This is a classical and essentially geometric problem. In fact, *the answer does not depend
on the distribution of the points* (as long as the distribution is centrally symmetric).

The following result is due to Wendel (link).

**Theorem.** If $X_1$, ..., $X_N$ are i.i.d. random points in $R^d$ whose distribution
is symmetric with respect to $0$ and assigns measure zero to every hyperplane
through $0$, then
$$\mathbb P(0\notin \mbox{conv}\{X_1,\dots,X_N\})=\frac{1}{2^{N-1}}\sum\limits_{k=0}^{d-1}{N-1 \choose k}.$$

The proof is straightforward. Let $\mu$ be the distribution of $X_k$, and set
$$ f(x_1,\dots,x_N) = \begin{cases} 1, & \mbox{if } x_1,\dots,x_N\ \mbox{ lie in an open halfspace of $\mathbb R^d$ with $0$ in the boundary}, \newline 0, & \mbox{else.} \end{cases}$$
Then due to the invariance of $\mu$ under reflection in the origin, we have that
$$\mathbb P(0\notin \mbox{conv}\{X_1,\dots,X_N\})=\int_{\mathbb R^d}\dots \int_{\mathbb R^d} \frac{1}{2^N}\sum\limits_{\varepsilon_i=\pm1}f(\varepsilon_1x_1,\dots,\varepsilon_Nx_N)\ \mu(dx_1)\dots\mu(dx_N).$$
Now, the sum
$$C(N,d)=\sum\limits_{\varepsilon_i=\pm1}f(\varepsilon_1x_1,\dots,\varepsilon_Nx_N)$$
can be interpreted as the number of connected components of the set $\mathbb R^d\backslash (H_1\cup\dots\cup H_N)$ induced by the hyperplanes $H_1$, ..., $H_N$ through $0$ which are in general position. But there is a classical calculation going back to to Steiner and Schläfli, which shows that
$$C(N,d)= 2\sum\limits_{k=0}^{d-1}{N-1 \choose k}.$$