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.
 A: There is no need to divide by $2$.  You need to multiply the probability of drawing a white ball on the first draw by the probability of drawing a green ball on the second draw.  Since the draws are independent and the first ball is replaced, the probability of drawing a green ball on the second draw given that a white ball was drawn on the first draw is just the probability of drawing a green ball from the urn.  Hence,
$$\Pr(\text{drawing white, then green}) = \Pr(W)\Pr(G \mid W) = \Pr(W)\Pr(G) = \frac{5}{20} \cdot \frac{8}{20} = \frac{1}{10}$$
A: First you calculate the probability of getting a white ball. $$P(white)=\frac{Number\ of\ white\ balls}{Total\ number\ of\ balls} = \frac{5}{20}$$ 
Then the probability of getting a green ball is $$P(green)=\frac{Number\ of\ green\ balls}{Total\ number\ of\ balls} = \frac{8}{20}$$ 
Because you put the ball back in the urn all the probabilities are independant
Thus $$P(White\ then\ Green) = P(White)*P_{White}(Green) = P(White)*P(Green) = \frac{5*8}{20*20} = \frac{1}{10}$$
A: 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 $= (5/20) * (8/20)$
