# Probability Problem: Drawing Balls Without Replacement ## Problem Restatement A box contains: - 4 red balls - 5 blue balls - 6 green balls Total balls: \(4 + 5 + 6 = 15\). Two balls are drawn **one after another without replacement**. Given that **at least one ball is green**, what is the probability that **both balls are green**? --- ## Step 1: Define Events Let: - \(G\): ball is green - \(A\): both drawn balls are green - \(B\): at least one drawn ball is green We need to find: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Note: \(A \Rightarrow B\), so \(A \cap B = A\). \[ P(A|B) = \frac{P(A)}{P(B)} \] --- ## Step 2: Calculate \(P(A)\) (Both Balls Are Green) Number of ways to choose 2 green balls: \[ \binom{6}{2} = 15 \] Total number of ways to choose 2 balls: \[ \binom{15}{2} = 105 \] So, \[ P(A) = \frac{15}{105} = \frac{1}{7} \] --- ## Step 3: Calculate \(P(B)\) (At Least One Ball Is Green) It's often easier to use complementary counting: - Number of ways **no** green balls are drawn (only red or blue): - Total non-green balls: \(4+5=9\) - Ways to draw 2 non-green balls: \(\binom{9}{2} = 36\) - Therefore, ways to draw **at least one** green ball: \[ \text{Total} - \text{No green} = 105 - 36 = 69 \] - So, \[ P(B) = \frac{69}{105} \] --- ## Step 4: Compute the Conditional Probability \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{1}{7}}{\frac{69}{105}} = \frac{1}{7} \times \frac{105}{69} = \frac{15}{69} = \frac{5}{23} \] --- ## Final Answer \[ \boxed{\frac{5}{23}} \] **Given at least one ball is green, the probability that both balls are green is \(\frac{5}{23}\).**

# Probability Problem: Drawing Balls Without Replacement ## Definitions and Concepts ### Basic Probability Concepts - **Probability**: A measure of the likelihood of an event occurring, expressed as a ratio of favorable outcomes to total outcomes. - **Conditional Probability**: The probability of an event \(A\) given that another event \(B\) has occurred, denoted as \(P(A|B)\). ### Events 1. **Event \(G\)**: A ball drawn is green. 2. **Event \(A\)**: Both drawn balls are green. 3. **Event \(B\)**: At least one drawn ball is green. ### Conditional Probability Formula To find the conditional probability of event \(A\) given event \(B\), we use: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] Since \(A\) implies \(B\) (if both balls are green, at least one is green), we have: \[ P(A \cap B) = P(A) \] Thus, the formula simplifies to: \[ P(A|B) = \frac{P(A)}{P(B)} \] --- ## Problem Restatement A box contains: - 4 red balls - 5 blue balls - 6 green balls Total balls: \(4 + 5 + 6 = 15\). Two balls are drawn **one after another without replacement**. Given that **at least one ball is green**, find the probability that **both balls are green**. --- ## Step-by-Step Solution ### Step 1: Calculate \(P(A)\) (Both Balls Are Green) - Number of ways to choose 2 green balls: \[ \binom{6}{2} = 15 \] - Total number of ways to choose 2 balls: \[ \binom{15}{2} = 105 \] - Thus, \[ P(A) = \frac{15}{105} = \frac{1}{7} \] ### Step 2: Calculate \(P(B)\) (At Least One Ball Is Green) Using complementary counting: - Total non-green balls: \(4 + 5 = 9\) - Ways to draw 2 non-green balls: \[ \binom{9}{2} = 36 \] - Therefore, ways to draw **at least one** green ball: \[ \text{Total} - \text{No green} = 105 - 36 = 69 \] - Thus, \[ P(B) = \frac{69}{105} \] ### Step 3: Compute the Conditional Probability Substituting values into the conditional probability formula: \[ P(A|B) = \frac{P(A)}{P(B)} = \frac{\frac{1}{7}}{\frac{69}{105}} = \frac{1}{7} \times \frac{105}{69} = \frac{15}{69} = \frac{5}{23} \] --- ## Final Answer \[ \boxed{\frac{5}{23}} \] **Given at least one ball is green, the probability that both balls are green is \(\frac{5}{23}\).**