The student T distribution would be used since we don't know the population variance (by extension the population standard deviation).
However, if n > 30, then the student T distribution is going to look very similar to the standard normal Z distribution. It won't be a perfect match, but it'll be close enough that it's more simple to use the Z distribution. With the Z distribution, you only have one set of values and you don't have to worry about the degrees of freedom.
Let E be the event that the first marble selected is green. Let F be the event that the second marble selected is green. A box contains 20 blue marbles, 16 green marbles and 14 red marbles P(F/E)=15/49 because if the first marble selected is green there are 49 in total and 15 are green. I think this is it.