Distance of a d dimensional point in a hypercube

Question

Assume a particle X in $d$-dimensional hypercube. Each dimension is independent of another and the particle's position is distributed uniformly in each.

A distance measure of $D = \frac{1}{2} \max\limits_{i\in 1\cdots d} \lvert \frac{1}{2} - X_i\rvert$ is computed

$P(D >a) = p(\frac{1}{2} \max\limits_{i\in 1\cdots d} \lvert \frac{1}{2} - X_i\rvert >a) = p(\max\limits_{i\in 1\cdots d} \lvert \frac{1}{2} - X_i\rvert > 2a)= 1- p(\max\limits_{i\in 1\cdots d} \lvert \frac{1}{2} - X_i\rvert \leq 2a) = 1 - \prod\limits_i p(\lvert \frac{1}{2} - X_i \rvert \leq 2a))$

$ = \prod\limits_i 2 p(X_i-\frac{1}{2} \leq 2a | X_i >\frac{1}{2}) = 1-(8a)^d$

Am I goin wrong somewhere?

Answer 1

The probability that the particle deviates from ${1 \over 2}$ on at least one dimension is 1 - probability that none of them do.

Answer 2

The event $D<a$ represents the fact that the particle is inside a hypercube centered at $(1/2,...,1/2)$ with edges of length $4a$. This hypercube has volume $(4a)^d$. Given the uniform distribution inside the original (unit, I suppose) hypercube, the probability of that event is the ratio of volumes: $$ P(D<a)=\frac{(4a)^d}{1^d}=(4a)^d. $$ Therefore $$ P(D>a)=1-(4a)^d. $$ (of course if $a\geq1/4$ the probability is $0$).