A bag contains x green balls , 7 blue balls and 8 red balls . When two balls are drawn from bag randomly ,then the probability of one ball being green and one ball being red is `(4)/(15)` .Find value of x (number of green balls cannot be more than 18 balls).

Options

A. 3
B. 5
C. 10
D. 9

Answer

To solve the problem, we need to find the value of \( x \) (the number of green balls) in a bag that contains \( x \) green balls, 7 blue balls, and 8 red balls, given that the probability of drawing one green ball and one red ball is \( \frac{4}{15} \).

### Step-by-Step Solution:

1. **Identify Total Number of Balls**:
The total number of balls in the bag is the sum of green, blue, and red balls:
\[ \text{Total balls} = x + 7 + 8 = x + 15 \]

2. **Calculate Total Outcomes for Drawing Two Balls**:
The total ways to choose 2 balls from \( x + 15 \) balls is given by the combination formula:
\[ \text{Total outcomes} = \binom{x + 15}{2} = \frac{(x + 15)(x + 14)}{2} \]

3. **Calculate Favorable Outcomes for One Green and One Red Ball**:
The number of ways to choose 1 green ball from \( x \) green balls and 1 red ball from 8 red balls is:
\[ \text{Favorable outcomes} = \binom{x}{1} \cdot \binom{8}{1} = x \cdot 8 = 8x \]

4. **Set Up the Probability Equation**:
According to the problem, the probability of drawing one green ball and one red ball is given as \( \frac{4}{15} \). Therefore, we can set up the equation:
\[ \frac{8x}{\frac{(x + 15)(x + 14)}{2}} = \frac{4}{15} \]

5. **Cross Multiply to Solve for \( x \)**:
Cross multiplying gives:
\[ 8x \cdot 15 = 4 \cdot \frac{(x + 15)(x + 14)}{2} \]
Simplifying this, we get:
\[ 120x = 2(x + 15)(x + 14) \]
Expanding the right side:
\[ 120x = 2(x^2 + 29x + 210) \]
This simplifies to:
\[ 120x = 2x^2 + 58x + 420 \]

6. **Rearranging the Equation**:
Rearranging gives us a quadratic equation:
\[ 2x^2 - 62x + 420 = 0 \]
Dividing the entire equation by 2:
\[ x^2 - 31x + 210 = 0 \]

7. **Factoring the Quadratic Equation**:
We need to factor the quadratic equation:
\[ (x - 21)(x - 10) = 0 \]
Thus, the solutions for \( x \) are:
\[ x = 21 \quad \text{or} \quad x = 10 \]

8. **Consider the Constraint**:
Since the number of green balls cannot be more than 18, we discard \( x = 21 \) and accept:
\[ x = 10 \]

### Conclusion:
The number of green balls \( x \) is \( 10 \).