PLease show work. Thank you Here\'s a second game, You have a deek of cards. each colored either red, green or blue. Suppose inilially there are two red. three green and four blue cards in the deck. A card, say Do {r, g, b}, is drawn which sets the point (color) to match. You are then allowed to draw t wo subsequent cards, say D1 and D2. When the point is successfully matched you win. otherwise you lose, (a) If each card is put back into the deck immediately after drawn, determine the probability that you win the game, (b) If each card is not put back into the deck after being drawn, determine the winning probability. Solution (a)We have three possible cases in the scenario (i)D0 is red (ii)D0 is green (iii)D0 is blue for the first case the situations in which i will win are as follows: (1)D1 is red and D2 is not (2)D1 is not red and D2 is red (3)D1 is red and D2 isred The probabilty of (i)(1) is (2/9)*(2/9)*(7/9) The probabilty of (i)(2) is (2/9)*(7/9)*(2/9) The probabilty of (i)(3) is (2/9)*(2/9)*(2/9) Similarly cases for (ii)&(iii) will give me the following probabilities The probabilty of (ii)(1) is (3/9)*(3/9)*(6/9) The probabilty of (ii)(2) is (3/9)*(6/9)*(3/9) The probabilty of (ii)(3) is (3/9)*(3/9)*(3/9) The probabilty of (iii)(1) is (4/9)*(4/9)*(5/9) The probabilty of (iii)(2) is (4/9)*(5/9)*(4/9) The probabilty of (iii)(3) is (4/9)*(4/9)*(4/9) Adding up all these probabilites (47/81) (b) As in case (a) we will have the exactly same 9 cases but in this case the probability of drawing a ball will also depend upon all the previous draws The probabilty of (i)(1) is (2/9)*(1/8)*(7/7) The probabilty of (i)(2) is (2/9)*(7/8)*(1/7) The probabilty of (i)(3) is (2/9)*(1/8)*(0/7) The probabilty of (ii)(1) is (3/9)*(2/8)*(6/7) The probabilty of (ii)(2) is (3/9)*(6/8)*(2/7) The probabilty of (ii)(3) is (3/9)*(2/8)*(1/7) The probabilty of (iii)(1) is (4/9)*(3/8)*(5/7) The probabilty of (iii)(2) is (4/9)*(5/8)*(3/7) The probabilty of (iii)(3) is (4/9)*(3/8)*(2/7) Adding up all these probabilites (125/252).