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Comb&Prob from GMATPrep

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DenisSh
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Comb&Prob from GMATPrep [#permalink] New post   04 Oct 2009, 23:30
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

84% (01:06) correct 16% (01:12) wrong based on 33 sessions
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Please help me to solve that problem.



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Re: Comb&Prob from GMATPrep [#permalink] New post   05 Oct 2009, 00:54
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Say the first two red and the last is green. Probability of that perm. is =
7/10*6/9*3/8=126/720=7/40
There are 3 different permutations=
RRG
RGR
GRR
So the total prob is. 3*7/40=21/40
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Re: Comb&Prob from GMATPrep [#permalink] New post   05 Oct 2009, 01:09
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The probability is equivalent to the total number of ways to select 2 red apples and 1 green apple versus the total number of ways to select 10 apples.

Total number of ways to select 2 red apples and 1 green apple: 7C2 * 3C1

Total number of ways to select 3 apples: 10C3

Probability = \(\frac{7C2 * 3C1}{10C3}\)

\(= \frac{(\frac{7*6}{2*1})*3}{(\frac{10*9*8}{3*2*1})}\)

\(=\frac{21}{40}\)

Therefore, the correct answer is D: 21/40
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Re: Comb&Prob from GMATPrep [#permalink] New post   03 May 2011, 08:11
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3* (7/10) *(6/9) * (1/3)
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Re: Comb&Prob from GMATPrep [#permalink] New post   05 May 2011, 17:12
=(7c2* 3c1)/(10c3)

= 21/40
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Re: Comb&Prob from GMATPrep [#permalink] New post   11 May 2011, 08:20
(7C2 * 3C1)/10C3

(7!/5!2! * 3)/(10!/3!7!)


= (7*6/2 * 3)/(10 * 9 * 8)/6

= 63/120 = 21/40

Answer - D
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Re: Comb&Prob from GMATPrep [#permalink] New post   11 May 2011, 15:41
For Red - 7C2
for Green - 3C1
=> 7C2*3C1/10C3
= 21/40



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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
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Re: Comb&Prob from GMATPrep [#permalink]
11 May 2011, 15:41
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