jmc

We can compute this a few ways, but the numbers seem small enough that we can go ahead and just compute the probability of A being selected all three days, and the probability of A being selected exactly 2 of the three days. Team A is selected on any given day with probability $\frac{2}{3}$, because there are $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$ possible pairs of teams, and 2 of them contain A. So, there is a ${\left(\frac{2}{3}\right)}^3=\frac{8}{27}$ chance of being selected all three days. Of being selected exactly twice, there is a $\frac{2}{3}\cdot \frac{2}{3}\cdot \frac{1}{3}\cdot \left(\genfrac{}{}{0ex}{}{3}{2}\right)=\frac{4}{9}$ chance. Adding these two yields $\frac{8}{27}+\frac{4}{9}=\frac{8+12}{27}=\boxed{\frac{20}{27}}$.