Does this question on conditional probability have an incorrect solution? Going through some exercises on probability, and I chanced upon this problem.

An urn contains 6 red balls and 3 blue balls. One ball is selected at
  random and is replaced by a ball of the other color. A second ball is
  then chosen. What is the conditional probability that the first ball
  selected is red, given that the second ball was red?

My solution seems correct. I do not know where I went wrong.
Let R1 be the event that a first red ball is selected.
Let R2 be the event that a second red ball is selected.

Then,
$P(R_2|R_1) = \dfrac{5}{9}\times\dfrac{6}{9}=\dfrac{30}{81}\Rightarrow P(R_1\cap R_2) = P(R_2|R_1)\times P(R_1)=\dfrac{30}{81}\times \dfrac{6}{9}=\dfrac{180}{729}$
$P(R_2|R_1^C) = \dfrac{7}{9}\times\dfrac{3}{9}=\dfrac{21}{81}\Rightarrow P(R_1^C\cap R_2) = P(R_2|R_1^C)\times P(R_1^C)=\dfrac{21}{81}\times \dfrac{3}{9}=\dfrac{63}{729}$
and thus, $P(R_2) = P(R_2 \cap R_1) + P(R_2 \cap R_1^C) = \dfrac{243}{729}$
$\therefore P(R_1|R_2)=\dfrac{P(R_1 \cap R_2)}{P(R_2)}=\dfrac{\dfrac{180}{729}}{\dfrac{243}{729}}=\dfrac{20}{27}$
The answer, however, is $\dfrac{10}{17}$, which seems like a typo to me, but I'm not sure. Thanks!

With problems like this I like to turn $P(R_1 | R_2)$ into $P(R_2 | R_1)$ if it is an easier calculation. We have,
$$\begin{align*}
P(R_1 | R_2)
&= \frac{P(R_1 \cap R_2)}{P(R_2)}\\\\
&= \frac{P(R_2 | R_1)P(R_1)}{P(R_2)}\\\\
&= \frac{\frac{5}{9}\cdot\frac{6}{9}}{\frac{6}{9}\cdot\frac{5}{9}+\frac{3}{9}\cdot\frac{7}{9}}\\\\
&\approx .558 \\\\
&= \frac{10}{17}
\end{align*}$$
Just some explanations for these values:
$P(R_2 | R_1) = \frac{5}{9}$ since if we select red first, then we only have $5$ reds of the $9$ the second time around. 
$P(R_2)=\frac{6}{9}\cdot\frac{5}{9}+\frac{3}{9}\cdot\frac{7}{9}$ since we can either select red and then red again or select blue and then red.
