A box contains 12 balls of which some are red in colour. If 6 more red balls are put in the box and a ball is drawn at random, the probability of drawing a red ball doubles than what it was before. Find the number of red balls in the box.

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step-by-Step Solution:

1. **Define the Variables**:
Let the number of red balls originally in the box be \( x \).

2. **Calculate the Total Balls After Adding More Red Balls**:
Initially, there are 12 balls in total. After adding 6 more red balls, the total number of balls becomes:
\[ 12 + 6 = 18 \]

3. **Calculate the Original Probability of Drawing a Red Ball**:
The probability of drawing a red ball originally is given by the formula:
\[ P(\text{Red})_{\text{old}} = \frac{\text{Number of Red Balls}}{\text{Total Number of Balls}} = \frac{x}{12} \]

4. **Calculate the New Probability of Drawing a Red Ball**:
After adding the 6 red balls, the new probability of drawing a red ball is:
\[ P(\text{Red})_{\text{new}} = \frac{x + 6}{18} \]

5. **Set Up the Equation Based on the Given Condition**:
According to the problem, the new probability is double the original probability:
\[ P(\text{Red})_{\text{new}} = 2 \times P(\text{Red})_{\text{old}} \]
Substituting the probabilities we found:
\[ \frac{x + 6}{18} = 2 \times \frac{x}{12} \]

6. **Simplify the Equation**:
First, simplify the right side:
\[ 2 \times \frac{x}{12} = \frac{2x}{12} = \frac{x}{6} \]
Now the equation becomes:
\[ \frac{x + 6}{18} = \frac{x}{6} \]

7. **Cross Multiply to Eliminate the Fractions**:
Cross multiplying gives:
\[ (x + 6) \cdot 6 = x \cdot 18 \]
This simplifies to:
\[ 6x + 36 = 18x \]

8. **Rearrange the Equation**:
Move all terms involving \( x \) to one side:
\[ 36 = 18x - 6x \]
This simplifies to:
\[ 36 = 12x \]

9. **Solve for \( x \)**:
Divide both sides by 12:
\[ x = \frac{36}{12} = 3 \]

10. **Conclusion**:
The number of red balls originally in the box is \( \boxed{3} \).