If $x$ lies between $3$ and $5$, $5$ inclusive, what is the probability that $\lfloor x^2\rfloor = \lfloor x\rfloor^2?$

Question

I am stuck with this problem for a while and couldn't think my way out of it. Help, please !!!

My thought process:
As per question $3 \lt x \leq 5$, then $\lfloor x\rfloor \in \{3,4,5\}$. Please correct me if I am wrong, but I think $\lfloor x\rfloor^2 \in\{9,16,25\}$.

Also, as per question, $3 \lt x \leq 5$ implies $9 \lt x^2 \leq 25$ and $\lfloor x^2\rfloor \in \{9,10,11,\ldots,25\}$. After this, I am not able to find the probability for this number range. Please help.

Thanks in advance!

Answer 1

You want to find the probability that $[x^2]=[x]^2\in\{9,16,25\}$ (since the latter is a perfect square). Now $$[x^2]=9=[x]^2\iff x\in(3,\sqrt{10})$$ (in the given range). Similarly $$[x^2]=16=[x]^2\iff x\in[4,\sqrt{17})$$ And finally, $$[x^2]=25=[x]^2\iff x=5$$ Assuming uniform distribution, the required probability is just the sum of the lengths of the intervals in which favorable cases occur divided by the length of the total interval. So it becomes $$P([x^2]=[x]^2)=\frac{(\sqrt{10}-3)+(\sqrt{17}-4)+(5-5)}{5-3}=\boxed{\frac{\sqrt{10}+\sqrt{17}-7}{2}}$$

Answer 2

Hint:

$[x^2]=9 \iff$ $3\le x < \sqrt{10}$
$[x^2]=16 \iff$ $4\le x < \sqrt{17}$
$[x^2]=25 \iff$ $5\le x < \sqrt{26}$