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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

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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

by VJesus12 » Wed May 20, 2020 6:15 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

[spoiler]OA=C[/spoiler]

Source: Official Guide
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Re: 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

by Brent@GMATPrepNow » Wed May 20, 2020 5:52 pm
VJesus12 wrote: ↑
Wed May 20, 2020 6:15 am
 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

[spoiler]OA=C[/spoiler]

Source: Official Guide
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.
Brent Hanneson - Creator of GMATPrepNow.com
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Re: 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

by Scott@TargetTestPrep » Sun May 24, 2020 7:21 am
VJesus12 wrote: ↑
Wed May 20, 2020 6:15 am
 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

[spoiler]OA=C[/spoiler]

Source: Official Guide
Solution:

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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