A bag has $r$ red balls and $b$ black balls. All balls are identical except for their colours. In a trial, a ball is randomly drawn from the bag, its colour is noted and the ball is placed back into the bag along with another ball of the same colour. Note that the number of balls in the bag will increase by one, after the trial. A sequence of four such trials is conducted. Which one of the following choices gives the probability of drawing a red ball in the fourth trial?

This experiment is a classical Pólya's Urn Model with replacement plus $1$ additional ball of the drawn color.

By the exchangeability property of Pólya's urn scheme, the probability of drawing a ball of a specified color at trial $n$ is equal to the initial proportion of that color in the urn, for every $n \ge 1$.

Let $R_k$ be the event that a red ball is drawn in the $k$-th trial.

1. For the first trial:

   $$P(R_1) = \frac{r}{r+b}$$

2. For the second trial, by the law of total probability:

   $$P(R_2) = P(R_2 \mid R_1)P(R_1) + P(R_2 \mid B_1)P(B_1)$$

   $$P(R_2) = \left(\frac{r+1}{r+b+1}\right)\left(\frac{r}{r+b}\right) + \left(\frac{r}{r+b+1}\right)\left(\frac{b}{r+b}\right) = \frac{r(r+1+b)}{(r+b)(r+b+1)} = \frac{r}{r+b}$$

By mathematical induction, for any $n$-th trial (including the $4$-th trial):

Therefore, option A is the correct choice.