jmc

We need exactly two flips to come up heads and three to come up tails. The odds that two flips come up heads is ${\left(\frac{2}{3}\right)}^2$ and the odds that the other three all come up tails is ${\left(\frac{1}{3}\right)}^3$. We then need to consider the distinct ways of positioning the heads among the 5 flips: we can put the first one in any of 5 places and the second one in any of the remaining 4 places, but they aren't distinct so we need divide by 2 for a total of $\frac{5\times 4}{2}=10$ ways. Thus the probability is ${\left(\frac{2}{3}\right)}^2\times {\left(\frac{1}{3}\right)}^3\times 10=\boxed{\frac{40}{243}}$.

Alternatively, we can view flipping this coin 5 times as being equivalent to the expansion of $(h+t{)}^5$ where $h=\frac{2}{3}$ and $t=\frac{1}{3}$. The value of the $h^n{t}^{5-n}$ term in this expansion will be the probability of getting exactly $n$ heads, so setting $n=2$ and applying the binomial theorem gives us $p=\left(\genfrac{}{}{0ex}{}{5}{2}\right){\left(\frac{2}{3}\right)}^2{\left(\frac{1}{3}\right)}^3=\boxed{\frac{40}{243}}$, which is the same answer we got using the other method.