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Let's solve the problem described in the image.

### Problem Recap:
- There are 2 balls that are randomly placed into 3 boxes.
- Each box can contain 0, 1, or 2 balls.
- The player can choose any 2 boxes to open. The player wins if both balls are in the selected boxes.

### Step 1: Calculate the Total Number of Possible Outcomes
Since the balls are identical, the possible distributions of the 2 balls into the 3 boxes can be represented as follows:
- Both balls in one box: (2,0,0), (0,2,0), (0,0,2)
- One ball in two different boxes: (1,1,0), (1,0,1), (0,1,1)

There are 6 distinct ways to distribute the balls across the three boxes.

### Step 2: Calculate the Favorable Outcomes
The player wins if they choose the two boxes that contain the balls. For each possible distribution, we need to determine how many of the combinations of two boxes will result in the player winning.

Let's analyze the favorable cases for each distribution:

1. **(2,0,0), (0,2,0), (0,0,2):**
   - The player must select the one box with both balls to win.
   - Number of favorable selections = 1

2. **(1,1,0), (1,0,1), (0,1,1):**
   - The player must select the two boxes that each contain one ball to win.
   - Number of favorable selections = 1

### Step 3: Calculate the Probability of Winning
Since each of the 6 distributions is equally likely, we can calculate the probability by counting the number of favorable outcomes and dividing by the total number of possibilities:

- **Total number of possible selections** = \(\binom{3}{2} = 3\)
- **Number of favorable outcomes:**
  - For the distributions where both balls are in the same box: \(3 \times 1 = 3\)
  - For the distributions where the balls are in different boxes: \(3 \times 1 = 3\)
  
Total favorable outcomes = \(3 + 3 = 6\)

- **Total number of outcomes** = \(6\) (total possible distributions) × \(3\) (ways to choose two boxes) = \(18\)

Thus, the probability of winning is:
\[
\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{18} = \frac{1}{3}
\]

### Final Answer:
The correct answer is **A. \( \frac{1}{3} \)**.

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Do you want further details, or do you have any questions? Here are some related questions you might find interesting:

1. What is the probability of the player losing the game?
2. How would the probability change if there were 3 balls instead of 2?
3. How does the probability change if the player can only open one box?
4. How would the problem change if the boxes were distinguishable (e.g., labeled A, B, C)?
5. What if the player can open all three boxes and needs to find exactly two balls to win?

**Tip:** Understanding the possible distributions and the importance of distinguishing identical vs. non-identical objects can help solve more complex probability problems.

Explore how to calculate the probability of winning a game with 2 balls randomly placed in 3 boxes. Understand the concept of probability, using the binomial coefficient to determine the number of possible outcomes and favorable outcomes. Suitable for Grades 10-12 students and anyone interested in probability theory.

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