Exponential distribution probability calculation

Question

If $X\sim \mathcal{Exp}(4)$ and $y=x^3$. I need to calculate $P(Y≤1)$. So All I gotta do is to use the CDF with $y=1$ and $\lambda=4$? since the only way for $y=1$ is for $x$ to be $1$ as well.

So the result I get is 0.981. am I correct or out of the way completely?

Answer 1

$$\mathbb{P}[Y \leq1]=\mathbb{P}[X^3 \leq1]=\mathbb{P}[X \leq1]=F_X(1)=1-e^{-4}\approx 0.981684$$