Let $x$ represent the number of green balls in the box. Thus, there are a total of $3 + x$ balls in the box.

Step 1: Calculate the total ways to draw 2 balls The number of ways to select 2 balls from $3 + x$ balls is: $\binom{3 + x}{2} = \frac{(3 + x)(2 + x)}{2}$

Step 2: Calculate the ways to have one blue and one green ball To have one ball of each color (one blue and one green), we choose 1 blue ball from the 3 blue balls and 1 green ball from the $x$ green balls. The number of ways to do this is: $\binom{3}{1} \times \binom{x}{1} = 3 \times x = 3x$

Step 3: Set up the probability equation According to the problem, the probability of drawing one blue and one green ball is $\frac{1}{2}$: $\frac{3x}{\binom{3 + x}{2}} = \frac{1}{2}$

Step 4: Substitute $\binom{3 + x}{2}$ and solve $\frac{3x}{\frac{(3 + x)(2 + x)}{2}} = \frac{1}{2}$ Multiply both sides by 2 to clear the fraction: $\frac{6x}{(3 + x)(2 + x)} = 1$ Now, cross-multiply: $6x = (3 + x)(2 + x)$ Expand the right side: $6x = 6 + 5x + x^2$ Rearrange the equation to form a standard quadratic equation: $x^2 - x - 6 = 0$ Step 5: Solve the quadratic equation The equation $x^2 - x - 6 = 0$ factors as: $(x - 3)(x + 2) = 0$ Thus, $x = 3$ or $x = -2$.

Since the number of green balls, $x$, must be a positive integer, we have: $x = 3$

Answer: The possible number of green balls in the box is 3.

Would you like further explanation or have any questions?

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Here are some related questions you might consider:

1. How would the probability change if there were 4 blue balls?
2. What if we added more colors to the box, like red balls?
3. How does the calculation adjust if we draw more than two balls?
4. What if the probability of drawing one blue and one green ball was different?
5. How would adding one more green ball change the total probability?

Tip: When working with probability, it's helpful to write out each step and label which outcomes correspond to each event.