Chance a random cube has last two digits 11

An integer $x$ is selected uniformly at random from $1,2,\dots,10^{12}$. What is the probability that the decimal expansion of $x^3$ ends with the digits $11$?

$1/100$. Only the last two digits of $x$ matter. For $x^3$ to end in $1$, $x$ must end in $1$; writing $x\equiv 10a+1$ gives $x^3\equiv30a+1\pmod{100}$, so $3a\equiv1\pmod{10}$, i.e. $a=7$. Exactly one residue, $71$, works modulo $100$, and $10^{12}$ is a multiple of $100$.