SOLUTION: An urn contains 15 balls identical in every respect except color. There are 4 red balls, 8 green balls, and 3 blue balls. (a) You draw two balls from the urn but replace the fir

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: An urn contains 15 balls identical in every respect except color. There are 4 red balls, 8 green balls, and 3 blue balls. (a) You draw two balls from the urn but replace the fir      Log On

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Question 364123: An urn contains 15 balls identical in every respect except
color. There are 4 red balls, 8 green balls, and 3 blue balls.
(a) You draw two balls from the urn but replace the first
ball before drawing the second. Find the probability
that the first ball is red and the second is green.
(b) Repeat part (a) but do not replace the first ball before
drawing the second

Answer by ewatrrr(10682) About Me  (Show Source):
You can put this solution on YOUR website!


Hi, 

15 balls f which 4 are red, 8 are green, and 3 are blue 

With replacement of the first ball

P(first ball is red and the second is green) = 4/15 * 8/15 = 32/225= .142



without replacement (14 total balls in the urn on the second draw)

P(first ball is red and the second is green) = 4/15 * 8/14 = 32/210 = .152