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In a box of 12 pens, a total of 3 are defective. If a customer buys 2

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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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New post 18 Jun 2019, 11:15
Though I understand that C is correct answer. Can someone please explain why B is wrong? If one selects 2 pencils out of 9 which are good then prob should be 2/9 ??

Confusion is because I reckon a sum which asked prob of selecting 2 red balls as 2/5 out of 5 red and 4 blue balls. Can someone pls clear this? Thank you.

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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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New post 19 Jun 2019, 07:43
GMATinsight wrote:
Bunuel 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


Kudos for a correct solution.


Method- 1
There are 9 fine pieces of pen and 3 defective in a lot of 12 pens

i.e. Probability of first pen NOT being defective = (9/12)
i.e. Probability of Second pen NOT being defective = (8/11) [11 pen remaining with 8 defective remaining considering that first was defective]

Probability of Both pen being NON-defective = (9/12)*(8/11) = 6/11

Answer: option C

Method- 2
There are 9 fine pieces of pen and 3 defective in a lot of 12 pens

No. of ways of choosing 2 NON defective out of 9 fine pieces of pen = 9C2
No. of ways of choosing 2 out of 12 pieces of pen = 12C2

Required probability that none of the chosen is defective = 9C2 / 12C2 = 36 / 66 = 6/11

Answer: option C


[11 pen remaining with 8 defective remaining considering that first was defective]....Is it a typo or i am missing something . Shouldn't it be [11 pen remaining with 8 fine piece remaining considering that first was non-defective]
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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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New post 10 Feb 2020, 09:20
1
hi can some kindly point out my mstake
3c2/12c2+(3c1*9c1/12c2)2

I understand that it should be 3c1/12*9c1/11 but how should i distinguish them. like which one to use where Kindly guide me
Zillion thanks in advance Bunuel ScottTargetTestPrep
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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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New post 10 Feb 2020, 11:56
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suganyam wrote:
hi can some kindly point out my mstake
3c2/12c2+(3c1*9c1/12c2)2

I understand that it should be 3c1/12*9c1/11 but how should i distinguish them. like which one to use where Kindly guide me



suganyam wrote:
hi can some kindly point out my mistake
3c2/12c2+(3c1*9c1/12c2)2

I understand that it should be 3c1/12*9c1/11 but how should i distinguish them. like which one to use where Kindly guide me
Zillion thanks in advance



Question:
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?

There are 9 non-defective pens out of 12

If you want to pick simultaneously VS you pick one at a time without replacement, the answers would be SAME

If you want to do this by picking simultaneously, you should interpret the question as:
Favorable: Pick 2 pens that are not defective i.e. pick 2 pens from 9 = 9C2 ways = 9*8/2! = 36 ways
Total: Pick 2 pens from 12 = 12C2 ways = 12*11/2! = 66 ways
=> Probability = 36/66 = 6/11

If you want to do this by picking one at a time, you should interpret the question as:
Pick 1st pen which is not defective AND Pick 2nd pen which is not defective (Note: AND means MULTIPLY)

Probability of picking 1st pen which is not defective = 9/12

Not, since we pick WITHOUT REPLACEMENT, we have 8 non-defective out of 11

Thus, probability of picking 2nd pen which is not defective = 8/11

Thus, required probability = 9/12 * 8/11 = 6/11
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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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New post 14 Feb 2020, 09:40
suganyam wrote:
hi can some kindly point out my mstake
3c2/12c2+(3c1*9c1/12c2)2



Your reasoning is not correct. Suppose you are choosing 2 pencils out of 9 pencils which are all good; i.e. there are no defective pencils in the box. The two pencils you choose will certainly be both non-defective. In this case, the probability that neither pencil is defective is 1 or equivalently, 100%. According to your reasoning, since there are 9 good pencils and we are choosing 2, the probability should have been 2/9.
Conversely, suppose that there are like a million defective pencils in the box and only 9 good pencils. Again according to your method, since there are 9 good pencils and we are choosing 2, the probability should have been 2/9 but in reality, you can observe that if there were a million defective pencils and only 9 good pencils, the probability of choosing two good pencils is very very low, almost zero.

For the other question you mentioned, if there were 5 red and 4 blue balls in a box and we were choosing 2 balls without replacement; the probability that both are red is (5/9)*(4/8) = 5/18 and the probability that both are blue is (4/9)*(3/8) = 1/6.
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Re: In a box of 12 pens, a total of 3 are defective. If a customer buys 2   [#permalink] 14 Feb 2020, 09:40

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