I can't figure out $P(3 \le X)$ from the density and distribution functions.
$$ f(x) = \begin{cases} cx^2 & 0 \le x \le 6 \\ 0 & \text{elsewhere} \\ \end{cases} ,$$
from which I determined $c$ to be $\frac{1}{72}$, giving me
$$ f(x) = \begin{cases} \frac{x^2}{72} & 0 \le x \le 6 \\ 0 & \text{elsewhere} \\ \end{cases} ,$$
and the distribution function to be
$$ F(x) = \begin{cases} 0 & x \le 0 \\ \frac{x^3}{216} & 0 \le x \le 6 \\ 1 & 6 \le x \\ \end{cases} .$$
I feel like there's an obvious solution but I cannot seem to figure it out.
Also I apologize for the poor layout I am completely new to all this
$P(3 \leq X)=1-P(X<3)=1-F_X(3)=1-\frac {3^{3}} {216}=\frac 7 8$