The correct answer is 1. Your reasoning is correct by repeated applications of Bayes' theorem that relates *joint* and *conditional probabilities*.

P(B1=Y, B2=Y) = P(B2=Y| B1=Y) P(B1=Y)

As you correctly point out in point out in 1, P(B1 = Y) = 4/8, and P(B2 = Y | B1 = Y) = 3/7, i.e. that the probability of ball 2 being yellow *conditioned upon* or *knowing that* the first ball is yellow is 3/7.

We can chain this rule together for multiple events:

P(B1=Y, B2=Y, B3=Y) = P(B3=Y| B1=Y, B2=Y) P(B1=Y, B2=Y)

= P(B3=Y| B1=Y, B2=Y) P(B2=Y |B1 = Y) P(B1 = Y)

And so on, yielding your answer (4/8)(3/7)(2/6)(1/5).

Your second answer is wrong because you count incorrectly. There are 4!4! ways to draw 4 yellow balls then 4 remaining balls. There are 8! ways to draws 8 balls. 4!4!/8! = 4\*3\*2\*~~4~~\*~~3~~\*~~2~~/(8\*7\*6\*5\*~~4~~\*~~3~~\*~~2~~) = (4/8)(3/7)(2/6)(1/5).