Answer:
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Explanation:
To solve this problem, we can use conditional probability. Let's define events:
- Event A: Drawing a red ball.
- Event B: Drawing a green ball.
The information given is that at least one ball is red. This means either the first ball is red, or the second ball is red, or both. We want to find the probability of drawing a green ball given that at least one ball is red, denoted as \( P(B|A) \).
\[ P(B|A) = \frac{P(B \cap A)}{P(A)} \]
Where:
- \( P(B|A) \) is the probability of drawing a green ball given that at least one ball is red.
- \( P(B \cap A) \) is the probability of drawing a green ball and a red ball.
- \( P(A) \) is the probability of drawing at least one red ball.
Since we know at least one ball is red, the probability of drawing at least one red ball is 1 (certain). Now, we need to consider the probability of drawing a green ball and a red ball.
If one ball is red, the other can be green. So, the probability of drawing a green ball and a red ball is the same as the probability of drawing a red ball and a green ball.
\[ P(B \cap A) = P(A \cap B) \]
Therefore,
\[ P(B|A) = \frac{2 \cdot P(A \cap B)}{1} \]
Now, if there are \( n \) balls in total, and at least one is red, there are \( n-1 \) possibilities for the other ball to be green.
\[ P(B|A) = \frac{2 \cdot (n-1)}{1 \cdot n} \]
Given that there are two colors (red and green), and we know at least one ball is red, the total number of balls is \( n = 2 \).
\[ P(B|A) = \frac{2 \cdot (2-1)}{1 \cdot 2} = \frac{2}{2} = 1 \]
Therefore, the probability that the other ball is green, given that at least one ball is red, is \( 1 \).