Why is P{$\cup_{k=2,4,6,...even}[\frac{1}{2^k},\frac{1}{2^k}+\frac{1}{2^{k+1}}]$}=$\sum_{k=2,4,6,..even}\frac{1}{2^{k+1}}$ Question: Given the uniform pdf on $[0,1]$,$f(x)=1$; x $\in [0,1] $.Define the cumulate distribution function (cdf) $F$ as the probability of the event {$x:x \le r$}as a function of $r \in \Re$: $F(r)=P((-\infty,r])=\int^{r}_{-\infty}f(x)dx$ Find the probability of the event:$G$={$w:w \in [\frac{1}{2^k},\frac{1}{2^k}+\frac{1}{2^{k+1}}]$ for some event $k$}=$\cup_{k=even}[\frac{1}{2^k},\frac{1}{2^k}+\frac{1}{2^{k+1}}]$} Solution: This is part of some question's solution,and i don't know why is the left formula equal to the right formula,can anyone explain it to me?thankyou! P{$\cup_{k=2,4,6,...even}[\frac{1}{2^k},\frac{1}{2^k}+\frac{1}{2^{k+1}}]$}=$\sum_{k=2,4,6,..even}\frac{1}{2^{k+1}}$

For uniform distribution on $[0,1]$ we have $P([a,b])=b-a$ whenever $0\leq a\leq b \leq1$. Just apply this to $a=\frac 1 {2^{k}}$ and $b=\frac 1 {2^{k}}+\frac 1 {2^{k+1}}$. [The intervals are disjoint when $k$ is restricted to even integers].