## Given Information - Probability that A speaks the truth, \( P(A) = .75 \) - Probability that B speaks the truth, \( P(B) = .60 \) - Both A and B independently state that a certain event occurred. - We are to find the probability that the event actually occurred, given that both said it occurred. --- ## Definitions and Concepts Used - **Conditional Probability:** The probability of an event \( E \) given another event \( F \) is \( P(E|F) = \frac{P(E \cap F)}{P(F)} \). - **Bayes' Theorem:** Used to find the probability of a cause given an effect. - **Independence:** The truthfulness of A and B are independent. Let \( E \) be the event that the event actually occurred. --- ## Solution Let \( S \) be the event that both A and B say the event occurred. If the event actually occurred (\( E \)), - Probability A says it occurred = .75 - Probability B says it occurred = .60 - Probability both say it occurred = \( .75 \times .60 = .45 \) If the event did **not** occur (\( \overline{E} \)), - Probability A says it occurred (i.e., lies) = \( 1 - .75 = .25 \) - Probability B says it occurred (i.e., lies) = \( 1 - .60 = .40 \) - Probability both say it occurred = \( .25 \times .40 = .10 \) Assume the event is equally likely to have occurred or not, so \( P(E) = P(\overline{E}) = .5 \). The total probability that both A and B say the event occurred: \[ P(S) = P(E) \cdot P(S|E) + P(\overline{E}) \cdot P(S|\overline{E}) \] \[ P(S) = .5 \times .45 + .5 \times .10 = .225 + .05 = .275 \] The probability that the event actually occurred, given both say it occurred: \[ P(E|S) = \frac{P(E) \cdot P(S|E)}{P(S)} = \frac{.5 \times .45}{.275} = \frac{.225}{.275} = \frac{45}{55} = \frac{9}{11} \] --- ## Final Answer The probability that the event actually occurred, given both A and B independently say it occurred, is \[ \boxed{\frac{9}{11}} \]

## Given Information - Probability that A speaks the truth, \( P(A) = 0.75 \) - Probability that B speaks the truth, \( P(B) = 0.60 \) - Both A and B independently state that a certain event occurred. - We need to find the probability that the event actually occurred, given that both said it occurred. --- ## Definitions and Concepts Used - **Conditional Probability:** The probability of an event \( E \) given another event \( F \) is \( P(E|F) = \frac{P(E \cap F)}{P(F)} \). - **Bayes' Theorem:** To find the probability of a cause given an effect. - **Independence:** The truthfulness of A and B are independent. Let \( E \) be the event that the event actually occurred. --- ## Solution Let \( S \) be the event that both A and B say the event occurred. ### Case 1: Event Actually Occurred (\( E \)) - Probability A says it occurred: \( P(S|E) = 0.75 \) - Probability B says it occurred: \( P(S|E) = 0.60 \) - Probability both say it occurred: \[ P(S|E) = P(A) \times P(B) = 0.75 \times 0.60 = 0.45 \] ### Case 2: Event Did Not Occur (\( \overline{E} \)) - Probability A says it occurred (lying): \( P(S|\overline{E}) = 1 - P(A) = 0.25 \) - Probability B says it occurred (lying): \( P(S|\overline{E}) = 1 - P(B) = 0.40 \) - Probability both say it occurred: \[ P(S|\overline{E}) = 0.25 \times 0.40 = 0.10 \] ### Total Probability of Both Saying the Event Occurred Assuming the event is equally likely to have occurred or not, \[ P(E) = P(\overline{E}) = 0.5 \] Calculate the total probability \( P(S) \): \[ P(S) = P(E) \cdot P(S|E) + P(\overline{E}) \cdot P(S|\overline{E}) \] \[ P(S) = 0.5 \times 0.45 + 0.5 \times 0.10 = 0.225 + 0.05 = 0.275 \] ### Probability that the Event Actually Occurred Given Both Say It Occurred Using Bayes' Theorem: \[ P(E|S) = \frac{P(E) \cdot P(S|E)}{P(S)} = \frac{0.5 \times 0.45}{0.275} \] Calculating the numerator: \[ 0.5 \times 0.45 = 0.225 \] Thus, \[ P(E|S) = \frac{0.225}{0.275} \approx 0.8181818181818182 \] Converting to four decimal places: \[ P(E|S) \approx 0.8182 \] --- ## Final Answer The probability that the event actually occurred, given both A and B independently say it occurred, is \[ \boxed{0.8182} \]

## Given Information - The probabilities of heads for the three coins are: - Coin 1: \( P(H|C_1) = 0.3 \) - Coin 2: \( P(H|C_2) = 0.5 \) - Coin 3: \( P(H|C_3) = 0.7 \) - A coin is chosen at random and tossed twice. - We need to find the probability that the chosen coin was Coin 3 given that two heads appeared. --- ## Definitions and Concepts Used - **Conditional Probability:** \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] - **Bayes' Theorem:** \[ P(C_i|H) = \frac{P(H|C_i) \cdot P(C_i)}{P(H)} \] - **Independence of Tosses:** The outcomes of the two tosses are independent. Let \( H \) be the event of getting two heads. --- ## Solution ### Step 1: Calculate \( P(H|C_i) \) for Each Coin - For Coin 1: \[ P(H|C_1) = P(H|C_1) \times P(H|C_1) = 0.3 \times 0.3 = 0.09 \] - For Coin 2: \[ P(H|C_2) = P(H|C_2) \times P(H|C_2) = 0.5 \times 0.5 = 0.25 \] - For Coin 3: \[ P(H|C_3) = P(H|C_3) \times P(H|C_3) = 0.7 \times 0.7 = 0.49 \] ### Step 2: Total Probability of Getting Two Heads \( P(H) \) Assuming each coin has an equal probability of being chosen: \[ P(C_1) = P(C_2) = P(C_3) = \frac{1}{3} \] Now calculate \( P(H) \): \[ P(H) = P(H|C_1) \cdot P(C_1) + P(H|C_2) \cdot P(C_2) + P(H|C_3) \cdot P(C_3) \] \[ P(H) = (0.09 \times \frac{1}{3}) + (0.25 \times \frac{1}{3}) + (0.49 \times \frac{1}{3}) = \frac{0.09 + 0.25 + 0.49}{3} = \frac{0.83}{3} = 0.27666666667 \] ### Step 3: Calculate \( P(C_3|H) \) Using Bayes' Theorem: \[ P(C_3|H) = \frac{P(H|C_3) \cdot P(C_3)}{P(H)} \] Substituting the values: \[ P(C_3|H) = \frac{0.49 \times \frac{1}{3}}{0.27666666667} \] Calculating the numerator: \[ 0.49 \times \frac{1}{3} = 0.16333333333 \] Thus, \[ P(C_3|H) = \frac{0.16333333333}{0.27666666667} \approx 0.5892857142857143 \] ### Final Answer The probability that the chosen coin was Coin 3 given that two heads appeared is \[ \boxed{0.5893} \]

## Given Information - The probabilities of heads for the three coins are: - Coin 1: \( P(H|C_1) = 0.3 \) - Coin 2: \( P(H|C_2) = 0.5 \) - Coin 3: \( P(H|C_3) = 0.7 \) - A coin is chosen at random and tossed twice. - We need to find the probability that the chosen coin was Coin 3 given that two heads appeared. --- ## Definitions and Concepts Used - **Conditional Probability:** This is the probability of an event occurring given that another event has already occurred. Mathematically, it is defined as: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] - **Bayes' Theorem:** This theorem allows us to update the probability estimate for an event based on new evidence. It is expressed as: \[ P(C_i|H) = \frac{P(H|C_i) \cdot P(C_i)}{P(H)} \] - **Independence of Events:** In this context, the outcomes of the two tosses are independent, meaning the result of one toss does not affect the other. Let \( H \) represent the event of getting two heads, and \( C_i \) represent the event of choosing coin \( i \). --- ## Solution Calculate the probability of getting two heads for each coin. For Coin 1: \[ P(H|C_1) = P(H|C_1) \times P(H|C_1) = 0.3 \times 0.3 = 0.09 \] For Coin 2: \[ P(H|C_2) = P(H|C_2) \times P(H|C_2) = 0.5 \times 0.5 = 0.25 \] For Coin 3: \[ P(H|C_3) = P(H|C_3) \times P(H|C_3) = 0.7 \times 0.7 = 0.49 \] Next, determine the total probability of getting two heads, \( P(H) \). Assuming each coin has an equal chance of being chosen, we have: \[ P(C_1) = P(C_2) = P(C_3) = \frac{1}{3} \] Now calculate \( P(H) \): \[ P(H) = P(H|C_1) \cdot P(C_1) + P(H|C_2) \cdot P(C_2) + P(H|C_3) \cdot P(C_3) \] \[ P(H) = (0.09 \times \frac{1}{3}) + (0.25 \times \frac{1}{3}) + (0.49 \times \frac{1}{3}) = \frac{0.09 + 0.25 + 0.49}{3} = \frac{0.83}{3} \approx 0.2767 \] Now we can find \( P(C_3|H) \) using Bayes' Theorem: \[ P(C_3|H) = \frac{P(H|C_3) \cdot P(C_3)}{P(H)} \] Substituting the values: \[ P(C_3|H) = \frac{0.49 \times \frac{1}{3}}{0.2767} = \frac{0.1633}{0.2767} \approx 0.5895 \] --- ## Final Answer The probability that the chosen coin was Coin 3 given that two heads appeared is \[ \boxed{0.5895} \]

## Given Information - The probabilities of heads for the three coins are: - Coin 1: \( P(H|C_1) = 0.3 \) - Coin 2: \( P(H|C_2) = 0.5 \) - Coin 3: \( P(H|C_3) = 0.7 \) - A coin is selected at random and flipped twice. - We want to find the probability that the selected coin was Coin 3 given that two heads were observed. --- ## Definitions and Concepts Used - **Conditional Probability:** This refers to the probability of an event occurring given that another event has already taken place. It can be expressed mathematically as: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] - **Bayes' Theorem:** This theorem provides a way to update the probability estimate for an event based on new evidence. It is mathematically represented as: \[ P(C_i|H) = \frac{P(H|C_i) \cdot P(C_i)}{P(H)} \] - **Independence of Events:** This concept implies that the outcome of one event does not affect the outcome of another. In this scenario, the results of the two flips are independent. Let \( H \) denote the occurrence of getting two heads, and let \( C_i \) denote the event of choosing coin \( i \). --- ## Solution Calculate the probability of obtaining two heads for each coin. For Coin 1: \[ P(H|C_1) = P(H|C_1) \times P(H|C_1) = 0.3 \times 0.3 = 0.09 \] For Coin 2: \[ P(H|C_2) = P(H|C_2) \times P(H|C_2) = 0.5 \times 0.5 = 0.25 \] For Coin 3: \[ P(H|C_3) = P(H|C_3) \times P(H|C_3) = 0.7 \times 0.7 = 0.49 \] Next, calculate the overall probability of getting two heads, \( P(H) \). Assuming each coin has an equal likelihood of being selected: \[ P(C_1) = P(C_2) = P(C_3) = \frac{1}{3} \] Now compute \( P(H) \): \[ P(H) = P(H|C_1) \cdot P(C_1) + P(H|C_2) \cdot P(C_2) + P(H|C_3) \cdot P(C_3) \] \[ P(H) = (0.09 \times \frac{1}{3}) + (0.25 \times \frac{1}{3}) + (0.49 \times \frac{1}{3}) = \frac{0.09 + 0.25 + 0.49}{3} = \frac{0.83}{3} \approx 0.2767 \] Next, determine \( P(C_3|H) \) using Bayes' Theorem: \[ P(C_3|H) = \frac{P(H|C_3) \cdot P(C_3)}{P(H)} \] Substituting the known values: \[ P(C_3|H) = \frac{0.49 \times \frac{1}{3}}{0.2767} = \frac{0.1633}{0.2767} \approx 0.5895 \] --- ## Final Answer The probability that the chosen coin was Coin 3, given that two heads were observed, is \[ \boxed{0.5895} \]

## Given Information - The probabilities of heads for the three coins are: - Coin 1: \( P(H|C_1) = 0.3 \) - Coin 2: \( P(H|C_2) = 0.5 \) - Coin 3: \( P(H|C_3) = 0.7 \) - A coin is randomly selected and flipped twice. - We seek the probability that the selected coin was Coin 3, given that two heads were observed. --- ## Definitions and Concepts Used - **Conditional Probability:** This measures the likelihood of an event occurring given that another event has already occurred. It is expressed as: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] - **Bayes' Theorem:** This theorem allows us to update the probability of a hypothesis based on new evidence. It is represented as: \[ P(C_i|H) = \frac{P(H|C_i) \cdot P(C_i)}{P(H)} \] - **Independence of Events:** This indicates that the outcome of one event does not influence the outcome of another. In this case, the two flips are independent. Let \( H \) represent the event of obtaining two heads, and let \( C_i \) denote the selection of coin \( i \). --- ## Solution First, calculate the probability of achieving two heads for each of the coins. For Coin 1: \[ P(H|C_1) = P(H|C_1) \times P(H|C_1) = 0.3 \times 0.3 = 0.09 \] For Coin 2: \[ P(H|C_2) = P(H|C_2) \times P(H|C_2) = 0.5 \times 0.5 = 0.25 \] For Coin 3: \[ P(H|C_3) = P(H|C_3) \times P(H|C_3) = 0.7 \times 0.7 = 0.49 \] Next, calculate the total probability of getting two heads, \( P(H) \). Assuming each coin is equally likely to be chosen: \[ P(C_1) = P(C_2) = P(C_3) = \frac{1}{3} \] Now compute \( P(H) \): \[ P(H) = P(H|C_1) \cdot P(C_1) + P(H|C_2) \cdot P(C_2) + P(H|C_3) \cdot P(C_3) \] \[ P(H) = (0.09 \times \frac{1}{3}) + (0.25 \times \frac{1}{3}) + (0.49 \times \frac{1}{3}) = \frac{0.09 + 0.25 + 0.49}{3} = \frac{0.83}{3} \approx 0.2767 \] Now apply Bayes' Theorem to find \( P(C_3|H) \): \[ P(C_3|H) = \frac{P(H|C_3) \cdot P(C_3)}{P(H)} \] Substituting the values: \[ P(C_3|H) = \frac{0.49 \times \frac{1}{3}}{0.2767} = \frac{0.1633}{0.2767} \approx 0.5895 \] --- ## Final Answer The probability that the selected coin was Coin 3, given that two heads were flipped, is \[ \boxed{0.5895} \]

## Given Information - Probability of event A: \( P(A) = 0.6 \) - Probability of event B: \( P(B) = 0.4 \) - Events A and B are independent. --- ## Definitions and Concepts Used - **Intersection of Events:** The probability that both events A and B occur is denoted as \( P(A \cap B) \). For independent events, this can be calculated as: \[ P(A \cap B) = P(A) \cdot P(B) \] - **Union of Events:** The probability that at least one of the events A or B occurs is denoted as \( P(A \cup B) \). For any two events, it is calculated as: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] --- ## Solution ### Calculate \( P(A \cap B) \) Using the formula for the intersection of independent events: \[ P(A \cap B) = P(A) \cdot P(B) = 0.6 \times 0.4 = 0.24 \] ### Calculate \( P(A \cup B) \) First, find \( P(A \cup B) \) using the union formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substituting the known values: \[ P(A \cup B) = 0.6 + 0.4 - 0.24 = 1.0 - 0.24 = 0.76 \] --- ## Final Answers - Probability that both events A and B occur: \[ P(A \cap B) = \boxed{0.24} \] - Probability that at least one of the events A or B occurs: \[ P(A \cup B) = \boxed{0.76} \]

## Given Information - Probability of event \( A \): \( P(A) = 0.6 \) - Probability of event \( B \): \( P(B) = 0.4 \) - Events \( A \) and \( B \) are independent. --- ## Definitions and Concepts Used - **Intersection of Events:** The probability that both events \( A \) and \( B \) occur simultaneously, denoted \( P(A \cap B) \). For independent events, this is calculated as: \[ P(A \cap B) = P(A) \times P(B) \] - **Union of Events:** The probability that at least one of the events \( A \) or \( B \) occurs, denoted \( P(A \cup B) \). This is calculated using the formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] --- ## Solution ### Calculate \( P(A \cap B) \) Using the independence of events \( A \) and \( B \): \[ P(A \cap B) = P(A) \times P(B) = 0.6 \times 0.4 = 0.24 \] ### Calculate \( P(A \cup B) \) Now, using the union formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substituting the known values: \[ P(A \cup B) = 0.6 + 0.4 - 0.24 = 1.0 - 0.24 = 0.76 \] --- ## Final Answers - Probability that both events \( A \) and \( B \) occur: \[ P(A \cap B) = \boxed{0.24} \] - Probability that at least one of the events \( A \) or \( B \) occurs: \[ P(A \cup B) = \boxed{0.76} \]

## Given Information - Probability of event \( A \): \( P(A) = 0.6 \) - Probability of event \( B \): \( P(B) = 0.4 \) - Events \( A \) and \( B \) are independent. --- ## Definitions and Concepts Used - **Intersection of Events:** The probability that both events \( A \) and \( B \) happen at the same time, denoted as \( P(A \cap B) \). When events are independent, this is calculated as: \[ P(A \cap B) = P(A) \cdot P(B) \] - **Union of Events:** The probability that either event \( A \) or event \( B \) occurs, denoted as \( P(A \cup B) \). This is calculated using the formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] --- ## Solution ### Calculate \( P(A \cap B) \) Since \( A \) and \( B \) are independent, we can find their intersection using the product of their probabilities: \[ P(A \cap B) = P(A) \cdot P(B) = 0.6 \cdot 0.4 = 0.24 \] ### Calculate \( P(A \cup B) \) Next, we can determine the probability that at least one of the events occurs: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substituting the known values into the equation: \[ P(A \cup B) = 0.6 + 0.4 - 0.24 = 1.0 - 0.24 = 0.76 \] --- ## Final Answers - The probability that both events \( A \) and \( B \) occur simultaneously: \[ P(A \cap B) = \boxed{0.24} \] - The probability that at least one of the events \( A \) or \( B \) occurs: \[ P(A \cup B) = \boxed{0.76} \]

## Given Information - Probability of event \( A \): \( P(A) = 0.6 \) - Probability of event \( B \): \( P(B) = 0.4 \) - Events \( A \) and \( B \) are independent. --- ## Definitions and Concepts Used - **Intersection of Events:** The probability that both events \( A \) and \( B \) happen concurrently, represented as \( P(A \cap B) \). For independent events, this probability is given by: \[ P(A \cap B) = P(A) \times P(B) \] - **Union of Events:** The probability that at least one of the events \( A \) or \( B \) occurs, represented as \( P(A \cup B) \). This can be calculated using the formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] --- ## Solution ### Calculate \( P(A \cap B) \) Since \( A \) and \( B \) are independent, we calculate the probability of both events occurring together: \[ P(A \cap B) = P(A) \cdot P(B) = 0.6 \cdot 0.4 = 0.24 \] ### Calculate \( P(A \cup B) \) Next, we compute the probability that either event \( A \) or event \( B \) occurs: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Inserting the known probabilities: \[ P(A \cup B) = 0.6 + 0.4 - 0.24 = 1.0 - 0.24 = 0.76 \] --- ## Final Answers - The probability that both events \( A \) and \( B \) occur: \[ P(A \cap B) = \boxed{0.24} \] - The probability that at least one of the events \( A \) or \( B \) occurs: \[ P(A \cup B) = \boxed{0.76} \]

## Given Information - Probability of event A: \( P(A) = 0.6 \) - Probability of event B: \( P(B) = 0.4 \) - Events A and B are independent. --- ## Definitions and Concepts Used - Intersection of Events: The probability that both events A and B occur at the same time, denoted as \( P(A \cap B) \). For independent events, this is calculated as: \[ P(A \cap B) = P(A) \times P(B) \] - Union of Events: The probability that at least one of the events A or B occurs, denoted as \( P(A \cup B) \). This is calculated using the formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] --- ## Solution To find \( P(A \cap B) \): Since A and B are independent, the intersection can be calculated as: \[ P(A \cap B) = P(A) \times P(B) = 0.6 \times 0.4 = 0.24 \] Next, to find \( P(A \cup B) \): Using the union formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substituting the known values: \[ P(A \cup B) = 0.6 + 0.4 - 0.24 = 1.0 - 0.24 = 0.76 \] --- ## Final Answers - Probability that both events A and B occur: \[ P(A \cap B) = \boxed{0.24} \] - Probability that at least one of the events A or B occurs: \[ P(A \cup B) = \boxed{0.76} \]

## Given Information Probability of event A: \( P(A) = 0.6 \) Probability of event B: \( P(B) = 0.4 \) Events A and B are independent. --- ## Definitions and Concepts Used Intersection of events: The probability that both events A and B occur simultaneously, denoted as \( P(A \cap B) \). For independent events, this is calculated as: \[ P(A \cap B) = P(A) \times P(B) \] Union of events: The probability that at least one of the events A or B occurs, denoted as \( P(A \cup B) \). This is calculated using the formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] --- ## Solution To calculate \( P(A \cap B) \), since A and B are independent: \[ P(A \cap B) = P(A) \times P(B) = 0.6 \times 0.4 = 0.24 \] Next, to determine \( P(A \cup B) \): \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substituting the known values: \[ P(A \cup B) = 0.6 + 0.4 - 0.24 = 1.0 - 0.24 = 0.76 \] --- ## Final Answers Probability that both events A and B occur: \[ P(A \cap B) = \boxed{0.24} \] Probability that at least one of the events A or B occurs: \[ P(A \cup B) = \boxed{0.76} \]

## Given Information Probability of event A is \( P(A) = 0.6 \). Probability of event B is \( P(B) = 0.4 \). Events A and B are independent. --- ## Definitions and Concepts Used Intersection of events refers to the probability that both events A and B happen at the same time, represented as \( P(A \cap B) \). For events that are independent, this is calculated using: \[ P(A \cap B) = P(A) \times P(B) \] Union of events indicates the probability of at least one of the events A or B occurring, denoted as \( P(A \cup B) \). This can be computed using the formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] --- ## Solution To find \( P(A \cap B) \), since A and B are independent: \[ P(A \cap B) = P(A) \times P(B) = 0.6 \times 0.4 = 0.24 \] Next, to find \( P(A \cup B) \): \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substituting the values we have: \[ P(A \cup B) = 0.6 + 0.4 - 0.24 = 1.0 - 0.24 = 0.76 \] --- ## Final Answers The probability that both events A and B occur together is: \[ P(A \cap B) = \boxed{0.24} \] The probability that at least one of the events A or B occurs is: \[ P(A \cup B) = \boxed{0.76} \]

## Given Information The probability of event A is \( P(A) = 0.6 \). The probability of event B is \( P(B) = 0.4 \). Events A and B are considered independent. --- ## Definitions and Concepts Used Intersection of events: This represents the probability that both events A and B occur at the same time, denoted \( P(A \cap B) \). For independent events, it is computed as: \[ P(A \cap B) = P(A) \times P(B) \] Union of events: This denotes the probability that at least one of the events A or B occurs, represented as \( P(A \cup B) \). The formula used to calculate this is: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] --- ## Solution To determine \( P(A \cap B) \): Given the independence of events A and B, we calculate their intersection as follows: \[ P(A \cap B) = P(A) \times P(B) = 0.6 \times 0.4 = 0.24 \] To find \( P(A \cup B) \): Using the union formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substituting the known values: \[ P(A \cup B) = 0.6 + 0.4 - 0.24 = 1.0 - 0.24 = 0.76 \] --- ## Final Answers The probability that both events A and B occur is: \[ P(A \cap B) = \boxed{0.24} \] The probability that at least one of the events A or B occurs is: \[ P(A \cup B) = \boxed{0.76} \]

## Given Information The probability of event A occurring is \( P(A) = 0.6 \). The probability of event B occurring is \( P(B) = 0.4 \). Events A and B are independent of one another. --- ## Definitions and Concepts Used Intersection of events: This is the probability that both events A and B occur together, denoted as \( P(A \cap B) \). For independent events, this can be calculated using the product of their probabilities: \[ P(A \cap B) = P(A) \times P(B) \] Union of events: This refers to the probability that either event A or event B occurs (or both), represented as \( P(A \cup B) \). The formula to calculate this is: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] --- ## Solution To compute the probability of both events A and B occurring simultaneously, we apply the formula for the intersection of independent events: \[ P(A \cap B) = P(A) \times P(B) \] Substituting the known probabilities: \[ P(A \cap B) = 0.6 \times 0.4 = 0.24 \] Next, we need to determine the probability that at least one of the events occurs. To do this, we will use the union formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Now, substituting the values we calculated previously along with the individual probabilities: \[ P(A \cup B) = 0.6 + 0.4 - 0.24 \] Simplifying the equation: \[ P(A \cup B) = 1.0 - 0.24 = 0.76 \] --- ## Final Answers The probability that both events A and B occur together is: \[ P(A \cap B) = \boxed{0.24} \] The probability that at least one of the events A or B occurs is: \[ P(A \cup B) = \boxed{0.76} \]

## Given Information The probability of event A is \( P(A) = 0.6 \). The probability of event B is \( P(B) = 0.4 \). Events A and B are independent. --- ## Definitions and Concepts Used Intersection of events: This is the probability that both events A and B happen simultaneously, represented as \( P(A \cap B) \). For independent events, this is calculated as: \[ P(A \cap B) = P(A) \times P(B) \] Union of events: This refers to the probability that either event A or event B occurs (or both), represented as \( P(A \cup B) \). The formula to compute this is: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] --- ## Solution To find the probability of both events A and B occurring at the same time, we utilize the formula for the intersection of independent events: \[ P(A \cap B) = P(A) \times P(B) \] Substituting the known probabilities into the equation: \[ P(A \cap B) = 0.6 \times 0.4 = 0.24 \] Next, we will calculate the probability that at least one of the events occurs. For this, we apply the union formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Now substituting the values we calculated along with the individual probabilities: \[ P(A \cup B) = 0.6 + 0.4 - 0.24 \] This simplifies to: \[ P(A \cup B) = 1.0 - 0.24 = 0.76 \] --- ## Final Answers The probability that both events A and B occur simultaneously is: \[ P(A \cap B) = \boxed{0.24} \] The probability that at least one of the events A or B occurs is: \[ P(A \cup B) = \boxed{0.76} \]

## Given Information The probability of event A is \( P(A) = 0.6 \). The probability of event B is \( P(B) = 0.4 \). Events A and B are independent. --- ## Definitions and Concepts Used Intersection of events: This represents the probability that both events A and B occur at the same time, denoted as \( P(A \cap B) \). For independent events, this probability is calculated as: \[ P(A \cap B) = P(A) \times P(B) \] Union of events: This indicates the probability that at least one of the events A or B occurs, represented as \( P(A \cup B) \). The formula to compute this is: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] --- ## Solution To determine the probability of both events A and B happening simultaneously, we apply the formula for the intersection of independent events: \[ P(A \cap B) = P(A) \times P(B) \] Substituting the known probabilities: \[ P(A \cap B) = 0.6 \times 0.4 = 0.24 \] Next, we need to calculate the probability that at least one of the events occurs. This can be done using the union formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Now, substituting the values we have: \[ P(A \cup B) = 0.6 + 0.4 - 0.24 \] This simplifies to: \[ P(A \cup B) = 1.0 - 0.24 = 0.76 \] --- ## Final Answers The probability that both events A and B occur is: \[ P(A \cap B) = \boxed{0.24} \] The probability that at least one of the events A or B occurs is: \[ P(A \cup B) = \boxed{0.76} \]

## Given Information The probability of event A is \( P(A) = 0.6 \). The probability of event B is \( P(B) = 0.4 \). Events A and B are considered independent. --- ## Definitions and Concepts Used - **Intersection of Events:** This refers to the probability that both events A and B occur at the same time, denoted as \( P(A \cap B) \). For independent events, this is calculated as: \[ P(A \cap B) = P(A) \times P(B) \] - **Union of Events:** This indicates the probability that at least one of the events A or B occurs, represented as \( P(A \cup B) \). The formula to calculate this is: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] --- ## Solution To find the probability that both events A and B happen simultaneously, we use the formula for the intersection because A and B are independent: \[ P(A \cap B) = P(A) \times P(B) \] Substituting in the known probabilities: \[ P(A \cap B) = 0.6 \times 0.4 = 0.24 \] Next, we calculate the probability that at least one of the events occurs. For this, we use the union formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Now we substitute the values we have: \[ P(A \cup B) = 0.6 + 0.4 - 0.24 \] This results in: \[ P(A \cup B) = 1.0 - 0.24 = 0.76 \] --- ## Final Answers The probability that both events A and B occur is: \[ P(A \cap B) = \boxed{0.24} \] The probability that at least one of the events A or B occurs is: \[ P(A \cup B) = \boxed{0.76} \]

## Given Information - Sample size (\( n \)): 40 phones - Sample mean (\( \bar{x} \)): 14 hours - Sample standard deviation (\( s \)): 1.5 hours - Level of significance (\( \alpha \)): 0.01 --- ## Definitions and Concepts Used - **Standard Error (SE):** The standard deviation of the sample mean distribution, calculated as: \[ SE = \frac{s}{\sqrt{n}} \] - **Margin of Error (ME):** The range within which the true population parameter is expected to fall, calculated as: \[ ME = z \cdot SE \] where \( z \) is the z-score corresponding to the desired confidence level. - **Confidence Interval (CI):** A range of values used to estimate the true population parameter. It is calculated as: \[ CI = \bar{x} \pm ME \] --- ## Solution ### a. Calculate the Standard Error First, we calculate the standard error: \[ SE = \frac{s}{\sqrt{n}} = \frac{1.5}{\sqrt{40}} \approx \frac{1.5}{6.3246} \approx 0.237 \] ### b. Calculate the Margin of Error Next, we find the z-score for a 99% confidence level. The z-score for \( \alpha = 0.01 \) (two-tailed) is approximately 2.576. Now we can calculate the margin of error: \[ ME = z \cdot SE = 2.576 \cdot 0.237 \approx 0.611 \] ### c. Construct a 99% Confidence Interval The confidence interval is: \[ CI = \bar{x} \pm ME = 14 \pm 0.611 \] Calculating the lower and upper limits: - Lower limit: \( 14 - 0.611 \approx 13.389 \) - Upper limit: \( 14 + 0.611 \approx 14.611 \) Thus, the 99% confidence interval is: \[ CI = (13.389, 14.611) \] ### d. Interpret the Confidence Interval The 99% confidence interval of (13.389, 14.611) suggests that we are 99% confident that the true average battery life of the new smartphone lies between approximately 13.39 hours and 14.61 hours. This interval provides an estimate based on the sample data, indicating that while we expect the average battery life to be around 14 hours, there is variability, and the true mean could be slightly less or more than this value.

## Given Information - Sample size (\( n \)): 64 components - Sample mean (\( \bar{x} \)): 980 hours - Sample standard deviation (\( s \)): 120 hours --- ## Definitions and Concepts Used - **Standard Error (SE):** This measures the variability of the sample mean and is calculated as: \[ SE = \frac{s}{\sqrt{n}} \] - **Confidence Interval (CI):** This is a range of values that is likely to contain the population mean. It can be calculated using the formula: \[ CI = \bar{x} \pm z \cdot SE \] where \( z \) is the z-score corresponding to the desired confidence level. --- ## Solution ### a. Calculate the 95% Two-Sided Confidence Interval 1. **Calculate the Standard Error:** \[ SE = \frac{s}{\sqrt{n}} = \frac{120}{\sqrt{64}} = \frac{120}{8} = 15 \] 2. **Determine the z-score for a 95% confidence level:** The z-score corresponding to a 95% confidence level (two-tailed) is approximately 1.96. 3. **Calculate the Margin of Error:** \[ ME = z \cdot SE = 1.96 \cdot 15 = 29.4 \] 4. **Construct the Confidence Interval:** \[ CI = \bar{x} \pm ME = 980 \pm 29.4 \] The lower limit is \( 980 - 29.4 = 950.6 \) The upper limit is \( 980 + 29.4 = 1009.4 \) Thus, the 95% two-sided confidence interval is: \[ CI = (950.6, 1009.4) \] ### b. Calculate the 95% One-Sided Confidence Interval 1. **Determine the z-score for a 95% one-sided confidence level:** The z-score for a 95% confidence level (one-tailed) is approximately 1.645. 2. **Calculate the Margin of Error for one-sided interval:** \[ ME = z \cdot SE = 1.645 \cdot 15 = 24.675 \] 3. **Construct the One-Sided Confidence Interval:** For a one-sided upper confidence interval, we add the margin of error to the sample mean: \[ CI = \bar{x} + ME = 980 + 24.675 = 1004.675 \] Therefore, the 95% one-sided confidence interval is: \[ CI = (−\infty, 1004.675) \] --- ## Final Answers - The 95% two-sided confidence interval for the population mean is: \[ \boxed{(950.6, 1009.4)} \] - The 95% one-sided confidence interval for the population mean is: \[ \boxed{(−\infty, 1004.675)} \]

## Given Information - Sample size (\( n \)): 64 components - Sample mean (\( \bar{x} \)): 980 hours - Sample standard deviation (\( s \)): 120 hours --- ## Definitions and Concepts Used - **Standard Error (SE):** This measures the variability of the sample mean and is calculated as: \[ SE = \frac{s}{\sqrt{n}} \] - **Confidence Interval (CI):** A range of values that likely contains the population mean. For one-sided intervals, it is calculated using: \[ CI_{\text{upper}} = \bar{x} + z \cdot SE \] \[ CI_{\text{lower}} = \bar{x} - z \cdot SE \] where \( z \) is the z-score corresponding to the desired confidence level. --- ## Solution ### a. Calculate the 95% Two-Sided Confidence Interval 1. **Calculate the Standard Error:** \[ SE = \frac{s}{\sqrt{n}} = \frac{120}{\sqrt{64}} = \frac{120}{8} = 15 \] 2. **Determine the z-score for a 95% confidence level:** The z-score corresponding to a 95% confidence level (two-tailed) is approximately 1.96. 3. **Calculate the Margin of Error:** \[ ME = z \cdot SE = 1.96 \cdot 15 = 29.4 \] 4. **Construct the Confidence Interval:** \[ CI = \bar{x} \pm ME = 980 \pm 29.4 \] Lower limit: \( 980 - 29.4 = 950.6 \) Upper limit: \( 980 + 29.4 = 1009.4 \) Thus, the 95% two-sided confidence interval is: \[ CI = (950.6, 1009.4) \] ### b. Calculate the 95% One-Sided Confidence Intervals 1. **Determine the z-score for a 95% one-sided confidence level:** The z-score for a 95% confidence level (one-tailed) is approximately 1.645. 2. **Calculate the Standard Error:** (Already calculated in part a) \[ SE = 15 \] 3. **Calculate the Upper One-Sided Confidence Interval:** \[ CI_{\text{upper}} = \bar{x} + z \cdot SE = 980 + 1.645 \cdot 15 = 980 + 24.675 = 1004.675 \] 4. **Calculate the Lower One-Sided Confidence Interval:** \[ CI_{\text{lower}} = \bar{x} - z \cdot SE = 980 - 1.645 \cdot 15 = 980 - 24.675 = 955.325 \] Thus, the 95% one-sided confidence intervals are: - Upper one-sided CI: \[ CI_{\text{upper}} = (−\infty, 1004.675) \] - Lower one-sided CI: \[ CI_{\text{lower}} = (955.325, +\infty) \] --- ## Final Answers - The 95% two-sided confidence interval for the population mean is: \[ \boxed{(950.6, 1009.4)} \] - The 95% upper one-sided confidence interval for the population mean is: \[ \boxed{(−\infty, 1004.675)} \] - The 95% lower one-sided confidence interval for the population mean is: \[ \boxed{(955.325, +\infty)} \]

## Given Data The number of hours studied by 8 students is: \[ (2, 3, 4, 2, 5, 3, 4, 2) \] --- ## Definitions and Concepts Used - **Mean:** The average of a set of numbers, calculated by summing all values and dividing by the count of values: \[ \text{Mean} = \frac{\sum x}{n} \] - **Median:** The middle value in a sorted list of numbers. If the count of numbers is odd, it is the middle number; if even, it is the average of the two middle numbers. - **Mode:** The value that appears most frequently in a data set. --- ## Solution ### Step 1: Calculate the Mean Sum of the hours studied: \[ \sum x = 2 + 3 + 4 + 2 + 5 + 3 + 4 + 2 = 25 \] Number of students: \[ n = 8 \] Calculating the mean: \[ \text{Mean} = \frac{25}{8} = 3.125 \] ### Step 2: Calculate the Median First, sort the data: \[ (2, 2, 2, 3, 3, 4, 4, 5) \] Since there are 8 values (even), the median is the average of the 4th and 5th values: \[ \text{Median} = \frac{3 + 3}{2} = 3 \] ### Step 3: Calculate the Mode The mode is the value that appears most frequently in the data set. In the sorted data: - \( 2 \) appears 3 times - \( 3 \) appears 2 times - \( 4 \) appears 2 times - \( 5 \) appears 1 time Thus, the mode is: \[ \text{Mode} = 2 \] --- ## Final Answers - Mean: \[ \boxed{3.125} \] - Median: \[ \boxed{3} \] - Mode: \[ \boxed{2} \]

## Given Data The number of hours studied by 8 students is: \[ (2, 3, 4, 2, 5, 3, 4, 2) \] --- ## Definitions and Concepts Used - Mean: The average of a set of values, calculated by dividing the sum of all values by the number of values: \[ \text{Mean} = \frac{\sum x}{n} \] - Median: The middle value in a sorted list of numbers. If the list has an odd count, it is the central number; if even, it is the average of the two middle numbers. - Mode: The value that appears most frequently in the data set. --- ## Solution ### Step 1: Calculate the Mean First, calculate the total sum of the hours studied: \[ \sum x = 2 + 3 + 4 + 2 + 5 + 3 + 4 + 2 = 25 \] Next, determine the number of students: \[ n = 8 \] Now, calculate the mean: \[ \text{Mean} = \frac{25}{8} = 3.125 \] ### Step 2: Calculate the Median To find the median, first sort the data in ascending order: \[ (2, 2, 2, 3, 3, 4, 4, 5) \] Since there are 8 values (an even number), the median is the average of the 4th and 5th values: \[ \text{Median} = \frac{3 + 3}{2} = 3 \] ### Step 3: Calculate the Mode Identify the mode by finding the value that occurs most frequently in the data set. In the sorted data: - \( 2 \) appears 3 times - \( 3 \) appears 2 times - \( 4 \) appears 2 times - \( 5 \) appears 1 time The mode is: \[ \text{Mode} = 2 \] --- ## Final Answers - Mean: \[ \boxed{3.125} \] - Median: \[ \boxed{3} \] - Mode: \[ \boxed{2} \]

## Given Data The hours studied by a group of 8 students are as follows: \[ (2, 3, 4, 2, 5, 3, 4, 2) \] --- ## Definitions and Concepts Used - Mean: This is the average value calculated by adding all the numbers together and dividing by the total count of values: \[ \text{Mean} = \frac{\sum x}{n} \] - Median: This is the middle number in a sorted sequence. For an even number of observations, it is the average of the two central numbers. - Mode: This is the value that occurs most frequently in the dataset. --- ## Solution ### Step 1: Calculate the Mean First, we need to find the total sum of the hours studied: \[ \sum x = 2 + 3 + 4 + 2 + 5 + 3 + 4 + 2 = 25 \] Next, we identify the number of students: \[ n = 8 \] Now, we calculate the mean: \[ \text{Mean} = \frac{25}{8} = 3.125 \] ### Step 2: Calculate the Median To find the median, we first arrange the data in ascending order: \[ (2, 2, 2, 3, 3, 4, 4, 5) \] With 8 values (an even count), the median is computed by averaging the 4th and 5th values: \[ \text{Median} = \frac{3 + 3}{2} = 3 \] ### Step 3: Calculate the Mode The mode is determined by identifying the most frequently occurring value in the dataset. In the sorted list: - \( 2 \) occurs 3 times - \( 3 \) occurs 2 times - \( 4 \) occurs 2 times - \( 5 \) occurs 1 time Thus, the mode is: \[ \text{Mode} = 2 \] --- ## Final Answers - Mean: \[ \boxed{3.125} \] - Median: \[ \boxed{3} \] - Mode: \[ \boxed{2} \]

## Given Data The hours studied by 8 students are: \[ (2, 3, 4, 2, 5, 3, 4, 2) \] --- ## Definitions and Concepts Used - Mean: The average value of the data set, found by dividing the total sum of the values by the number of values: \[ \text{Mean} = \frac{\sum x}{n} \] - Median: The middle value in a sorted list of numbers. If the list has an even number of entries, the median is the average of the two central values. - Mode: The most frequently occurring value within the dataset. --- ## Solution ### Step 1: Calculate the Mean First, we calculate the total sum of the hours studied: \[ \sum x = 2 + 3 + 4 + 2 + 5 + 3 + 4 + 2 = 25 \] The number of students is: \[ n = 8 \] Now we compute the mean: \[ \text{Mean} = \frac{25}{8} = 3.125 \] ### Step 2: Calculate the Median To find the median, we first sort the data in ascending order: \[ (2, 2, 2, 3, 3, 4, 4, 5) \] Since there are 8 observations (an even number), the median is calculated by averaging the 4th and 5th values: \[ \text{Median} = \frac{3 + 3}{2} = 3 \] ### Step 3: Calculate the Mode The mode is the value that appears most frequently in the dataset. In the sorted list: - \( 2 \) appears 3 times - \( 3 \) appears 2 times - \( 4 \) appears 2 times - \( 5 \) appears 1 time Thus, the mode is: \[ \text{Mode} = 2 \] --- ## Final Answers - Mean: \[ \boxed{3.125} \] - Median: \[ \boxed{3} \] - Mode: \[ \boxed{2} \]

## Given Data The hours studied by 8 students are: \[ (2, 3, 4, 2, 5, 3, 4, 2) \] --- ## Definitions and Concepts Used - Mean: This is the average of the data set, calculated by summing all the values and dividing by the number of values: \[ \text{Mean} = \frac{\sum x}{n} \] - Median: The middle value in a sorted list of numbers. For an even count of numbers, it is the average of the two central values. - Mode: The number that appears most frequently in the dataset. --- ## Solution ### Step 1: Calculate the Mean To find the mean, first calculate the total sum of the hours studied: \[ \sum x = 2 + 3 + 4 + 2 + 5 + 3 + 4 + 2 = 25 \] Next, note the number of students: \[ n = 8 \] Now, compute the mean: \[ \text{Mean} = \frac{25}{8} = 3.125 \] ### Step 2: Calculate the Median To determine the median, sort the data in ascending order: \[ (2, 2, 2, 3, 3, 4, 4, 5) \] With 8 values (an even count), the median is the average of the 4th and 5th values: \[ \text{Median} = \frac{3 + 3}{2} = 3 \] ### Step 3: Calculate the Mode Identify the mode by finding the value that occurs most frequently in the dataset. In the sorted list: - \( 2 \) occurs 3 times - \( 3 \) occurs 2 times - \( 4 \) occurs 2 times - \( 5 \) occurs 1 time Thus, the mode is: \[ \text{Mode} = 2 \] --- ## Final Answers - Mean: \[ \boxed{3.125} \] - Median: \[ \boxed{3} \] - Mode: \[ \boxed{2} \]

## Given Information A standard die has 6 faces, numbered from 1 to 6. --- ## Definitions and Concepts Used - **Probability (P)**: The likelihood of an event occurring, calculated as: \[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \] --- ## Solution ### A. Probability of Rolling an Even Number The even numbers on a die are: 2, 4, and 6. The total number of favorable outcomes for rolling an even number is 3. Total possible outcomes when rolling a die = 6. Calculating the probability: \[ P(\text{Even}) = \frac{3}{6} = \frac{1}{2} \] ### B. Probability of Rolling a Number Greater than 4 The numbers greater than 4 on a die are: 5 and 6. The total number of favorable outcomes for rolling a number greater than 4 is 2. Calculating the probability: \[ P(\text{Greater than 4}) = \frac{2}{6} = \frac{1}{3} \] --- ## Final Answers - Probability of rolling an even number: \[ \boxed{\frac{1}{2}} \] - Probability of rolling a number greater than 4: \[ \boxed{\frac{1}{3}} \]

## Given Information Two coins are tossed simultaneously. --- ## Definitions and Concepts Used - **Probability (P)**: The likelihood of an event occurring, defined as: \[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \] - **Sample Space (S)**: The set of all possible outcomes when tossing two coins is: \[ S = \{HH, HT, TH, TT\} \] --- ## Solution ### A. Probability of Getting Two Heads The favorable outcome for getting two heads is: - Outcome: \( HH \) Total possible outcomes when tossing two coins = 4 (as listed in the sample space). Calculating the probability: \[ P(\text{Two Heads}) = \frac{1}{4} = \frac{1}{4} \] ### B. Probability of Getting at Least One Tail The outcomes that include at least one tail are: - Outcomes: \( HT, TH, TT \) Counting the favorable outcomes: - Total favorable outcomes = 3 Calculating the probability: \[ P(\text{At least one Tail}) = \frac{3}{4} \] --- ## Final Answers - Probability of getting two heads: \[ \boxed{\frac{1}{4}} \] - Probability of getting at least one tail: \[ \boxed{\frac{3}{4}} \]

## Given Information A standard die has 6 faces, numbered from 1 to 6. --- ## Definitions and Concepts Used - **Probability (P)**: The likelihood of an event occurring, calculated as: \[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \] --- ## Solution ### A. Probability of Rolling an Even Number The even numbers on a die are: 2, 4, and 6. The total number of favorable outcomes for rolling an even number is 3. Total possible outcomes when rolling a die = 6. Calculating the probability: \[ P(\text{Even}) = \frac{3}{6} = \frac{1}{2} \] ### B. Probability of Rolling a Number Greater than 4 The numbers greater than 4 on a die are: 5 and 6. The total number of favorable outcomes for rolling a number greater than 4 is 2. Calculating the probability: \[ P(\text{Greater than 4}) = \frac{2}{6} = \frac{1}{3} \] --- ## Final Answers - Probability of rolling an even number: \[ \boxed{\frac{1}{2}} \] - Probability of rolling a number greater than 4: \[ \boxed{\frac{1}{3}} \]

## Given Data The number of hours studied by 8 students is: \[ (2, 3, 4, 2, 5, 3, 4, 2) \] --- ## Definitions and Concepts Used - **Mean**: The average of a set of numbers, calculated by summing all values and dividing by the count of values: \[ \text{Mean} = \frac{\sum x}{n} \] - **Median**: The middle value in a sorted list of numbers. If the count of numbers is odd, it is the middle number; if even, it is the average of the two middle numbers. - **Mode**: The value that appears most frequently in a data set. --- ## Solution ### Step 1: Calculate the Mean First, sum the hours studied: \[ \sum x = 2 + 3 + 4 + 2 + 5 + 3 + 4 + 2 = 25 \] Now, determine the number of students: \[ n = 8 \] Next, calculate the mean: \[ \text{Mean} = \frac{25}{8} = 3.125 \] ### Step 2: Calculate the Median To find the median, first sort the data in ascending order: \[ (2, 2, 2, 3, 3, 4, 4, 5) \] Since there are 8 values (even), the median is the average of the 4th and 5th values: \[ \text{Median} = \frac{3 + 3}{2} = 3 \] ### Step 3: Calculate the Mode The mode is the value that occurs most frequently in the data set. In the sorted data: - \( 2 \) appears 3 times - \( 3 \) appears 2 times - \( 4 \) appears 2 times - \( 5 \) appears 1 time Thus, the mode is: \[ \text{Mode} = 2 \] --- ## Final Answers - Mean: \[ \boxed{3.125} \] - Median: \[ \boxed{3} \] - Mode: \[ \boxed{2} \]