A bag has 6 blue balls ,'x' red balls and 5 green balls .If two balls are picked randomly ,then probability of 1 being red and 1 being green is `(9)/(38)` Find value of x.

Options

A. 5
B. 9
C. 4
D. 7

Answer

To solve the problem, we need to find the value of \( x \) given the probability of picking 1 red ball and 1 green ball from a bag containing 6 blue balls, \( x \) red balls, and 5 green balls.

### Step-by-Step Solution:

1. **Identify the total number of balls**:
The total number of balls in the bag is the sum of blue, red, and green balls.
\[ \text{Total balls} = 6 + x + 5 = x + 11 \]

2. **Determine the total ways to pick 2 balls**:
The total number of ways to choose 2 balls from \( x + 11 \) balls is given by the combination formula:
\[ \text{Total ways} = \binom{x + 11}{2} = \frac{(x + 11)(x + 10)}{2} \]

3. **Calculate the number of favorable outcomes (1 red and 1 green)**:
The number of ways to choose 1 red ball from \( x \) red balls and 1 green ball from 5 green balls is:
\[ \text{Favorable outcomes} = \binom{x}{1} \times \binom{5}{1} = x \times 5 = 5x \]

4. **Set up the probability equation**:
The probability of picking 1 red and 1 green ball is given by:
\[ P(\text{1 red, 1 green}) = \frac{\text{Favorable outcomes}}{\text{Total ways}} = \frac{5x}{\frac{(x + 11)(x + 10)}{2}} = \frac{10x}{(x + 11)(x + 10)} \]

5. **Equate the probability to the given value**:
We know that this probability equals \( \frac{9}{38} \):
\[ \frac{10x}{(x + 11)(x + 10)} = \frac{9}{38} \]

6. **Cross-multiply to solve for \( x \)**:
Cross-multiplying gives:
\[ 10x \cdot 38 = 9 \cdot (x + 11)(x + 10) \]
Simplifying this:
\[ 380x = 9(x^2 + 21x + 110) \]
\[ 380x = 9x^2 + 189x + 990 \]
Rearranging the equation:
\[ 9x^2 + 189x + 990 - 380x = 0 \]
\[ 9x^2 - 191x + 990 = 0 \]

7. **Use the quadratic formula to find \( x \)**:
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 9 \), \( b = -191 \), and \( c = 990 \):
\[ b^2 - 4ac = (-191)^2 - 4 \cdot 9 \cdot 990 \]
\[ = 36481 - 35640 = 841 \]
\[ x = \frac{191 \pm \sqrt{841}}{2 \cdot 9} = \frac{191 \pm 29}{18} \]
This gives us two possible solutions:
\[ x = \frac{220}{18} = \frac{110}{9} \quad \text{(not an integer)} \]
\[ x = \frac{162}{18} = 9 \quad \text{(valid)} \]

### Final Answer:
The value of \( x \) is \( 9 \).