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In a box of 12 pens, a total of 3 are defective. If a custom
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02 Sep 2020, 09:43
02 Sep 2020, 09:43
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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
A. 1/6
B. 2/9
C. 6/11
D. 9/16
E. 3/4
Re: In a box of 12 pens, a total of 3 are defective. If a custom
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02 Sep 2020, 09:44
02 Sep 2020, 09:44
Expert Reply
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Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
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Re: In a box of 12 pens, a total of 3 are defective. If a custom
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02 Sep 2020, 10:18
02 Sep 2020, 10:18
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Carcass 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
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.
