Probability of colour with putting back twice. I'm having a bit of trouble with following probability-related question:
A container contains $5$ red and $10$ black balls. Take a ball out of the container at random and note its color. After drawing each ball, the ball gets put back and one extra ball of the same color gets added.
1) Given that the first $n$ drawn balls are all black, calculate the probability (say $a_{n}$)  that the $(n+1)$st ball will also be black. What is $\lim_{x\to\infty} a_{n}$ equal to?
2) Given that the second until the $(n+1)$st ball (inclusive) are all black, calculate the probability that (say $b_{n}$) that the first drawn ball was also black. Also calculate $\lim_{n\to\infty} b_{n}$.
I succeeded at answering the first subquestion. As the result I get that $$a_{n} = \dfrac{10+n}{15+n}$$ and thus, $$\lim_{n\to\infty} a_{n} = 1$$
However, I have much more trouble answering the second question, and would like your help.
Thanks!
EDIT:
The end solution of the limit in subquestion (2) is also given, though not the way of getting to that result. The result is: $\lim_{n\to\infty} b_{n} = 1$.
 A: Let $B_n$ be event "all balls from the second until the $n$-th (inclusive) are black", and $B$ and $R$ be events "first ball was black" and "first ball was red".
Then $b_n = P(B | B_n) = \frac{P(B \cap B_n)}{P(B \cap B_n) + P(R \cap B_n)} = 1 - \frac{P(R \cap B_n)}{P(B \cap B_n) + P(R \cap B_n)}$.
$P(B \cap B_n) = P(B) \cdot P(B_1 | B) \cdot P(B_2 | B, B_1) \cdot \ldots \cdot P(B_n | B, B_{n - 1}) = a_0 \cdot a_1 \ldots \cdot a_n = \prod\limits_{i=0}^n \frac{10 + i}{15 + i} = \frac{10 \cdot 11 \cdot 12 \cdot 13 \cdot 14}{(n + 11) \cdot (n + 12) \cdot (n + 13) \cdot (n + 14)\cdot(n + 15)}$.
Similarly $P(R \cap B_n) = \frac{5}{15} \cdot \frac{10}{16} \cdot \frac{11}{17} \cdot \ldots \cdot \frac{10 + n - 1}{15 + n} = \frac{5 \cdot 10 \cdot 11 \cdot 12 \cdot 13 \cdot 14}{(10 + n) \cdot (11 + n) \cdot (12 + n) \cdot (13 + n) \cdot (14 + n) \cdot (15 + n)}$.
From this we have $\frac{P(R \cap B_n)}{P(B \cap B_n) + P(R \cap B_n)} = \frac{\frac{5}{10 + n}}{1 + \frac{5}{10 + n}} = \frac{5}{15 + n} \to 0$, so $b_n \to 1$.
