Let $U_1, U_2, \dots,U_n \sim \operatorname{Unif}(-1,1)$ be i.i.d. uniform random variables.
(a) Find the probability that the random vector $(U_1, U_2, \dots,U_n)$ is in the unit sphere in $\mathbb{R}^n.$
(b) Let c be a constant with $0 < c < 1$, and let $X_n$ count how many of the $U_j$ satisfy $\mid U_j \mid > c$. What is the distribution of $X_n$?
(c) For $c=\frac{1}{\sqrt{2}}$, use the result of Part (c) to give a simple, short derivation of what happens to the probability from (a) as $n \rightarrow \infty$.
My solutions
(a) The volume of the unit sphere in $\mathbb{R}^n$ is give by $$ V_n = \frac{\pi^{n/2}}{\Gamma(n/2 + 1)}. $$ The volume of the hypercube in $\mathbb{R}^n$ with edge length of $2$ equals $2^n$. As each point in the hypercube is equally likely, the probability in question should be $V_n/2^n$.
(b) For each $U_j$, $P(\mid U_j \mid > c) = 2(1-c)$. Thus, $X_n \sim \operatorname{Bin}(n, 2(1-c))$.
c) For $c=\frac{1}{\sqrt{2}}$, $X_n \sim \operatorname{Bin}(n, 2-\sqrt{2})$. Whenever $X_n \geq 2$, the point will not lie within the unit sphere, and the probability for it being on the surface is zero, thus,
\begin{align} P(\text{point in unit sphere in } \mathbb{R}^n) &< P(X\leq 2) \\ &= \sum_{k=0}^1 \binom{n}{k} (2-\sqrt{2})^k (\sqrt{2}-1)^{n-k}. \end{align}
Thus,
$$ \lim_{n \rightarrow \infty} P(\text{point in unit sphere in } \mathbb{R}^n) = 0 $$
Is this correct?