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Probability of not selecting something

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crackgmat007 Legendary Member Default Avatar
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Probability of not selecting something

Post Thu May 07, 2009 9:50 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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

    Can someone explain the concept pls? Thanks.

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    VP_Jim GMAT Instructor Default Avatar
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    Post Thu May 07, 2009 10:14 am
    I like to think of these questions with slightly different wording:

    What is the probability of buying a non-defective pen and then another non-defective pen?

    Initially, there are 9 out of 12 non defective, so the probability of buying a non defective pen for the first pick is 9/12 = 3/4

    Now, for the second pen, one of the "good" pens is now gone. So, 8 of the remaining 11 pens are not defective.

    So, (3/4)*(8/11) = 6/11 = C

    Hope this helps!

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    crackgmat007 Legendary Member Default Avatar
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    Post Thu May 07, 2009 10:57 am
    Definitely. Thanks for solving.

    In order to understand the logic better, I tried to find the probability of both pens being defective first and deduce the probability of both pens not being defective.

    Probability of first pen being defective - desired =3, total =12, hence 3/12 =>1/4

    For second pen being defective - desired =2, total = 11, hence 2/11

    Both pens being defective = (1/4)*(2/11)=>1/22

    But based on this probability, it seems like we cannot arrive at probability of both pens not being defective. Is there something I am missing? Pls let me know. thanks much!

    VP_Jim GMAT Instructor Default Avatar
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    Post Thu May 07, 2009 12:26 pm
    What you did is find the probability of not getting two defective pens - in other words, you have included the possibility of getting one defective but the other not defective - when the question is asking for the probability of getting NO defective pens.

    In math terms, what you did is:

    1 - (prob. of both defective)

    Which is not the same thing as:

    (1 - prob. of defective) x (1- prob. of defective)

    Does that make sense?

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    crackgmat007 Legendary Member Default Avatar
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    Post Thu May 07, 2009 1:33 pm
    I see your point. Thanks for the clarification

    Steppanyaki Newbie | Next Rank: 10 Posts
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    Post Sat May 09, 2009 7:35 am
    But couldn't you theoretically find the probability of choosing non defective by taking 1 and subtracting it with the probability of choosing defective? I get so confused about the right or wrong time to apply this rule.

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