Probability of choosing a real number between 0 and 1

Question

When randomly choosing real numbers in range (0,1).
What's the probability that if we chose x, the first decimal digit in $x^2$ is 3?
In other words, if we chose x from (0,1), what the probability that $x^2$ looks like $0.3\square\square\square\square.....$ ?

The only thing is I'm sure about is that X~U(0,1) which means $Pr[X\le x]=F(x)=x$
So if we wanted to choose a number which looks like $0.1\square\square\square\square.....$ this means $0.1\le x<0.2 \to Pr[0.1\le X< 0.2]=F(0.2)-F(0.1)=0.1$

Answer 1

For $x>0$, $$0.3\leq x^2 <0.4 \;\iff \; \sqrt{0.3}\leq x < \sqrt{0.4}$$ Then $$ \mathbb P(0.3\leq x^2 <0.4)=\mathbb P(\sqrt{0.3}\leq x < \sqrt{0.4}) = \sqrt{0.4} -\sqrt{0.3}\approx 0.084732975. $$

Answer 2

$x$ must be in the range $[\sqrt{0.3},\sqrt{0.4})$

So the size of the range is $\sqrt{0.4}-\sqrt{0.3}=\frac{2\sqrt{10}-\sqrt{30}}{10}$ (because $\sqrt{0.4}=\frac{\sqrt2}{\sqrt5}\cdot\frac{\sqrt5}{\sqrt5}=\frac{\sqrt{10}}{5}$ and similarly $\sqrt{0.3}=\frac{\sqrt{3}}{\sqrt{10}}\cdot\frac{\sqrt{10}}{\sqrt{10}}=\frac{\sqrt{30}}{10}$)

This is your probability because the range of possible values between $0$ and $1$ is $1$