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# In a box of 12 pens, a total of 3 are defective. If a customer buys 2

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In a box of 12 pens, a total of 3 are defective. If a customer buys 2 [#permalink]

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08 Nov 2008, 19:45
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Question Stats:

80% (01:51) correct 20% (01:16) wrong based on 97 sessions

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In a box of 12 pens, a total of 3 are defective. If a customer buys 2 pens selected at random from the box, what is the probability that neither pen will be defective?

A. 1/6
B. 2/9
C. 6/11
D. 9/16
E. 3/4

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-a-box-of-12-pens-a-total-of-3-are-defective-if-a-customer-buys-207395.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 28 Jul 2016, 05:18, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: In a box of 12 pens, a total of 3 are defective. If a customer buys 2 [#permalink]

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08 Nov 2008, 19:59
1
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# defective pens = 3
# good pens = 9

Probability of the 1st pen being good = # of favorable outcomes / # of total outcomes = 9/12

Probability of the 2nd pen being good = # of remaining favorable outcomes / # of total remaining outcomes = 8 / 11

Total probability = 9/12 * 8/11 = 6/11

My pick C
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Re: In a box of 12 pens, a total of 3 are defective. If a customer buys 2 [#permalink]

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08 Nov 2008, 20:51
1
KUDOS
gorden wrote:
In a box of 12 pens, a total of 3 are defective. If a customer buys 2 pens selected at random from the box, what is the probability that neither pen will be defective?

A. 1/6
B. 2/9
C. 6/11
D. 9/16
E. 3/4

I have another way to solve by using combinations.
First, there are 9C2 ways you can select 2 good pencils from 9 good ones.
Second, there are 12C2 ways you select 2 pencils from 12 ones in the box.
Then, the probability that neither pen will be defective is: 9C2/12C2=6/11

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Re: In a box of 12 pens, a total of 3 are defective. If a customer buys 2 [#permalink]

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09 Jul 2011, 21:36
Thanks all for the answer. I'm doing this question and do not know why I was wrong.
This is the way how I solve this: calculating the probability both of pens he pick are defective then take 1 subtract the result.
1st is defective = 3/12
2nd is defective = 2/11
--> pro. of chance both is defective is: 3/12 * 2/11 = 21/22
--> pro. of chane neither is defective is: 1 - 21/22 = 1/22

(Yep, There is no result like this in the choice. I just want to know why I was wrong.)
Thanks much!
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Re: In a box of 12 pens, a total of 3 are defective. If a customer buys 2 [#permalink]

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13 Jul 2011, 05:04
1
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Expert's post
dizim wrote:
Thanks all for the answer. I'm doing this question and do not know why I was wrong.
This is the way how I solve this: calculating the probability both of pens he pick are defective then take 1 subtract the result.
1st is defective = 3/12
2nd is defective = 2/11
--> pro. of chance both is defective is: 3/12 * 2/11 = 21/22
--> pro. of chane neither is defective is: 1 - 21/22 = 1/22

(Yep, There is no result like this in the choice. I just want to know why I was wrong.)
Thanks much!

You are not getting the right answer because there is another case: Probability that only one pen is defective.
Probability that both are defective + Probability that only one is defective + Probability that neither is defective = 1
You need to calculate the probability that only is defective as well and subtract that from 1 too.
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Re: In a box of 12 pens, a total of 3 are defective. If a customer buys 2 [#permalink]

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13 Jul 2011, 06:51
Thanks VeritasPrepKarishma!
Kudos for you ^^
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Re: In a box of 12 pens, a total of 3 are defective. If a customer buys 2 [#permalink]

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27 Jul 2016, 16:18
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In a box of 12 pens, a total of 3 are defective. If a customer buys 2 [#permalink]

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27 Jul 2016, 23:24
9C2 / 12C2
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Re: In a box of 12 pens, a total of 3 are defective. If a customer buys 2 [#permalink]

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28 Jul 2016, 05:19
gorden wrote:
In a box of 12 pens, a total of 3 are defective. If a customer buys 2 pens selected at random from the box, what is the probability that neither pen will be defective?

A. 1/6
B. 2/9
C. 6/11
D. 9/16
E. 3/4

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-a-box-of-12-pens-a-total-of-3-are-defective-if-a-customer-buys-207395.html
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Re: In a box of 12 pens, a total of 3 are defective. If a customer buys 2   [#permalink] 28 Jul 2016, 05:19
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