The p.d.f. of a continuous random variable X is $\mathrm{f}(x)=\left\{\begin{array}{cc} \frac{x^2}{18} & \text { if }-3 < x < 3 \\ 0 & \text { otherwise } \end{array}\right.$ Then $\mathrm{P}[|\mathrm{X}| < 2]=$

The probability density function is defined by $\mathrm{f}(x) = \begin{cases} \frac{x^2}{18} & \text{if } -3 < x < 3 \\ 0 & \text{otherwise} \end{cases}$.

Since $|X| < 2$ implies $-2 < X < 2$ and this interval lies within $(-3, 3)$, the probability is

$\mathrm{P}[|X| < 2] = \int_{-2}^{2} \frac{x^2}{18} \, dx$.

Integrating yields

$\frac{1}{18} \left[ \frac{x^3}{3} \right]_{-2}^{2} = \frac{1}{18} \left( \frac{8}{3} - \frac{-8}{3} \right) = \frac{1}{18} \cdot \frac{16}{3} = \frac{16}{54} = \frac{8}{27}$.

This corresponds to option C.