Is the probability that a spherically symmetric random number is in a cone always the same, no matter the specific distribution?

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Definitions:

Let $C_n$ be a generalized cone (but with the interior included, i.e., the $C_n$ is solid, not just a surface) in $\mathbb{R}^n$, with vertex at the origin. The following may not be crucial, but in the case I'm working with $C_n$ is furthermore open, non-convex and simply-connected.

Now let us pick a random number from a spherically symmetric multivariate distribution, i.e., some random variable $\Theta X$, where $\Theta$ is a random vector distributed uniformly over an $n$-sphere and $X$ is a random variable with support over $(0,s)$, where $s$ can either be some finite, positive number or positive infinity.

For a given $n$, there is some probability that $\Theta X\in C_n$, call it $p(n)$.

Question:

Is $p(n)$ the same, no matter the choice of $X$ (when $X$ is defined in accordance with the above)?

Context:

This seems to be the case when $C_n$ is the space of $n\times n$ Hurwitz-stable matrices (and $\Theta$ is drawn from an $n^2$-sphere) and I thought this could be explained by this space being a cone (as defined above), but I'm not sure how one would show this.

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Since $X$ is a positive scalar, $X\Theta\in C_n$ exactly when $\Theta\in C_n$, by any reasonable definition of "generalized cone".

Therefore $P(X\Theta\in C_n)=P(\Theta\in C_n)$ because it's the same event -- and the latter doesn't even mention $X$.