SOLUTION: An urn contains two red balls and two green balls. A ball is drawn randomly. If the ball is red it is replaced to the urn as before; it it is green, it is replaced, along with five

Algebra ->  Probability-and-statistics -> SOLUTION: An urn contains two red balls and two green balls. A ball is drawn randomly. If the ball is red it is replaced to the urn as before; it it is green, it is replaced, along with five      Log On


   

Question 1028796: An urn contains two red balls and two green balls. A ball is drawn randomly. If the ball is red it is replaced to the urn as before; it it is green, it is replaced, along with five more red balls, so that there are now a total of nine balls in the urn. Then a second ball is drawn randomly.
1.Find the probability that the second ball is red
2. Find the conditional probability that the first ball was red given that the second one is red
Answer by Natolino1983(23) About Me  (Show Source):
You can put this solution on YOUR website!
1) P(second ball is red) =P(second ball is red/first is Red)×P(first is red) + P(second ball is red/first is green)×P(first is green) = 2/4×2/4 +7/9×2/4 = 23/36.
2) P(first is red/second ball is red)= P(second ball is red/first is red)×P(first is red)/P(second ball is red) =(7/9)×(2/4)/(23/36) = 14/23.