Question 1189328: a bag contains 5 white and 7 black balls. If 3 are drawn from the bag, What is the probability that:
a) All are white
b) 2 are white and one is black
c) All are of sam colour
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
Sorry -- in my original response I had the numbers of white and black balls switched. Here are the corrected calculations:
(1) Total number of ways of choosing 3 of the 5+7=12 balls: C(12,3)
(2) Number of ways of choosing all 3 white (and 0 black): C(5,3)*C(7,0) (not C(7,3)*C(5,0))
(3) Number of ways of choosing 2 white and 1 black: C(5,2)*C(7,1) (not C(7,2)*C(5,1))
(4) Number of ways of choosing all 3 white OR all 3 black: C(7,3)*C(5,0) + C(7,0)*C(5,3)
Note in the above calculations that "and" means the numbers of ways are multiplied, while "or" means the numbers of ways are added.
ANSWERS:
a) (2) divided by (1)
b) (3) divided by (1)
c) (4) divided by (1)
You can do the calculations
Answer by ikleyn(52813)  (Show Source):
You can put this solution on YOUR website! .
A bag contains 5 white and 7 black balls. If 3 are drawn from the bag  ,
what is the probability that
a) All are white
b) 2 are white and one is black
c) All are of same color
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In the post by @greenestamps, the formulas (2) and (3) should be corrected in this way
(2) Number of ways of choosing all 3 white (and 0 black): C(5,3)*C(7,0)
(3) Number of ways of choosing 2 white and 1 black: C(5,2)*C(7,1)
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