If the function `f(x) =(x^(2))/(3), - 1lt x lt 2`
= 0 , otherwise is a p.d.f of X, then `P(X lt 0)` is

Options

A. `(1)/(9)`
B. `(2)/(9)`
C. `(3)/(9)`
D. `(4)/(9)`

Answer

To solve the problem, we need to find the probability \( P(X < 0) \) given the probability density function (p.d.f.) \( f(x) = \frac{x^2}{3} \) for the interval \( -1 < x < 2 \) and \( f(x) = 0 \) otherwise.

### Step 1: Understand the p.d.f.
The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} \frac{x^2}{3} & \text{for } -1 < x < 2 \\ 0 & \text{otherwise} \end{cases} \]

### Step 2: Set up the probability expression
To find \( P(X < 0) \), we need to evaluate the integral of \( f(x) \) from \( -1 \) to \( 0 \):
\[ P(X < 0) = \int_{-\infty}^{0} f(x) \, dx \]
Since \( f(x) = 0 \) for \( x < -1 \), we can simplify this to:
\[ P(X < 0) = \int_{-1}^{0} f(x) \, dx \]

### Step 3: Substitute the p.d.f. into the integral
Now we substitute \( f(x) \) into the integral:
\[ P(X < 0) = \int_{-1}^{0} \frac{x^2}{3} \, dx \]

### Step 4: Factor out the constant
We can factor out the constant \( \frac{1}{3} \):
\[ P(X < 0) = \frac{1}{3} \int_{-1}^{0} x^2 \, dx \]

### Step 5: Calculate the integral
Now we calculate the integral \( \int_{-1}^{0} x^2 \, dx \):
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Evaluating this from \( -1 \) to \( 0 \):
\[ \left[ \frac{x^3}{3} \right]_{-1}^{0} = \frac{0^3}{3} - \frac{(-1)^3}{3} = 0 - \left(-\frac{1}{3}\right) = \frac{1}{3} \]

### Step 6: Substitute back into the probability expression
Now substituting back into the expression for \( P(X < 0) \):
\[ P(X < 0) = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9} \]

### Final Answer
Thus, the probability \( P(X < 0) \) is:
\[ \boxed{\frac{1}{9}} \]