SOLUTION: Can someone let me know if i'm on the right track here? A bag contains 8 green, 4 red, 5 yellow, and 6 blue marbles. Four marbles are drawn from the box at the same time. A.

Algebra ->  Probability-and-statistics -> SOLUTION: Can someone let me know if i'm on the right track here? A bag contains 8 green, 4 red, 5 yellow, and 6 blue marbles. Four marbles are drawn from the box at the same time. A.      Log On


   

Question 635451: Can someone let me know if i'm on the right track here?
A bag contains 8 green, 4 red, 5 yellow, and 6 blue marbles. Four marbles are drawn from the box at the same time.
A. What is the probability that all four are the same color?
B. What is the probability that none of the marbles are blue?
C. What is the probability that you get one marble of each color?
D. What is the probability that you draw more than 2 green?
I'm not sure if I am on the right track here or not...
A. P(all same color) = 8C4+4C4+5C4+6C4/23C4 = 91/8855
B. P (non are blue) = 17C4/23C4= 2380/8855
C. P (one of each) = 8*4*5*6/23C4 = 960/8855
D. P (More than 2 green) = 8C3*15C1=840 (3 yellow)
8C4*15C0=70 (4 yellow)
840+70/23C4= 910/8855
Any help would be greatly appreciated
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

A bag contains 23 marbles: 8 green, 4 red, 5 yellow, and 6 blue marbles

Four marbles are drawn from the box at the same time. 23C4 = 8855 total possiblities

A. P(all same color) = 8C4+4C4+5C4+6C4/23C4 = 91/8855  Yes!

B. P (non are blue) = 17C4/23C4= 2380/8855   Yes!

C. P (one of each) =+8C1%2A4C1%2A5C1%2A6C1%2F23C4%29 = 8*4*5*6/(23C4)= 960/8855  Yes!

D. P (More than 2 green) = 8C3*15C1=840 (3 yellow)

 8C4*15C0=70 (4 yellow)

 840+70/23C4= 910/8855  Yes!

Good work!