2020 Fermat Problem 20

A cube has six faces. Each face has some dots on it. The numbers of dots on the six faces are 2, 3, 4, 5, 6, and 7. Harry removes one of the dots at random, with each dot equally likely to be removed. When the cube is rolled, each face is equally likely to be the top face. What is the probability that the top face has an odd number of dots on it?

\textbf{(A)}\ \frac{4}{7}\quad \textbf{(B)}\ \frac{1}{2}\quad \textbf{(C)}\ \frac{13}{27}\quad \textbf{(D)}\ \frac{11}{21}\quad \textbf{(E)}\ \frac{3}{7}

If there are no answer choices shown, enter a numerical answer.

Full credit to this problem is given to the CEMC, you may view all Fermat contests here.