smc

We have ${\log }_{10}(d+1)-{\log }_{10}d={\log }_{10}\left(\frac{d+1}{d}\right)$, so the probability of choosing $2$ is ${\log }_{10}\left(\frac{3}{2}\right)$. The probability that the digit chosen is in the set must therefore be $$\begin{array}{rl}2{\log }_{10}\left(\frac{3}{2}\right)=& {\log }_{10}\left({\left(\frac{3}{2}\right)}^2\right)\\ =& {\log }_{10}\left(\frac{9}{4}\right)\\ =& {\log }_{10}9-{\log }_{10}4\\ =& \left({\log }_{10}9-{\log }_{10}8\right)+\left({\log }_{10}8-{\log }_{10}7\right)+\left({\log }_{10}7-{\log }_{10}6\right)\\ & +\left({\log }_{10}6-{\log }_{10}5\right)+\left({\log }_{10}5-{\log }_{10}4\right),\end{array}$$ which, by definition, is the probability that the digit chosen is in the set $\boxed{\{4,5,6,7,8\}}$.