Which answer is correct for this product rule based probability problem The question reads like this:

A bag contains 5 black and 3 red balls. A ball is taken out of the bag and is not returned to it. If this process is repeated three times, then what is the probability of drawing a black ball in the next draw?

Solution 1
I think the three balls can be drawn in 4 ways 


*

*All the three balls are black = 5/8 .4/7 .3/6 then the probability of next black ball can be drawn is 2/5.

*2 black and 1 red = 5/8 . 4/7 . 3/6 then the probability of next black ball can be drawn is 3/5.

*1 black and 2 red = 5/8 . 3/7 . 2/6 then the probability of next black ball can be drawn is 4/5.

*All red = 3/8 . 2/7 . 1/6 then the probability of next black ball can be drawn is 1.


And probability of each way is 1/4.
Hence desired probability $=\frac{1}{4}\times\frac{2}{5}+\frac{1}{4}\times\frac{3}{5}+\frac{1}{4}\times\frac{4}{5}+\frac{1}{4}\times1=\frac{7}{10} = 0.7$
Solution 2
Probability of 4th ball being black when


*

*All first first 3 draws are black balls $=\frac{5}{8}\times\frac{4}{7}\times\frac{3}{6}\times\frac{2}{5}=\frac{120}{1680}$

*First three draws contain all red balls $=\frac{3}{8}\times\frac{2}{7}\times\frac{1}{6}\times\frac{5}{5}=\frac{30}{1680}$

*First three draws contain 2 red and 1 black ball $=\frac{3}{8}\times\frac{2}{7}\times\frac{5}{6}\times\frac{4}{5}\times 3=\frac{360}{1680}$

*First three draws contain 1 red and 2 black ball $=\frac{3}{8}\times\frac{5}{7}\times\frac{4}{6}\times\frac{3}{5}\times 3=\frac{540}{1680}$


Summing all above probabilities gives $=\frac{5}{8}=0.625$
I cant decide which one is correct. I feel the second one is just far more precise. Is it?

You are overcomplicating things.   There are eight balls, and any one of which could be the fourth ball drawn with equal probability.   Five of these balls are black.   The probability that a black ball is drawn on the fourth draw is thus: $$5/8$$

This agrees with your second solution, which correctly considers that there are three ways to arrange two balls of one colour and one of another.

To see this in action, lets examine ways to draw balls from a bag of two black and two red balls, and count those where the third ball is black.  By my method we should expect $2/4$ of the possible arrangements to have a black ball drawn in third place. Let's see...
$$\begin{array}{cc}\bullet\bullet\circ\circ \\\bullet\circ\bullet\circ &\star\\ \bullet\circ\circ\bullet \\ \circ \bullet\bullet\circ & \star \\ \circ\bullet\circ\bullet \\ \circ\circ\bullet\bullet & \star\end{array}$$
