I have twenty marbles in a bowl: eight are red, seven are white, and five are blue. Let's suppose I draw four marbles without replacement.
A. What is the probability that none of the four is red?
B. What is the probability that I draw at least one of each color?
20 marbles: 8 red, 7 white, 5 blue; draw 4 w/o replacement
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stapel_eliz
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I was really hoping somebody better qualified would jump in on this one.... 
A) You're choosing four of the twenty. There are C(20,4) ways to choose four of twenty.
You want to make sure that, of the eight red, you pick zero. There are C(8,0) ways to do this.
You want to make sure that, of the other twelve you pick some random four. There are C(12,4) ways to do this.
Then the probability of picking four, none of which is red, should (I believe) be:
. . . . .\, C(8,0)}{C(20,4)})
...or about 0.1021671827..., or about 10.2%.
B) Um.... Lemme get back to you on this....
A) You're choosing four of the twenty. There are C(20,4) ways to choose four of twenty.
You want to make sure that, of the eight red, you pick zero. There are C(8,0) ways to do this.
You want to make sure that, of the other twelve you pick some random four. There are C(12,4) ways to do this.
Then the probability of picking four, none of which is red, should (I believe) be:
. . . . .
...or about 0.1021671827..., or about 10.2%.
B) Um.... Lemme get back to you on this....
Re: 20 marbles: 8 red, 7 white, 5 blue; draw 4 w/o replacement
I did the same sort of reasoning for each of the individual cases (2 red, 1 white, 1 blue; 1 red, 2 white, 1 blue; and 1 red, 1 white, 2 blue) and got the right answer. Thank you!