A Polya urn has two balls, one red, and one blue. One of these is chosen
uniformly at random. It is put back, with another of the same color. Again, a ball is chosen
uniformly at random, and put back, with another of the same color. This process continues.
(i)Given that the second draw is blue, what is the probability that the first draw is blue?
(ii)Given that the first draw is blue, what is the probability that the fifth draw is blue?
Here is what I have: Let $A=${First draw is blue} and $B=${Second Draw is blue}. For part (i) we get $P(A|B)=\frac{P(B|A)P(B)}{P(A)}{=\frac{(1/2)(1/2)}{(1/2)}}=1/2$. For part (ii) we have $\frac{P(\text{both 1st and 5th are blue})}{P(\text{1st is blue})}=\frac{2^3/2^8}{2^4/2^8}=\frac{2^3}{2^4}=1/2$. I'm not sure if (ii) is correct.
Let $B_i$ be the event that a blue ball is drawn on the $i$th turn; let $R_i$ be the event that a red ball is drawn on the $i$th turn.
Observe that $\Pr(B_1) = \Pr(R_1) = 1/2$ since we are equally likely to draw either color on the first draw. If the first ball drawn is blue, then another blue ball is placed in the urn, so there are three balls in the urn, of which two are blue. Hence, $\Pr(B_2 \mid B_1) = 2/3$. If the first ball drawn is red, then another red ball is placed in the urn, so there are three balls in the urn, of which one is blue. Hence, $\Pr(B_2 \mid R_1) = 1/3$. With that in mind, we can now calculate the probability that the first ball drawn is blue given that the second ball drawn is blue.
The probability that the first ball drawn is blue ball given that the second ball drawn is blue is \begin{align*} \Pr(B_1 \mid B_2) & = \frac{\Pr(B_1 \cap B_2)}{\Pr(B_2)}\\ & = \frac{\Pr(B_1)\Pr(B_2 \mid B_1)}{\Pr(B_1)\Pr(B_2 \mid B_1) + \Pr(R_1)\Pr(B_2 \mid R_1)}\\ & = \frac{\frac{1}{2} \cdot \frac{2}{3}}{\frac{1}{2} \cdot \frac{2}{3} + \frac{1}{2} \cdot \frac{1}{3}}\\ & = \frac{2}{3} \end{align*}
Let's examine the probability that the third ball drawn is blue given that the first ball drawn is blue. There are two favorable events: all three balls that are drawn are blue or the first and third balls drawn are blue while the second ball drawn is red.
$$\Pr(B_3 \mid B_1) = \frac{\Pr(B_1)\Pr(B_2 \mid B_1)\Pr(B_3 \mid B_1 \cap B_2) + \Pr(B_1)\Pr(R_2 \mid B_1)\Pr(B_3 \mid B_1 \cap R_2)}{\Pr(B_1)}$$ I think you will find it instructive to complete that calculation. It should give you an idea of what $\Pr(B_5 \mid B_1)$ is.
When you try to justify that conclusion, keep in mind that some event must happen between the time the first blue ball is drawn and the last blue ball is drawn.