Calculate probability of binary expansion $x = \sum\limits_{n = 1}^\infty X_n2^{-n}$

Question

Let $x \sim \text{Unif}[0,1]$ be a continuous uniform random variable

Then the binary expansion of $x$ is given by $$x = \sum\limits_{n = 1}^\infty X_n2^{-n}$$

I am required to find the probability that $X_1 = 0$, $X_2 = 0$, $X_1 = 1$, $X_2 = 0$.

Attempt:

Using equivalent events, we can express

$\Pr[X_1 = 0] = \Pr[0 \leq x \leq 1/2] = 1/2$

The second equality is true because $X_1 = 0$ implies that the sum $x$ can be at most $1/2$, and at least $0$ in the case that all $X_{n}$s' are zero.

Thus similarly,

$\Pr[X_2 = 0] = \Pr[0 \leq x \leq 3/4] = 3/4$

$\Pr[X_1 = 1] = \Pr[1/2 \leq x \leq 1] = 1/2$

$\Pr[X_2 = 1] = \Pr[1/4 \leq x \leq 1] = 3/4$

I evaluated those probabilities based on the length of the intervals and the fact that $x$ is uniform in $[0,1]$.

Can someone check if these probabilies are correct? Thanks in advance!

Answer 1

It is certainly wrong since $P(X_2=0)+P(X_2=1) >1$.

\begin{align}P(X_2=0)&=P(X_1=0, X_2=0)+P(X_1=1, X_2=0) \\ &=P\left(0 \leq X\leq \frac14\right)+P\left(\frac12 \leq X\leq \frac34\right)\end{align}