Jtdylktuy

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.

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,
$$Pr(text{drawing white, then green}) = Pr(W)Pr(G mid W) = Pr(W)Pr(G) = frac{5}{20} cdot frac{8}{20} = frac{1}{10}$$

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 $$P(White then Green) = P(White)*P_{White}(Green) = P(White)*P(Green) = frac{5*8}{20*20} = frac{1}{10}$$