There are *x* red balls, *x - 2* green balls, and *x + 2* blue balls in a bag. The probability of drawing 2 red balls consecutively without replacement is $\frac{1}{12}$. What is the total number of balls in the bag?

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**Solution:**

1. Let the number of red balls be *x*.
   - Number of green balls = *x - 2*
   - Number of blue balls = *x + 2*

2. Total number of balls in the bag:
   $$	ext{Total} = x + (x - 2) + (x + 2) = 3x$$

3. Probability of drawing 2 red balls without replacement is given by:
   $$P = \frac{\binom{x}{2}}{\binom{3x}{2}} = \frac{1}{12}$$

4. Calculate the combinations:
   $$\binom{x}{2} = \frac{x(x - 1)}{2}$$
   $$\binom{3x}{2} = \frac{3x(3x - 1)}{2}$$

5. Substitute these into the probability equation:
   $$\frac{\frac{x(x - 1)}{2}}{\frac{3x(3x - 1)}{2}} = \frac{1}{12}$$

6. Simplify the equation:
   $$\frac{x(x - 1)}{3x(3x - 1)} = \frac{1}{12}$$

7. Cancel *x* from numerator and denominator (since *x* > 0):
   $$\frac{x - 1}{3(3x - 1)} = \frac{1}{12}$$

8. Cross-multiply:
   $$12(x - 1) = 3(3x - 1)$$

9. Expand both sides:
   $$12x - 12 = 9x - 3$$

10. Rearrange terms:
    $$12x - 9x = -3 + 12$$
    $$3x = 9$$

11. Solve for *x*:
    $$x = 3$$

12. Total number of balls:
    $$3x = 3 	imes 3 = 9$$

**Answer:** 9

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