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25 Jan 2008, 10:44
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5 blue marbles, 3 red marbles and 4 purple marbles are placed in a bag. If 4 marbles are drawn without replacement, what is the probability that the result will not be 2 blue and 2 purple marbles?
A) 4/33
B) (5/36)^2
C) 1/2
D) (31/36)^2
E) 29/33
A) 4/33
B) (5/36)^2
C) 1/2
D) (31/36)^2
E) 29/33
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25 Jan 2008, 13:11
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\(p=\frac{C^5_2*C^4_2*P^4_4}{P^{12}_4}=\frac{5!*4!*4!*8!}{3!*2!*2!*2!*12!}=\frac{5*4*4*3*4*3}{2*12*11*10*9}=\frac{4}{33}\)
\(p=\frac{C^5_2*C^4_2*P^4_4}{P^{12}_4}=\frac{5!*4!*4!*8!}{3!*2!*2!*2!*12!}=\frac{5*4*4*3*4*3}{2*12*11*10*9}=\frac{4}{33}\)
