20 marbles: 8 red, 7 white, 5 blue; draw 4 w/o replacement

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lawrence
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Joined: Wed Apr 01, 2009 10:48 pm

20 marbles: 8 red, 7 white, 5 blue; draw 4 w/o replacement

Postby lawrence » Wed Jul 29, 2009 9:04 pm

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?

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stapel_eliz
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Postby stapel_eliz » Thu Jul 30, 2009 10:21 pm

I was really hoping somebody better qualified would jump in on this one.... :shock:

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.... :oops:

lawrence
Posts: 15
Joined: Wed Apr 01, 2009 10:48 pm

Re: 20 marbles: 8 red, 7 white, 5 blue; draw 4 w/o replacement

Postby lawrence » Thu Aug 06, 2009 3:22 pm

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!


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