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04 Oct 2009, 23:30
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
84% (01:06) correct
16%
(01:12)
wrong
based on 33
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Please help me to solve that problem.
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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
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
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
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
