) A bag contains 12 blue marbles, 10 red marbles, and 8 green marbles. Three marbles are selected at random from the bag. Determine the probability of each event.

Question

a) All three marbles are green. b) Exactly two marbles are blue. c) One marble is blue, one is red, and one is green

Answer 1

B=12, R = 10, G = 8

P(3G) = (8/30)(7/29)(6/28) = 336/24360 = about .01379 = 1.379% to get 3 greens in 3 draws

P(2B) = (12/30)(11/29) = 132/870 = about .1517 = 15.17% to get 2 blues in 2 draws

but that's not the answer, as it's 3 draws

P(2B out of 3) = P(BB~B) + P(B~BB) + (~BBB) = 3(12/30)(11/29)(18/28) = 3(12)(11)(18)/(30)(29)(28) = 7128/24360 = about .2926 = 29.26%

P(BRG or BGR or RGB or RBG or GRB or GBR)

= (12/30)(10/29)(8/28) + (12/30)(8/29)(10/28) + (10/30)(8/29)(12/28) + (10/30)(12/29)(8/28) + (8/30)(10/29)(12/28) + (8/30)(12/29)(10/28)

= 6(12)(10)(8)/30)(29)(28)

= 5760/24360

= about 23.65

= about 23.65% probability of one of each color in 3 draws