There are *x* black balls, *x - 2* yellow balls, and *x + 2* green balls in a bag A. The probability of selecting two black balls when two balls are drawn at random from the bag is $\frac{1}{12}$. Find the total number of balls in the bag.

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**Solution:**

1. Let the number of black balls be *x*.

2. Then, the number of yellow balls is *x - 2* and the number of green balls is *x + 2*.

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

4. The probability of drawing two black balls when two balls are picked randomly is given by:
   $$	ext{Probability} = \frac{\binom{x}{2}}{\binom{3x}{2}} = \frac{1}{12}$$

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

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

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

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

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

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

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

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

13. Total balls in the bag:
    $$3x = 3 	imes 3 = 9$$

**Answer:** 9

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