Is probability that $x<2$ equal to the probability that $x^2<4$ given $1Assuming $x$ is a real number uniformly distributed over the interval $(1,3).$
so $x^2$ is also uniformly distributed over the interval $(1,9)${As for every $x=a\in (1,3) $ there exists $x^2=a^2\in (1,9)$}.
Probability that $x<2$ would be $\frac{1}{2}$ as $x$ can be in $(1,2)$ where sample set of $x$ is $(1,3)$, while probability that $x^2<4$ is $\frac{3}{8}$ as $x^2$ can be in $(1,4)$ where sample set of $x^2$ is $(1,9).$
So why is the probability that $x<2$ different from $x^2<4$ if both are identical?

If $X$ is uniformly distributed on $[1,3]$, then $P(X^2\le a^2)=P(X\le a)=\frac{a-1}2$. Therefore, $P(X^2\le a)=\frac{\sqrt{a}-1}2$. Thus, the PDF of $X^2$ is $\frac1{4\sqrt{a}}$.
That is, $X^2$ is not uniformly distributed on $[1,9]$.
