A function is defined as `f(x)={{:(0","" for "xgt2),((2x+3)/(18)" for "2lexle4),(0" for "xgt4):}`
Then `P(2ltXlt3)` is

Options

A. `5/9`
B. `4/9`
C. `7/9`
D. `2/9`

Explanation

To solve the problem, we need to find the probability \( P(2 < X < 3) \) for the given piecewise function:

\[ f(x) = \begin{cases} 0 & \text{for } x > 4 \\ \frac{2x + 3}{18} & \text{for } 2 \leq x \leq 4 \\ 0 & \text{for } x < 2 \end{cases} \]

### Step-by-step Solution:

1. **Identify the relevant section of the function**: Since we are looking for the probability between \( 2 \) and \( 3 \), we will use the function defined for \( 2 \leq x \leq 4 \):
\[ f(x) = \frac{2x + 3}{18} \]

2. **Set up the probability integral**: The probability \( P(2 < X < 3) \) can be found by integrating the function \( f(x) \) from \( 2 \) to \( 3 \):
\[ P(2 < X < 3) = \int_{2}^{3} f(x) \, dx = \int_{2}^{3} \frac{2x + 3}{18} \, dx \]

3. **Integrate the function**: We will integrate \( \frac{2x + 3}{18} \):
\[ \int \frac{2x + 3}{18} \, dx = \frac{1}{18} \int (2x + 3) \, dx = \frac{1}{18} \left( x^2 + 3x \right) + C \]

4. **Evaluate the definite integral from 2 to 3**:
\[ P(2 < X < 3) = \frac{1}{18} \left[ (3^2 + 3 \cdot 3) - (2^2 + 3 \cdot 2) \right] \]
\[ = \frac{1}{18} \left[ (9 + 9) - (4 + 6) \right] \]
\[ = \frac{1}{18} \left[ 18 - 10 \right] \]
\[ = \frac{1}{18} \cdot 8 = \frac{8}{18} = \frac{4}{9} \]

5. **Final answer**: Therefore, the probability \( P(2 < X < 3) \) is:
\[ \boxed{\frac{4}{9}} \]