Having Trouble Formalizing a Combinatorial Argument

Question

Let $\{X_i\}_{i = 1} ^d$ be symmetric linearly independent i.i.d random vectors in $\mathbb{R}^n$. I wish to show $$ \mathbb{P}\{ 0 \in \mathrm{conv}\{ X_i \}_{i \in I}| 0 \in \mathrm{aff}\{ X_i \}_{i \in I} \} = \frac{1}{2^{d - 1}}. $$ My Attempt: \begin{align*} \mathbb{P}\{ 0 \in \mathrm{conv}\{ X_i \}_{i \in I}| 0 \in \mathrm{aff}\{ X_i \}_{i \in I} \} &= \mathbb{P}\{ 0 \in \mathrm{conv}\{ X_i \}_{i \in I}| 0 \in \mathrm{aff}\{ X_i \}_{i \in I} \} \\ &= \mathbb{P}\{ 0 \in \mathrm{conv}\{ X_i \}_{i \in I} | 0 \in \mathrm{aff}\{ X_i \}_{i \in I}, \mathrm{coef}(\mathrm{aff}(X_i)) \geq 0 \} \\ &+ \mathbb{P}\{ 0 \in \mathrm{conv}\{ -X_i \}_{i \in I} | 0 \in \mathrm{aff}\{ X_i \}_{i \in I}, \mathrm{coef}(\mathrm{aff}(X_i)) \leq 0 \} \\ &= 2\mathbb{P}\{ 0 \in \mathrm{conv}\{ X_i \}_{i \in I} | 0 \in \mathrm{aff}\{ X_i \}_{i \in I}, \mathrm{coef}(\mathrm{aff}(X_i)) \geq 0 \} \\ &= \frac{1}{2^{d - 1}}. \end{align*} I am not sure if I can pass to the second equality as I claimed here.