Given $5$ white balls, $8$ green balls and $7$ red balls. Find the probability of drawing a white ball then a green one. Given $5$ white balls, $8$ green balls and $7$ red balls in an urn. Find out the probability to draw a white ball and then a green one if the drawing is done consecutively and after drawing the ball is returned into the urn.
What I ended up with as an answer is $1/20$ by taking the chance for drawing a white ball and multiplying it by the chance to get a green ball and then dividing by two, since I only want the case where the white ball is first, which I assume is half the cases.

Total number of balls $= 20$.
Probability of drawing a white ball $= 5/20.$
Now, the total number of balls is $19.$
Probability of drawing a green ball is $= 8/19.$
Thus, the required probability without replacement $= (5/20) * (8/19)$
and the required probability with replacement $= (5/20) * (8/20)$
