# Given Data - Number of red balls (\(R\)) = 4 - Number of blue balls (\(B\)) = 6 - Total balls = 4 + 6 = 10 - Two balls are drawn **one by one without replacement**. We are to find: Given that **at least one ball is red**, what is the probability that **both balls are red**? --- # Definitions and Terms Used - **Conditional Probability**: The probability of event A given that event B has occurred, denoted \(P(A|B)\). - **Without Replacement**: Once a ball is drawn, it is not put back into the box. - Let: - \(A\): Event that **both balls drawn are red** - \(B\): Event that **at least one ball drawn is red** - We need to find \(P(A|B)\). --- # Solution By definition of conditional probability: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] But here, if both balls are red, then certainly at least one is red, so \(A \subseteq B\). Therefore, \[ P(A|B) = \frac{P(A)}{P(B)} \] **Finding \(P(A)\):** Probability that both balls drawn are red: \[ P(A) = \frac{\text{Ways to choose 2 red balls}}{\text{Ways to choose any 2 balls}} \] Ways to choose 2 red balls: \({4 \choose 2} = 6\) Ways to choose any 2 balls: \({10 \choose 2} = 45\) So, \[ P(A) = \frac{6}{45} = \frac{2}{15} \] **Finding \(P(B)\):** Probability that **at least one ball is red** when drawing two balls. It's easier to use the complement: - Probability that **no ball is red** (i.e., both are blue), - Subtract from 1. Ways to choose 2 blue balls: \({6 \choose 2} = 15\) So, \[ P(\text{both blue}) = \frac{15}{45} = \frac{1}{3} \] \[ P(B) = 1 - \frac{1}{3} = \frac{2}{3} \] **Final Calculation:** \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{2}{15}}{\frac{2}{3}} = \frac{2}{15} \times \frac{3}{2} = \frac{3}{15} = \frac{1}{5} \] --- # Final Answer \[ \boxed{\frac{1}{5}} \] Thus, **the probability that both balls are red given that at least one ball is red is \( \frac{1}{5} \)**.

# Given Data - Number of red balls (\(R\)) = 4 - Number of blue balls (\(B\)) = 6 - Total number of balls = \(R + B = 10\) - Two balls are drawn **one by one without replacement**. We need to find: The probability that **both balls are red**, given that **at least one ball is red**. --- # Definitions and Terms Used - **Conditional Probability**: The likelihood of an event occurring given that another event has already occurred, denoted as \(P(A|B)\). - **Without Replacement**: The drawn ball is not returned to the box after being selected. - Let: - \(A\): Event that **both balls drawn are red**. - \(B\): Event that **at least one ball drawn is red**. - Our goal is to compute \(P(A|B)\). --- # Solution Using the formula for conditional probability: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Since \(A\) (both balls are red) implies \(B\) (at least one ball is red), we have: \[ P(A|B) = \frac{P(A)}{P(B)} \] **Calculating \(P(A)\):** The probability of drawing two red balls: \[ P(A) = \frac{\text{Ways to choose 2 red balls}}{\text{Ways to choose any 2 balls}} \] The number of ways to choose 2 red balls is given by \({4 \choose 2} = 6\). The total number of ways to choose any 2 balls from 10 is \({10 \choose 2} = 45\). Thus, \[ P(A) = \frac{6}{45} = \frac{2}{15} \] **Calculating \(P(B)\):** To find the probability of drawing at least one red ball, we use the complement (probability of drawing no red balls): - The only case for this is when both balls drawn are blue. The number of ways to choose 2 blue balls is \({6 \choose 2} = 15\). Then, \[ P(\text{both blue}) = \frac{15}{45} = \frac{1}{3} \] Thus, \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{1}{3} = \frac{2}{3} \] **Final Calculation:** \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{2}{15}}{\frac{2}{3}} = \frac{2}{15} \times \frac{3}{2} = \frac{3}{15} = \frac{1}{5} \] --- # Final Answer \[ \boxed{\frac{1}{5}} \] Therefore, the probability that both balls drawn are red, given that at least one ball drawn is red, is \( \frac{1}{5} \).

# Given Data - Number of red balls (\(R\)) = 4 - Number of blue balls (\(B\)) = 6 - Total number of balls = \(R + B = 10\) - Two balls are drawn **one by one without replacement**. We need to find: The probability that **both balls are red**, given that **at least one ball is red**. --- # Definitions and Terms Used - **Conditional Probability**: The probability of an event given that another event has occurred, represented as \(P(A|B)\). - **Without Replacement**: After drawing a ball, it is not returned to the box for subsequent draws. - Let: - \(A\): Event that **both balls drawn are red**. - \(B\): Event that **at least one ball drawn is red**. - We are tasked with finding \(P(A|B)\). --- # Solution Using the relationship for conditional probability: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Since drawing two red balls implies that at least one is red, we can express this as: \[ P(A|B) = \frac{P(A)}{P(B)} \] **Calculating \(P(A)\):** To determine the probability of both balls being red: \[ P(A) = \frac{\text{Ways to choose 2 red balls}}{\text{Ways to choose any 2 balls}} \] Calculating: - Ways to choose 2 red balls: \({4 \choose 2} = 6\) - Total ways to choose any 2 balls: \({10 \choose 2} = 45\) Thus, \[ P(A) = \frac{6}{45} = \frac{2}{15} \] **Calculating \(P(B)\):** To find the probability of drawing at least one red ball, we can use the complement (drawing no red balls): - The only scenario for this is drawing two blue balls. Calculating: - Ways to choose 2 blue balls: \({6 \choose 2} = 15\) Thus, \[ P(\text{both blue}) = \frac{15}{45} = \frac{1}{3} \] Then, \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{1}{3} = \frac{2}{3} \] **Final Calculation:** We can now substitute back into our expression for conditional probability: \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{2}{15}}{\frac{2}{3}} = \frac{2}{15} \times \frac{3}{2} = \frac{3}{15} = \frac{1}{5} \] --- # Final Answer \[ \boxed{\frac{1}{5}} \] Thus, the probability that both balls drawn are red given that at least one ball drawn is red is \( \frac{1}{5} \).

# Given Data - Number of red balls (\(R\)) = 4 - Number of blue balls (\(B\)) = 6 - Total number of balls = \(R + B = 10\) - Two balls are drawn **one by one without replacement**. We need to find: The probability that **both balls are red**, given that **at least one ball is red**. --- # Definitions and Terms Used - **Conditional Probability**: The probability of an event occurring given that another event has already occurred, expressed as \(P(A|B)\). - **Without Replacement**: A method where drawn items are not returned to the original set before the next item is drawn. - Let: - \(A\): Event that **both drawn balls are red**. - \(B\): Event that **at least one ball drawn is red**. - Our objective is to compute \(P(A|B)\). --- # Solution To find \(P(A|B)\), we use the conditional probability formula: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Since the occurrence of event \(A\) (both balls are red) guarantees that event \(B\) (at least one red ball) has occurred, we can simplify this to: \[ P(A|B) = \frac{P(A)}{P(B)} \] **Calculating \(P(A)\):** To find the probability that both balls drawn are red, we determine the number of favorable outcomes over the total possible outcomes: \[ P(A) = \frac{\text{Ways to choose 2 red balls}}{\text{Ways to choose any 2 balls}} \] Calculating the combinations: - Ways to choose 2 red balls: \({4 \choose 2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6\) - Total ways to choose any 2 balls from 10: \({10 \choose 2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45\) Thus, we find: \[ P(A) = \frac{6}{45} = \frac{2}{15} \] **Calculating \(P(B)\):** To find the probability of drawing at least one red ball, we can utilize the complement rule, which is often simpler. Thus, we calculate the probability of the complementary event (both balls are blue): - Number of ways to choose 2 blue balls: \({6 \choose 2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15\) Now, the probability of both balls being blue is: \[ P(\text{both blue}) = \frac{15}{45} = \frac{1}{3} \] Consequently, the probability of at least one ball being red is: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{1}{3} = \frac{2}{3} \] **Final Calculation:** Utilizing our earlier results for \(P(A)\) and \(P(B)\): \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{2}{15}}{\frac{2}{3}} = \frac{2}{15} \times \frac{3}{2} = \frac{3}{15} = \frac{1}{5} \] --- # Final Answer \[ \boxed{\frac{1}{5}} \] Hence, the probability that both balls drawn are red, given that at least one ball drawn is red, is \( \frac{1}{5} \). This result illustrates the impact of conditional probability and highlights the importance of understanding event relationships in probability theory.

# Given Data - Number of red balls (\(R\)) = 4 - Number of blue balls (\(B\)) = 6 - Total number of balls = \(R + B = 10\) - Two balls are drawn **one by one without replacement**. We need to find: The probability that **both balls are red**, given that **at least one ball is red**. --- # Definitions and Terms Used - **Conditional Probability**: The probability of one event occurring given that another event has occurred, denoted as \(P(A|B)\). - **Without Replacement**: Once a ball is drawn, it is not returned to the box for subsequent draws. - Let: - \(A\): Event that **both balls drawn are red**. - \(B\): Event that **at least one ball drawn is red**. - We need to calculate \(P(A|B)\). --- # Solution To find \(P(A|B)\), we use the conditional probability formula: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Since if both balls are red (event \(A\)), then at least one must also be red (event \(B\)), we have: \[ P(A|B) = \frac{P(A)}{P(B)} \] **Calculating \(P(A)\):** The event \(A\) occurs when both drawn balls are red. The formula for this probability is: \[ P(A) = \frac{\text{Ways to choose 2 red balls}}{\text{Ways to choose any 2 balls}} \] Finding the combinations: - Ways to choose 2 red balls: \({4 \choose 2} = 6\) - Total ways to choose any 2 balls from 10: \({10 \choose 2} = 45\) Thus, \[ P(A) = \frac{6}{45} = \frac{2}{15} \] **Calculating \(P(B)\):** To find the probability of drawing at least one red ball, we can use the complement approach: - The only scenario where no red balls are drawn is if both drawn balls are blue. Calculating the number of ways to choose 2 blue balls: \[ P(\text{both blue}) = \frac{{6 \choose 2}}{45} = \frac{15}{45} = \frac{1}{3} \] Thus, \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{1}{3} = \frac{2}{3} \] **Final Calculation:** Substituting back into the conditional probability formula: \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{2}{15}}{\frac{2}{3}} = \frac{2}{15} \times \frac{3}{2} = \frac{3}{15} = \frac{1}{5} \] --- # Final Answer \[ \boxed{\frac{1}{5}} \] Therefore, the probability that both balls drawn are red, given that at least one ball drawn is red, is \( \frac{1}{5} \). This demonstrates the application of conditional probability in solving real-world problems involving uncertainty.

# Given Data - Number of red balls (\(R\)) = 4 - Number of blue balls (\(B\)) = 6 - Total number of balls = \(R + B = 10\) - Two balls are drawn **one by one without replacement**. We need to find: The probability that **both balls are red**, given that **at least one ball is red**. --- # Definitions and Terms Used - **Conditional Probability**: The probability of an event occurring given that another event has already occurred, denoted as \(P(A|B)\). - **Without Replacement**: A situation where once an item is drawn, it is not returned to the original set for further draws. - Let: - \(A\): Event that **both balls drawn are red**. - \(B\): Event that **at least one ball drawn is red**. - Our goal is to calculate \(P(A|B)\). --- # Solution To determine \(P(A|B)\), we use the formula for conditional probability: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Since if both balls are red, it is guaranteed that at least one is red, we can express this relationship as: \[ P(A|B) = \frac{P(A)}{P(B)} \] **Calculating \(P(A)\):** The event \(A\) occurs when both drawn balls are red. The probability can be calculated as follows: \[ P(A) = \frac{\text{Ways to choose 2 red balls}}{\text{Ways to choose any 2 balls}} \] Calculating the combinations: - Ways to choose 2 red balls: \({4 \choose 2} = 6\) - Total ways to choose any 2 balls from the 10 available: \({10 \choose 2} = 45\) Thus, we find: \[ P(A) = \frac{6}{45} = \frac{2}{15} \] **Calculating \(P(B)\):** To find the probability of drawing at least one red ball, we can also employ the complement rule. The complement of event \(B\) is the scenario where both balls drawn are blue. We calculate this probability as follows: - Number of ways to choose 2 blue balls: \({6 \choose 2} = 15\) Now, the probability of both being blue is: \[ P(\text{both blue}) = \frac{15}{45} = \frac{1}{3} \] Consequently, the probability of at least one ball being red is: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{1}{3} = \frac{2}{3} \] **Final Calculation:** With our results for \(P(A)\) and \(P(B)\), we can substitute back into our formula for conditional probability: \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{2}{15}}{\frac{2}{3}} = \frac{2}{15} \times \frac{3}{2} = \frac{3}{15} = \frac{1}{5} \] --- # Final Answer \[ \boxed{\frac{1}{5}} \] Thus, the probability that both balls drawn are red, given that at least one ball drawn is red, is \( \frac{1}{5} \). This example illustrates the significance of conditional probability in determining outcomes based on prior knowledge, emphasizing analytical reasoning in probability theory. Understanding these concepts is essential in various applications, including statistics, risk assessment, and decision-making processes.

# Given Data - Number of red balls (\(R\)) = 4 - Number of blue balls (\(B\)) = 6 - Total number of balls = \(R + B = 10\) - Two balls are drawn **one by one without replacement**. We need to find: The probability that **both balls are red**, given that **at least one ball is red**. --- # Definitions and Terms Used - **Conditional Probability**: The likelihood of an event occurring given that another event has already occurred, represented as \(P(A|B)\). - **Without Replacement**: A method of drawing where each selected item is not returned to the original set before subsequent selections. - Let: - \(A\): Event that **both drawn balls are red**. - \(B\): Event that **at least one ball drawn is red**. - We aim to compute \(P(A|B)\). --- # Solution To derive \(P(A|B)\), we utilize the formula for conditional probability: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Since if both balls are red (event \(A\)), we inherently know that at least one ball is red (event \(B\)), this simplifies to: \[ P(A|B) = \frac{P(A)}{P(B)} \] **Calculating \(P(A)\):** The event \(A\) occurs when both drawn balls are red. The calculation for this probability is as follows: \[ P(A) = \frac{\text{Ways to choose 2 red balls}}{\text{Ways to choose any 2 balls}} \] Calculating combinations: - Ways to select 2 red balls: \({4 \choose 2} = 6\) - Total ways to choose any 2 balls from 10: \({10 \choose 2} = 45\) Thus, \[ P(A) = \frac{6}{45} = \frac{2}{15} \] **Calculating \(P(B)\):** To find the probability of drawing at least one red ball, we can use the complement method. The complement of event \(B\) is the scenario where both drawn balls are blue: - Number of ways to choose 2 blue balls: \({6 \choose 2} = 15\) Now, calculating the probability of both balls being blue: \[ P(\text{both blue}) = \frac{15}{45} = \frac{1}{3} \] Thus, the probability of at least one ball being red is: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{1}{3} = \frac{2}{3} \] **Final Calculation:** Combining our results for \(P(A)\) and \(P(B)\): \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{2}{15}}{\frac{2}{3}} = \frac{2}{15} \times \frac{3}{2} = \frac{3}{15} = \frac{1}{5} \] --- # Final Answer \[ \boxed{\frac{1}{5}} \] Therefore, the probability that both balls drawn are red, given that at least one ball drawn is red, is \( \frac{1}{5} \). This solution exemplifies the application of conditional probability in practical scenarios, showcasing how understanding relationships between events can lead to clearer insights in probability theory. Such concepts are fundamental in various fields, including statistics, finance, and decision-making.

# Given Data - Number of red balls (\(R\)) = 4 - Number of blue balls (\(B\)) = 6 - Total number of balls = \(R + B = 10\) - Two balls are drawn **one by one without replacement**. We need to find: The probability that **both balls are red**, given that **at least one ball is red**. --- # Definitions and Terms Used - **Conditional Probability**: The probability of an event occurring given that another event has occurred, denoted as \(P(A|B)\). - **Without Replacement**: A situation where a drawn item is not returned to the original set before the next draw. - Let: - \(A\): Event that **both balls drawn are red**. - \(B\): Event that **at least one ball drawn is red**. - Our objective is to compute \(P(A|B)\). --- # Solution To calculate \(P(A|B)\), we will use the conditional probability formula: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Since event \(A\) (both balls are red) guarantees event \(B\) (at least one ball is red), we can simplify: \[ P(A|B) = \frac{P(A)}{P(B)} \] **Calculating \(P(A)\)**: The probability that both drawn balls are red can be determined as follows: \[ P(A) = \frac{\text{Ways to choose 2 red balls}}{\text{Ways to choose any 2 balls}} \] Calculating the combinations: - Ways to select 2 red balls: \({4 \choose 2} = 6\) - Total ways to choose any 2 balls from 10: \({10 \choose 2} = 45\) Thus, we have: \[ P(A) = \frac{6}{45} = \frac{2}{15} \] **Calculating \(P(B)\)**: To find the probability of drawing at least one red ball, we can use the complement rule. The complement of event \(B\) is the situation where both balls drawn are blue: - Number of ways to choose 2 blue balls: \({6 \choose 2} = 15\) Now, calculating the probability of both balls being blue: \[ P(\text{both blue}) = \frac{15}{45} = \frac{1}{3} \] Thus, the probability of at least one ball being red is: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{1}{3} = \frac{2}{3} \] **Final Calculation**: Now that we have both probabilities, we can substitute them back into our formula for conditional probability: \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{2}{15}}{\frac{2}{3}} = \frac{2}{15} \times \frac{3}{2} = \frac{3}{15} = \frac{1}{5} \] --- # Final Answer \[ \boxed{\frac{1}{5}} \] Thus, the probability that both balls drawn are red, given that at least one ball drawn is red, is \( \frac{1}{5} \). This solution highlights the importance of conditional probability in analyzing outcomes based on previously known conditions, which is essential in many fields such as statistics, gaming theory, and risk management.

# Given Data - Number of red balls (\(R\)) = 4 - Number of blue balls (\(B\)) = 6 - Total number of balls = \(R + B = 10\) - The process of drawing two balls occurs **one by one without replacement**. We need to find: The probability that **both balls drawn are red**, given that **at least one ball drawn is red**. --- # Definitions and Terms Used - **Conditional Probability**: This is the probability of an event occurring given that another event has already occurred, represented mathematically as \(P(A|B)\). - **Without Replacement**: This means that once a ball is drawn, it is not placed back into the box before the next ball is drawn. - Let: - \(A\): The event that **both balls drawn are red**. - \(B\): The event that **at least one ball drawn is red**. - Our goal is to compute \(P(A|B)\), the probability of \(A\) given \(B\). --- # Solution To find \(P(A|B)\), we will utilize the formula for conditional probability: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Since if both balls are red, it is guaranteed that at least one ball is red, we can simplify the equation to: \[ P(A|B) = \frac{P(A)}{P(B)} \] **Calculating \(P(A)\)**: The probability that both drawn balls are red can be calculated by determining the number of successful outcomes over the total possible outcomes: \[ P(A) = \frac{\text{Ways to choose 2 red balls}}{\text{Ways to choose any 2 balls}} \] Calculating the combinations: - The number of ways to choose 2 red balls from 4: \[ {4 \choose 2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \] - The total number of ways to choose any 2 balls from 10: \[ {10 \choose 2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \] Thus, we find: \[ P(A) = \frac{6}{45} = \frac{2}{15} \] **Calculating \(P(B)\)**: To find the probability of drawing at least one red ball, we can use the complement method. The complement of event \(B\) is the scenario where both balls drawn are blue: - The number of ways to choose 2 blue balls from 6: \[ {6 \choose 2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \] Now, the probability of both balls being blue is: \[ P(\text{both blue}) = \frac{15}{45} = \frac{1}{3} \] Thus, we can determine the probability of at least one ball being red: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{1}{3} = \frac{2}{3} \] **Final Calculation**: Now that we have both probabilities calculated, we can substitute them back into our formula for conditional probability: \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{2}{15}}{\frac{2}{3}} = \frac{2}{15} \times \frac{3}{2} = \frac{3}{15} = \frac{1}{5} \] --- # Final Answer \[ \boxed{\frac{1}{5}} \] In conclusion, the probability that both balls drawn are red, given that at least one ball drawn is red, is \( \frac{1}{5} \). This scenario illustrates the concept of conditional probability effectively, showing how prior information (in this case, the knowledge that at least one ball is red) influences the likelihood of subsequent events. Understanding conditional probability is crucial in various fields, including statistics, economics, and decision-making frameworks, as it allows for more informed predictions based on available data. By embracing these statistical principles, one can develop a stronger foundation for analyzing uncertain situations and making sound judgments based on mathematical reasoning.

# Given Data - Number of red balls (\(R\)) = 4 - Number of blue balls (\(B\)) = 6 - Total number of balls = \(R + B = 10\) - The process of drawing two balls occurs **one by one without replacement**. We need to find: The probability that **both balls drawn are red**, given that **at least one ball drawn is red**. --- # Definitions and Terms Used - **Conditional Probability**: This is defined as the probability of an event occurring given that another event has occurred, mathematically represented as \(P(A|B)\). - **Without Replacement**: This means that once a ball is drawn from the box, it is not put back before the next ball is drawn. - Let: - \(A\): The event that **both balls drawn are red**. - \(B\): The event that **at least one ball drawn is red**. - The goal is to compute \(P(A|B)\), which is the probability of event \(A\) occurring given that event \(B\) has occurred. --- # Solution To calculate \(P(A|B)\), we will use the formula for conditional probability: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Since if both balls are red, it is guaranteed that at least one ball is red, we can simplify this to: \[ P(A|B) = \frac{P(A)}{P(B)} \] **Calculating \(P(A)\)**: The probability that both drawn balls are red can be determined by finding the number of successful outcomes over the total possible outcomes: \[ P(A) = \frac{\text{Ways to choose 2 red balls}}{\text{Ways to choose any 2 balls}} \] Calculating the combinations: - The number of ways to choose 2 red balls from 4: \[ {4 \choose 2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \] - The total number of ways to choose any 2 balls from 10: \[ {10 \choose 2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \] Thus, we find: \[ P(A) = \frac{6}{45} = \frac{2}{15} \] **Calculating \(P(B)\)**: To find the probability of drawing at least one red ball, we can use the complement method. The complement of event \(B\) is the scenario where both balls drawn are blue: - The number of ways to choose 2 blue balls from 6: \[ {6 \choose 2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \] Now, the probability of both balls being blue is: \[ P(\text{both blue}) = \frac{15}{45} = \frac{1}{3} \] Thus, we can determine the probability of at least one ball being red: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{1}{3} = \frac{2}{3} \] **Final Calculation**: Now that we have both probabilities calculated, we can substitute them back into our formula for conditional probability: \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{2}{15}}{\frac{2}{3}} = \frac{2}{15} \times \frac{3}{2} = \frac{3}{15} = \frac{1}{5} \] --- # Final Answer \[ \boxed{\frac{1}{5}} \] In conclusion, the probability that both balls drawn are red, given that at least one ball drawn is red, is \( \frac{1}{5} \). This scenario illustrates the concept of conditional probability effectively, showing how prior information (in this case, the knowledge that at least one ball is red) influences the likelihood of subsequent events. ### Additional Insights Understanding conditional probability is crucial in various fields such as statistics, finance, and decision-making frameworks, as it allows for more informed predictions based on available data. The ability to analyze events in this manner can also extend to more complex scenarios involving multiple variables and outcomes. In practical applications, this knowledge is used in risk management and strategic planning, where decision-makers must weigh the probabilities of various outcomes based on existing conditions. Furthermore, it emphasizes the importance of data-driven decision-making, where understanding relationships between different events leads to better outcomes and insights. This problem serves as a foundational example in probability theory, providing a clear illustration of how to apply mathematical reasoning to real-world situations. Engaging with such problems not only enhances analytical skills but also fosters a deeper appreciation for the role of chance in everyday life.

# Given Data - Number of red balls (\(R\)) = 4 - Number of blue balls (\(B\)) = 6 - Total number of balls = \(R + B = 10\) - The process of drawing two balls occurs **one by one without replacement**. We need to find: The probability that **both balls drawn are red**, given that **at least one ball drawn is red**. --- # Definitions and Terms Used - **Conditional Probability**: This is defined as the probability of an event occurring given that another event has occurred, mathematically represented as \(P(A|B)\). - **Without Replacement**: This means that once a ball is drawn from the box, it is not put back before the next ball is drawn. - Let: - \(A\): The event that **both balls drawn are red**. - \(B\): The event that **at least one ball drawn is red**. - The goal is to compute \(P(A|B)\), which is the probability of event \(A\) occurring given that event \(B\) has occurred. --- # Solution To calculate \(P(A|B)\), we will use the formula for conditional probability: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Since if both balls are red, it is guaranteed that at least one ball is red, we can simplify the equation to: \[ P(A|B) = \frac{P(A)}{P(B)} \] **Calculating \(P(A)\)**: The probability that both drawn balls are red can be determined by finding the number of successful outcomes over the total possible outcomes: \[ P(A) = \frac{\text{Ways to choose 2 red balls}}{\text{Ways to choose any 2 balls}} \] Calculating the combinations: - The number of ways to choose 2 red balls from 4: \[ {4 \choose 2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \] - The total number of ways to choose any 2 balls from 10: \[ {10 \choose 2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \] Thus, we find: \[ P(A) = \frac{6}{45} = \frac{2}{15} \] **Calculating \(P(B)\)**: To find the probability of drawing at least one red ball, we can use the complement method. The complement of event \(B\) is the scenario where both balls drawn are blue: - The number of ways to choose 2 blue balls from 6: \[ {6 \choose 2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \] Now, the probability of both balls being blue is: \[ P(\text{both blue}) = \frac{15}{45} = \frac{1}{3} \] Thus, we can determine the probability of at least one ball being red: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{1}{3} = \frac{2}{3} \] **Final Calculation**: Now that we have both probabilities calculated, we can substitute them back into our formula for conditional probability: \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{2}{15}}{\frac{2}{3}} = \frac{2}{15} \times \frac{3}{2} = \frac{3}{15} = \frac{1}{5} \] --- # Final Answer \[ \boxed{\frac{1}{5}} \] Thus, the probability that both balls drawn are red, given that at least one ball drawn is red, is \( \frac{1}{5} \). This calculation highlights the significance of conditional probability in evaluating scenarios based on prior knowledge, which is applicable in various fields such as gambling, finance, and everyday decision-making. ### Further Considerations Understanding conditional probability is essential for interpreting real-life situations where outcomes depend on previously established conditions. This concept is foundational in statistics and probability theory and is crucial for making informed decisions in uncertain environments. In practical applications, such as market research or risk assessment, decision-makers can leverage conditional probabilities to forecast trends or outcomes based on specific conditions. Additionally, this understanding aids in developing strategies that maximize favorable outcomes while mitigating risks. Engaging with such problems enhances analytical skills and encourages a deeper appreciation for the role of chance in various domains.

# Given Data - Number of red balls (\(R\)) = 4 - Number of blue balls (\(B\)) = 6 - Total number of balls = \(R + B = 10\) - The process of drawing two balls occurs **one by one without replacement**. We need to find: The probability that **both balls drawn are red**, given that **at least one ball drawn is red**. --- # Definitions and Terms Used - **Conditional Probability**: This is defined as the probability of an event occurring given that another event has occurred, mathematically represented as \(P(A|B)\). - **Without Replacement**: This means that once a ball is drawn from the box, it is not put back before the next ball is drawn. - Let: - \(A\): The event that **both balls drawn are red**. - \(B\): The event that **at least one ball drawn is red**. - Our goal is to compute \(P(A|B)\), which is the probability of event \(A\) occurring given that event \(B\) has occurred. --- # Solution To calculate \(P(A|B)\), we will use the formula for conditional probability: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Since if both balls are red, it is guaranteed that at least one ball is red, we can simplify the equation to: \[ P(A|B) = \frac{P(A)}{P(B)} \] **Calculating \(P(A)\)**: The probability that both drawn balls are red can be determined by finding the number of successful outcomes over the total possible outcomes: \[ P(A) = \frac{\text{Ways to choose 2 red balls}}{\text{Ways to choose any 2 balls}} \] Calculating the combinations: - The number of ways to choose 2 red balls from 4: \[ {4 \choose 2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \] - The total number of ways to choose any 2 balls from 10: \[ {10 \choose 2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \] Thus, we find: \[ P(A) = \frac{6}{45} = \frac{2}{15} \] **Calculating \(P(B)\)**: To find the probability of drawing at least one red ball, we can use the complement method. The complement of event \(B\) is the scenario where both balls drawn are blue: - The number of ways to choose 2 blue balls from 6: \[ {6 \choose 2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \] Now, the probability of both balls being blue is: \[ P(\text{both blue}) = \frac{15}{45} = \frac{1}{3} \] Thus, we can determine the probability of at least one ball being red: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{1}{3} = \frac{2}{3} \] **Final Calculation**: Now that we have both probabilities calculated, we can substitute them back into our formula for conditional probability: \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{2}{15}}{\frac{2}{3}} = \frac{2}{15} \times \frac{3}{2} = \frac{3}{15} = \frac{1}{5} \] --- # Final Answer \[ \boxed{\frac{1}{5}} \] Thus, the probability that both balls drawn are red, given that at least one ball drawn is red, is \( \frac{1}{5} \). ### Practical Implications This problem exemplifies how conditional probability can be applied in decision-making scenarios. For instance, in quality control within manufacturing, knowing certain conditions can drastically change the assessment of product quality. By mastering concepts like conditional probability, one can enhance their analytical skills, leading to better forecasting and strategic planning in various professional fields, including finance, healthcare, and operations management. This understanding also enables individuals to interpret results in studies and surveys more effectively, providing insights into how certain factors influence outcomes. For example, in market research, knowing the probability of a consumer making a purchase based on prior interest can guide marketing strategies effectively. By engaging with these mathematical concepts, one not only improves their problem-solving abilities but also gains a deeper appreciation for how probability shapes decision-making processes in everyday life. Each of these applications underscores the relevance of probability theory in making informed choices in uncertain environments.

# Given Data - Number of red balls (\(R\)) = 4 - Number of blue balls (\(B\)) = 6 - Total number of balls = \(R + B = 10\) - The process of drawing two balls occurs **one by one without replacement**. We need to find: The probability that **both balls drawn are red**, given that **at least one ball drawn is red**. --- # Definitions and Terms Used - **Conditional Probability**: The probability of an event occurring given that another event has occurred, expressed as \(P(A|B)\). - **Without Replacement**: A situation where a drawn item is not returned to the original set before the next item is drawn. - Let: - \(A\): The event that **both drawn balls are red**. - \(B\): The event that **at least one ball drawn is red**. - Our goal is to compute \(P(A|B)\). --- # Solution To calculate \(P(A|B)\), we will utilize the formula for conditional probability: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Since if both balls are red (event \(A\)), it is certain that at least one ball is red (event \(B\)), we can rewrite this as: \[ P(A|B) = \frac{P(A)}{P(B)} \] **Calculating \(P(A)\)**: The probability that both drawn balls are red can be determined as follows: \[ P(A) = \frac{\text{Ways to choose 2 red balls}}{\text{Ways to choose any 2 balls}} \] Calculating the combinations: - The number of ways to choose 2 red balls from 4: \[ {4 \choose 2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \] - The total number of ways to choose any 2 balls from 10: \[ {10 \choose 2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \] Thus, we find: \[ P(A) = \frac{6}{45} = \frac{2}{15} \] **Calculating \(P(B)\)**: To find the probability of drawing at least one red ball, we can use the complement method. The complement of event \(B\) is the scenario where both balls drawn are blue: - The number of ways to choose 2 blue balls from 6: \[ {6 \choose 2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \] Now, the probability of both balls being blue is: \[ P(\text{both blue}) = \frac{15}{45} = \frac{1}{3} \] Thus, the probability of at least one ball being red is: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{1}{3} = \frac{2}{3} \] **Final Calculation**: Now that we have both probabilities calculated, we can substitute them back into our formula for conditional probability: \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{2}{15}}{\frac{2}{3}} = \frac{2}{15} \times \frac{3}{2} = \frac{3}{15} = \frac{1}{5} \] --- # Final Answer \[ \boxed{\frac{1}{5}} \] Thus, the probability that both balls drawn are red, given that at least one ball drawn is red, is \( \frac{1}{5} \). ### Implications and Applications Understanding conditional probability is vital in various fields, such as finance, healthcare, and decision-making. This knowledge allows individuals to assess risks and make informed decisions based on available information. For instance, in healthcare, knowing the probability of a patient having a disease given a positive test result can guide treatment decisions. In finance, understanding the likelihood of market trends based on historical data helps investors make better choices. Moreover, this concept is foundational in statistical analysis, enabling researchers to interpret data accurately. By developing a strong grasp of conditional probability, one can enhance their analytical skills, leading to improved problem-solving capabilities across diverse scenarios. In conclusion, this problem serves as an excellent example of how mathematical concepts are applied to real-world situations, emphasizing the importance of probability theory in everyday life. Engaging with such examples fosters a deeper understanding of chance and uncertainty, which is crucial in both personal and professional contexts.

# Given Data - Number of red balls (R) = 4 - Number of blue balls (B) = 6 - Total number of balls = R + B = 10 - The process of drawing two balls occurs one by one without replacement. We need to find: The probability that both balls drawn are red, given that at least one ball drawn is red. --- # Definitions and Terms Used - Conditional Probability: The probability of an event occurring given that another event has occurred, expressed as P(A|B). - Without Replacement: Once a ball is drawn, it is not returned to the original set before the next ball is drawn. - Let: - A: The event that both drawn balls are red. - B: The event that at least one ball drawn is red. - The goal is to compute P(A|B), the probability of event A occurring given that event B has occurred. --- # Solution To calculate P(A|B), we will use the formula for conditional probability: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Since if both balls are red, it is guaranteed that at least one ball is red, we can simplify the equation to: \[ P(A|B) = \frac{P(A)}{P(B)} \] Calculating P(A): The probability that both drawn balls are red can be determined by finding the number of successful outcomes over the total possible outcomes: \[ P(A) = \frac{\text{Ways to choose 2 red balls}}{\text{Ways to choose any 2 balls}} \] Calculating the combinations: - The number of ways to choose 2 red balls from 4: \[ {4 \choose 2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \] - The total number of ways to choose any 2 balls from 10: \[ {10 \choose 2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \] Thus, we find: \[ P(A) = \frac{6}{45} = \frac{2}{15} \] Calculating P(B): To find the probability of drawing at least one red ball, we can use the complement method. The complement of event B is the scenario where both balls drawn are blue: - The number of ways to choose 2 blue balls from 6: \[ {6 \choose 2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \] Now, the probability of both balls being blue is: \[ P(\text{both blue}) = \frac{15}{45} = \frac{1}{3} \] Thus, the probability of at least one ball being red is: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{1}{3} = \frac{2}{3} \] Final Calculation: Now that we have both probabilities calculated, we can substitute them back into our formula for conditional probability: \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{2}{15}}{\frac{2}{3}} = \frac{2}{15} \times \frac{3}{2} = \frac{3}{15} = \frac{1}{5} \] --- # Final Answer \[ \boxed{\frac{1}{5}} \] The probability that both balls drawn are red, given that at least one ball drawn is red, is \(\frac{1}{5}\). ### Observations This problem demonstrates the usefulness of conditional probability in evaluating scenarios based on prior knowledge. Understanding conditional probability is essential across various fields, enabling better decision-making by assessing risks and outcomes. For instance, in medical diagnostics, knowing the probability of a condition based on test results can guide treatment choices. In marketing, analyzing consumer behavior based on previous purchases can inform strategies. The application of these concepts extends to statistical modeling and data analysis, where recognizing relationships between events informs predictions and conclusions. Engaging with such problems enhances analytical abilities and fosters a deeper understanding of uncertainty in real-world situations.

# Given Data - Number of red balls \(R = 4\) - Number of blue balls \(B = 6\) - Total number of balls \(= R + B = 10\) - Two balls are drawn **one by one without replacement**. We need to find: The probability that **both balls are red**, given that **at least one ball drawn is red**. --- # Definitions and Terms Used - **Conditional Probability**: This refers to the probability of an event occurring given that another event has already occurred, denoted as \(P(A|B)\). - **Without Replacement**: A method where a drawn ball is not returned to the box before the next draw. - Let: - \(A\): Event that **both balls drawn are red**. - \(B\): Event that **at least one ball drawn is red**. - We aim to compute \(P(A|B)\). --- # Solution To find \(P(A|B)\), we use the formula for conditional probability: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Since drawing two red balls guarantees that at least one of them is red, we simplify this to: \[ P(A|B) = \frac{P(A)}{P(B)} \] **Calculating \(P(A)\)**: To find the probability that both balls drawn are red: \[ P(A) = \frac{\text{Ways to choose 2 red balls}}{\text{Ways to choose any 2 balls}} \] Calculating the combinations: - The number of ways to choose 2 red balls from 4: \[ {4 \choose 2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \] - The total number of ways to choose any 2 balls from 10: \[ {10 \choose 2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \] Thus, we have: \[ P(A) = \frac{6}{45} = \frac{2}{15} \] **Calculating \(P(B)\)**: To find the probability of drawing at least one red ball, we can compute the complement where both balls are blue: - The number of ways to choose 2 blue balls from 6: \[ {6 \choose 2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \] The probability of both balls being blue is: \[ P(\text{both blue}) = \frac{15}{45} = \frac{1}{3} \] Therefore, the probability of at least one ball being red is: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{1}{3} = \frac{2}{3} \] **Final Calculation**: Substituting the values into the formula for conditional probability: \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{2}{15}}{\frac{2}{3}} = \frac{2}{15} \times \frac{3}{2} = \frac{3}{15} = \frac{1}{5} \] --- # Final Answer \[ \boxed{\frac{1}{5}} \] The probability that both balls drawn are red, given that at least one ball drawn is red, is \(\frac{1}{5}\). ### Insights and Applications This example of conditional probability illustrates how prior knowledge affects the assessment of outcomes. Mastery of this concept is crucial in various fields, including statistics, finance, and risk management. In practical scenarios, such as market analysis, understanding the likelihood of consumer behavior based on previous purchases helps in strategizing marketing efforts. Similarly, in health care, evaluating the probability of a disease given a specific test result informs treatment decisions. Engaging with problems like this fosters critical thinking and enhances one's ability to navigate uncertain situations by utilizing probabilistic reasoning. This foundation is essential not only in academic contexts but also in real-world applications, reinforcing the importance of statistical literacy in decision-making processes.

# Given Data - Number of red balls \(R = 4\) - Number of blue balls \(B = 6\) - Total number of balls \(= R + B = 10\) - Two balls are drawn **one by one without replacement**. We need to find: The probability that **both balls drawn are red**, given that **at least one ball drawn is red**. --- # Definitions and Terms Used - **Conditional Probability**: The probability of an event occurring given that another event has occurred, denoted as \(P(A|B)\). - **Without Replacement**: A method in which once a ball is drawn, it is not returned to the original set for the next draw. - Let: - \(A\): Event that **both balls drawn are red**. - \(B\): Event that **at least one ball drawn is red**. - The aim is to compute \(P(A|B)\). --- # Solution To find \(P(A|B)\), we use the formula for conditional probability: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Since if both balls are red (event \(A\)), it is guaranteed that at least one ball is red (event \(B\)), we can simplify this to: \[ P(A|B) = \frac{P(A)}{P(B)} \] **Calculating \(P(A)\)**: The probability that both drawn balls are red can be determined as follows: \[ P(A) = \frac{\text{Ways to choose 2 red balls}}{\text{Ways to choose any 2 balls}} \] Calculating the combinations: - The number of ways to choose 2 red balls from 4: \[ {4 \choose 2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \] - The total number of ways to choose any 2 balls from 10: \[ {10 \choose 2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \] Thus, we find: \[ P(A) = \frac{6}{45} = \frac{2}{15} \] **Calculating \(P(B)\)**: To find the probability of drawing at least one red ball, we can utilize the complement method. The complement of event \(B\) is the scenario where both balls drawn are blue: - The number of ways to choose 2 blue balls from 6: \[ {6 \choose 2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \] The probability of both balls being blue is: \[ P(\text{both blue}) = \frac{15}{45} = \frac{1}{3} \] Thus, the probability of at least one ball being red is: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{1}{3} = \frac{2}{3} \] **Final Calculation**: Now that we have both probabilities calculated, we can substitute them back into our formula for conditional probability: \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{2}{15}}{\frac{2}{3}} = \frac{2}{15} \times \frac{3}{2} = \frac{3}{15} = \frac{1}{5} \] --- # Final Answer \[ \boxed{\frac{1}{5}} \] The probability that both balls drawn are red, given that at least one ball drawn is red, is \(\frac{1}{5}\). ### Observations and Applications This example illustrates the application of conditional probability in evaluating outcomes based on prior knowledge. Understanding the principles of conditional probability is essential across various fields, including finance, statistics, and decision-making. For instance, in finance, investors can use conditional probabilities to assess the likelihood of a stock performing well based on market conditions. In healthcare, understanding probabilities related to symptoms and diseases helps in making accurate diagnoses. Engaging with problems such as this not only enhances analytical skills but also fosters a deeper understanding of the role of chance and uncertainty in real-world scenarios. This foundation is crucial for effective problem-solving and informed decision-making in both personal and professional contexts.

# Given Data - Number of red balls \(R = 4\) - Number of blue balls \(B = 6\) - Total number of balls \(= R + B = 10\) - Two balls are drawn **one by one without replacement**. We need to find: The probability that **both balls drawn are red**, given that **at least one ball drawn is red**. --- # Definitions and Terms Used - **Conditional Probability**: This is defined as the probability of an event occurring given that another event has occurred, represented as \(P(A|B)\). - **Without Replacement**: This means that once a ball is drawn, it is not placed back into the box before the next draw. - Let: - \(A\): The event that **both balls drawn are red**. - \(B\): The event that **at least one ball drawn is red**. - Our goal is to compute \(P(A|B)\). --- # Solution To find \(P(A|B)\), we will utilize the formula for conditional probability: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Since if both balls are red, at least one must also be red, we can simplify this to: \[ P(A|B) = \frac{P(A)}{P(B)} \] **Calculating \(P(A)\)**: The probability that both drawn balls are red can be calculated using: \[ P(A) = \frac{\text{Ways to choose 2 red balls}}{\text{Ways to choose any 2 balls}} \] Calculating the combinations: - Ways to choose 2 red balls from 4: \[ {4 \choose 2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \] - Total ways to choose any 2 balls from 10: \[ {10 \choose 2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \] Thus, we find: \[ P(A) = \frac{6}{45} = \frac{2}{15} \] **Calculating \(P(B)\)**: To find the probability of drawing at least one red ball, we can use the complement approach. The complement of event \(B\) is the situation where both balls drawn are blue: - Ways to choose 2 blue balls from 6: \[ {6 \choose 2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \] Now, the probability of both balls being blue is: \[ P(\text{both blue}) = \frac{15}{45} = \frac{1}{3} \] Therefore, the probability of at least one ball being red is: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{1}{3} = \frac{2}{3} \] **Final Calculation**: Now that we have computed both probabilities, we can substitute them into our formula for conditional probability: \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{2}{15}}{\frac{2}{3}} = \frac{2}{15} \times \frac{3}{2} = \frac{3}{15} = \frac{1}{5} \] --- # Final Answer \[ \boxed{\frac{1}{5}} \] The probability that both balls drawn are red, given that at least one ball drawn is red, is \(\frac{1}{5}\). ### Insights and Practical Applications This example highlights how conditional probability can be used to evaluate outcomes based on prior knowledge about events. Understanding this concept is crucial in various fields such as finance, healthcare, and behavioral sciences. In finance, for instance, conditional probabilities can inform investors about the likelihood of a stock's performance based on historical trends and market conditions. In healthcare, probabilities related to symptoms and test results can guide diagnosis and treatment strategies. Engaging with problems like this enhances analytical skills and promotes a deeper understanding of uncertainty in real-world situations. This foundation supports effective problem-solving and informed decision-making in both personal and professional contexts.