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 $0.1\le x<0.2 \to Pr[0.1\le X< 0.2]=F(0.2)-F(0.1)=0.1$
Best Answer
For $x>0$, $$0.3\leq x^2 <0.4 \;\iff \; \sqrt{0.3}\leq x < \sqrt{0.4}$$ Then $$ \mathbb P(0.3\leq x^2 <0.4)=\mathbb P(\sqrt{0.3}\leq x < \sqrt{0.4}) = \sqrt{0.4} -\sqrt{0.3}\approx 0.084732975. $$