English at home. Of those who speak another language at home, over 50 percent speak Spanish. (Source: U.S. Bureau of the Census, 1990 Census)

Let: E = speak English at home; E' = speak another language at home; S = speak Spanish at home

Finish each probability statement by matching the correct answer.

Probability Statements Answers
a. P(E') =
i. 0.8723
b. P(E) =
ii. > 0.50
c. P(S) =
iii. 0.1277
d. P(S 1 E') =
iv. > 0.0639

Table 3.4


Exercise 7
Given events G and H: P(G) = 0.43 ; P(H) = 0.26 ; P(H and G) = 0.14

  1. Find P(H or G)
  2. Find the probability of the complement of event (H and G)
  3. Find the probability of the complement of event (H or G)

Exercise 8
United Blood Services is a blood bank that serves more than 500 hospitals in 18 states. According to their website, http://www.unitedbloodservices.org/humanbloodtypes.html, a person with type O blood and a negative Rh factor (Rh- ) can donate blood to any person with any blood type. Their data show that 43% of people have type O blood and 15% of people have Rh- factor; 52% of people have type O or Rh— factor.

  1. Find the probability that a person has both type O blood and the Rh- factor
  2. Find the probability that a person does NOT have both type O blood and the Rh- factor.

Exercise 9
At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose that 32% of courses have a research paper and a final exam. Let F be the event that a course has a final exam. Let R be the event that a course requires a research paper.

  1. Find the probability that a course has a final exam or a research project.
  2. Find the probability that a course has NEITHER of these two requirements.

Exercise 10
A college finds that 10% of students have taken a distance learning class and that 40% of students are part time students. Of the part time students, 20% have taken a distance learning class. Let D = event that a student takes a distance learning class and E = event that a student is a part time student

  1. Find P(D and E)
  2. Find P(E I D)
  3. Find P(D or E)
  4. Using an appropriate test, show whether D and E are independent.
  5. Using an appropriate test, show whether D and E are mutually exclusive. 

Source: Barbara Illowsky and Susan Dean, https://archive.org/details/CollaborativeStatisticsHomeworkBook/page/n48/mode/1up
Creative Commons License This work is licensed under a Creative Commons Attribution 4.0 License.

Solutions to Exercises


    1. {Gl, G2, G3, G4, G5, Yl, Y2, Y3}
    2. \(\dfrac{5}{8}\)
    3. \(\dfrac{2}{3}\)
    4. \(\dfrac{2}{8}\)
    5. \(\dfrac{6}{8}\)
    6. No


    1. {(HHH) , (HHT) , (HTH) , (HTT) , (THH) , (THT) , (TTH) , (TTT)}
    2. \(\dfrac{4}{8}\)
    3. Yes

  3. 0


    1. 0
    2. 0
    3. 0.63


    1. \(\dfrac{43}{215}\)
    2. \(\dfrac{120}{215}\)
    3. \(\dfrac{20}{215}\)
    4. \(\dfrac{12}{172}\)
    5. \(\dfrac{115}{215}\)


    1. iii
    2. i
    3. iv
    4. ii


    1.  P(H or G) = P(H) + P(G) - P(H and G) = 0.26 + 0.43 - 0.14 = 0.55
    2. P( NOT (H and G) ) = 1 - P(H and G) = 1 - 0.14 = 0.86
    3. P( NOT (H or G) ) = 1 - P(H or G) = 1 - 0.55 = 0.45


    1. P (Type O or Rh-) = P(Type O) + P(Rh-) - P(Type O and Rh-)

      0.52 = 0.43 + 0.15 - P(Type O and Rh-); solve to find P(Type O and Rh-) = 0.06

      6% of people have type O Rh— blood

    2. P( NOT (Type O and Rh-) ) = 1 - P(Type O and Rh-) = 1 - 0.06 = 0.94

      94% of people do not have type O Rh— blood


    1. P(R or F) = P(R) + P(F) - P(R and F) = 0.72 + 0.46 - 0.32 = 0.86
    2. P( Neither R nor F ) = 1 - P(R or F) = 1 - 0.86 = 0.14


    1. P(D and E) = P(D I E)P(E) = (0.20)(0.40) = 0.08
    2. P(E I D) = P(D and E) / P(D) = 0.08/0.10 = 0.80
    3. P(D or E) = P(D) + P(E) - P(D and E) = 0.10 + 0.40 - 0.08 = 0.42
    4. Not Independent: P(D I E) = 0.20 which does not equal P(D) = .10
    5. Not Mutually Exclusive: P(D and E) = 0.08 ; if they were mutually exclusive then we would need to have P(D and E) = 0, which is not true here.