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# Pick 2 out of 10 (w/ combinatorics)

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Manager
Joined: 12 May 2009
Posts: 54
Followers: 2

Kudos [?]: 98 [0], given: 18

Pick 2 out of 10 (w/ combinatorics) [#permalink]

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04 Nov 2009, 17:31
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hi there,

I'm working on my last weak area before I am going to take the GMAT: PROBABILITY AND COMBINATORICS. So I apologize in advance if some of the questions I'm going to ask you guys are too easy.

QST: In a box with 10 blocks, 3 of which are red, what is the probability of picking a red block on each of your first two tries? Assume that you do NOT replace the first block after you have picked it.

ANS: I'm aware of the traditional way to solve problems like this (probability trees). By doing so, the answer is 3/10 * 2/9 = 1/15. Pretty straight forward.

Now, I want to also be able to solve this kind of problem with a counting method (combinatorics). So I thought I can solve it like this:

(3*2)/(10C2)

Unfortunately, I get the wrong result (2/15). This is double the OA (1/15). Any idea how to use the combination formula correctly in this scenario?

Thanks
S.
Math Expert
Joined: 02 Sep 2009
Posts: 36598
Followers: 7095

Kudos [?]: 93455 [0], given: 10563

Re: Pick 2 out of 10 (w/ combinatorics) [#permalink]

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05 Nov 2009, 06:34
enfinity wrote:
Hi there,

I'm working on my last weak area before I am going to take the GMAT: PROBABILITY AND COMBINATORICS. So I apologize in advance if some of the questions I'm going to ask you guys are too easy.

QST: In a box with 10 blocks, 3 of which are red, what is the probability of picking a red block on each of your first two tries? Assume that you do NOT replace the first block after you have picked it.

ANS: I'm aware of the traditional way to solve problems like this (probability trees). By doing so, the answer is 3/10 * 2/9 = 1/15. Pretty straight forward.

Now, I want to also be able to solve this kind of problem with a counting method (combinatorics). So I thought I can solve it like this:

(3*2)/(10C2)

Unfortunately, I get the wrong result (2/15). This is double the OA (1/15). Any idea how to use the combination formula correctly in this scenario?

Thanks
S.

Total # of favorable outcomes: 2 red block out of three: 3C2=3
Total # of outcomes: any 2 blocks out of 10: 10C2=45
P=3/45=1/15

If you are putting in denominator the number of selections of 2 out of 10, you should also put in nominator number of selections of 2 out of 3.

OR
# of selections of any first block out of 10=10C1, # of selection of any second out of 9=9C1=9, total 10C1*9C1=90
# of selections of the first red block out of three=3C1=3, # of selections of second red block out of three=2C1=2, total 3C1*2C1=6.
P=6/90=1/15

Hope it helps.

It would be better to post the questions in PS or DS forums to get more and quick replies.
_________________
Re: Pick 2 out of 10 (w/ combinatorics)   [#permalink] 05 Nov 2009, 06:34
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