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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, 18:45

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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?

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, 19:51

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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

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, 20:36

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!

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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27 Jul 2016, 15:18

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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?

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