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Probability Question

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Source: — Data Sufficiency |
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by smithpa2 » Sat Sep 27, 2008 2:47 pm
This is how I approached the problem:

Let's say you first pick a red, then another red, then a green apple.
The probability would be (7/10)(6/9)(3/8)=7/40

You must next consider the number of different arrangements, and I believe this is where most make mistakes.

RRG
RGR
GRR

or if you like combinations,

The number of ways to pick 2 red apples: 3C2 or 3
Or the number of ways to pick 1 green apple: 3C1 or 3...same thing.

There you have 3 possible arrangement, hence (3)(7/10)=21/40.
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by ddm » Sat Sep 27, 2008 6:22 pm
The number of ways to pick 2 red apples: 3C2 or 3

Number of ways to pick 2 red apples should be 7c2?

or am i making some mistake?
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by 4meonly » Sun Sep 28, 2008 3:19 am
ddm wrote:
The number of ways to pick 2 red apples: 3C2 or 3

Number of ways to pick 2 red apples should be 7c2?

or am i making some mistake?
Number of ways to pick 2 red apples from 3

To simplify, multiply 7/40 by 3
because you have 3 opportunities to get gree apple - i can be 1st, 2nd and 3rd in the group of three


smithpa2's post:

RRG
RGR
GRR
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by cubicle_bound_misfit » Sun Sep 28, 2008 4:56 am
the answer is

(7c2*3c1)/10c3

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

which comes down to 21/40
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