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Probability of not selecting something

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by crackgmat007 » Thu May 07, 2009 9:50 am
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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

Can someone explain the concept pls? Thanks.
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by VP_Jim » 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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by crackgmat007 » 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!
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by VP_Jim » 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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by crackgmat007 » Thu May 07, 2009 1:33 pm
I see your point. Thanks for the clarification
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by Steppanyaki » 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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by Scott@TargetTestPrep » Wed Dec 13, 2017 3:52 pm
crackgmat007 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
Since there are 3 defective pens from 12, the probability of selecting the first non-defective pen is 9/12 and the probability of selecting the second non-defective pen is 8/11. Thus, the probability that a customer buys 2 non-defective pens is 9/12 x 8/11 = 3/4 x 8/11 = 24/44 = 6/11.

Answer: C

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by Brent@GMATPrepNow » Wed Dec 13, 2017 4:01 pm
crackgmat007 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
P(neither pen is defective) = P(1st pen selected is NOT defective AND 2nd pen selected is NOT defective)
= P(1st pen selected is NOT defective) x P(2nd pen selected is NOT defective)
= 9/12 x 8/11
= 6/11
= C

ASIDE: How did I get 9/12 and 8/11?
For the first selection, 9 of the 12 pens are good.
For the second selection, we must assume that the first selection resulted in a GOOD pen. This means there are now 11 pens remaining, and 8 of them are GOOD.

Cheers,
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by [email protected] » Mon Jun 11, 2018 6:26 pm
Hi All,

We're told that in a box of 12 pens, a total of 3 are defective and that a customer buys 2 pens selected at random from the box. We're asked for the probability that neither pen will be defective. This question can be solved in a couple of different ways; treating it as a step-by-step Probability question is likely the easiest option for most Test Takers, but you can also solve it using the Combination Formula.

Since we are choosing 2 pens from a total of 12 pens, there are 12c2 = 12!/(2!)(10!) = (12)(11)/(2)(1) = 66 possible groups of 2 pens

We want to choose 2 pens that are NOT defective. Since there are 9 of those non-defective pens to choose from, there are 9c2 = 9!/(2!)(7!) = (9)(8)/(2)(1) = 36 groups of 2 pens that are not defective.

Thus, the probability of puling two non-defective pens is 36/66 = 6/11.

Final Answer: C

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