Conditional Probability with balls in an urn 
Two balls, each equally likely to be colored either red or blue, are put in an urn. At each stage one of the balls is randomly chosen, its color is noted, and it is then returned to the urn. If the first two balls chosen are colored red, what is the probability that
  (a) both balls in the urn are colored red; (b)  the next ball chosen will be red?

I'm wondering if my method for part (a) is correct:
Let $P(R)$ be the probability of picking a red ball: $P(R)=\frac{1}{2}$
Let $P(B)$ be the probability of picking a blue ball: $P(B)=\frac{1}{2}$
Let $P(C)$ be the probability of the condition, i.e., picking two red balls consecutively: $P(C)=P(C|H_1)P(H_1)+P(C|H_2)P(H_2)+P(C|H_3)P(H_3) $
where $P(H_1)$ is the probability both balls in the urn are red: $P(H_1)=(P(R))^2=(\frac{1}{2})^2=\frac{1}4$
$P(H_2)$ is the probability that both balls in the urn are blue: $P(H_2)=(P(B))^2=(\frac{1}{2})^2=\frac{1}4$
$P(H_3)$ is the probability that one ball is red and one is blue inside the urn: $P(H_3)=1-(P(H_1)+P(H_2))=1-\frac{1}2=\frac{1}2$ since the sum of the mutually exclusive hypotheses or events must sum to 1.
Thus $P(C)=1\times\frac{1}4+\frac{1}4\times0+\frac{1}2\times\frac{1}4=\frac{3}8$
and $P(H_1|C)= \frac{P(C \bigcap H_1)}{P(C)}=\frac{P(C|H_1)P(H_1)}{P(C)}=\frac{1\times\frac{1}4}{\frac{3}8}=\frac{2}3$
For part (b), I know that $P(R|C)$, the probability of picking a red ball given that the first two balls picked were red is to be found
$P(R|C)= \frac{P(C \bigcap R)}{P(C)}=\frac{P(C|R)P(R)}{P(C)}$
How can I find $P(C|R)$?
 A: The probability that the contents of the urn are two red is indeed $\frac{1}{4}$, as is the probability of two blue, and the probability of mixed is therefore $\frac{1}{2}$. The derivation could have been done more quickly.
Question (a) asks for the probability both are red given that the two drawn balls are red. Let $R$ be the event both are red, and $D$ be the event both drawn balls are red.  We want $\Pr(R|D)$. By the usual formula this is $\frac{\Pr(R\cap D)}{\Pr(D)}$.
To find $\Pr(D)$, note that if both balls are red (probability $\frac{1}{2}$), then the probability of $D$ is $1$, while if one ball is red and the other is not (probability $\frac{1}{2}$) then the probability of $D$ is $\frac{1}{4}$. Thus the probability of $D$ is $\left(\frac{1}{4}\right)(1)+\left(\frac{1}{2}\right)\left(\frac{1}{4}\right)$. This is $\frac{3}{8}$.
The probability of $R\cap D$ is $\frac{1}{4}$. So the ratio is indeed $\frac{2}{3}$. 
For (b), you can use the calculation of (a). With probability $\frac{2}{3}$ we are drawing from a double red, and we will get red with probability $1$. With probability $\frac{1}{3}$ the urn is a mixed one, and the probability of drawing a red is $\frac{1}{2}$, for a total of $\frac{2}{3}\cdot 1+\frac{1}{3}\cdot\frac{1}{2}$. 
One can also solve (b) without using the result of (a). With I hope self-explanatoru notation, we want $\Pr(RRR|RR)$. The probabilities needed in the conditional probability formula are easily computed. We have $\Pr(RRR\cap RR)=\Pr(RRR)=\frac{1}{4}\cdot 1+\frac{1}{2}\cdot\frac{1}{8}=\frac{5}{16}$. Divide by $\frac{3}{8}$. 
