Finding CDF of $Y$ when $Y=1/X$.

Question

$X$ be a uniform random variable on $[1,3]$. $Y=1/X$. What is the CDF of $Y$?

My attempt so far: CDF of $Y$ by definition:

$= P(Y\le y)\\ = P (1/X \le y)\\ = P (X\ge 1/y)$

I don't know how to proceed.

Answer 1

My attempt so far: CDF of Y by definition: $= P(Y\leq y) \\ = P (1/X \leq y) \\ = P (X\geq 1/y)$

Correct. You also know $X\sim\mathcal{U}[1,3]$ and so $P(X\leq x)= \tfrac{x-1}2\cdot\mathbf 1_{1\leqslant x\lt 3}+\mathbf 1_{3\leqslant x}$