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

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20 Oct 2015, 02:18
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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

Kudos for a correct solution.

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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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20 Oct 2015, 03:27
18
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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

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

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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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20 Oct 2015, 03:03
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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.

My Solution:

Total Pens = 12 Nos

Defective pens = 3

Remaining = 9, Therefore, selecting 2 no defective pens from 9 = 9C2 ways, and selecting 2 pens from 12 = 12C2 ways

Probability of neither pen will be defective = 9C2/12C2 = 36/66 = 6/11 Answer is C

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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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20 Oct 2015, 03:32
Total number of pens = 12
Defective pens = 3
Non- defective pens= 9

Probablity of selecting 2 non defective pens = 9C2/ 12C2 = 6/11

Alternatively , we can use probablity = No of favorable outcomes / No of total outcomes
= 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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20 Oct 2015, 03:41
1
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.

as already stated, we can solve it with oposite probalbility:
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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20 Oct 2015, 08:59
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.

If 3 pens are defective, 9 are good. Selecting 2 good pens and no defective one:

Probability = $$\frac{9}{12}$$ * $$\frac{8}{11}$$ = $$\frac{6}{11}$$

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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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21 Feb 2016, 10:38
If 3 out of the 12 pens are defective, 9 out of the 12 pens are not defective.

If a customer first select 1 pen at random from a box, the probability that this pen is not defective is $$\frac{9}{12}.$$ If the customer then selects, at random, the second pen from the box, the probability that this pen is not defective is $$\frac{8}{11}$$. Note that the customer already took 1 pen out of the box without replacement, hence, there are only 11 pens left in the box instead of 12 and the probability of selecting a pen that is not defective is 8 out of 11 and NOT 9 out of 12. (I fell for this mistake)

Therefore, the probability of taking 2 pens at random from the box that are not defective is: $$\frac{9}{12} * \frac{8}{11}= \frac{18}{33} = \frac{6}{11}.$$
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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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17 Aug 2016, 17:51
1
1
In case anyone is having trouble seeing the alternative solution (although longer, may help conceptualize how probabilities work) here it goes:

12 Pens - 3 Defective - 2 Selected - P (neither defective)

Find all probabilities of defective pens:
P (both defective) + P (1 defective, 1 not defective) + P (1 not defective, 1 defective)

P (both defective) = 3/12 * 2/11 = 1/4 * 2/11 = 2/44
P (1 defective, 1 not defective) = 3/12 * 9/11 = 1/4 * 9/11 = 9/44
P (1 not defective, 1 defective) = 9/12 * 3/11 = 3/4 * 3/11 = 9/44

P (of all possible cases where one may obtain a defective pen) = 2/44 + 9/44 + 9/44 = 20/44 = 5/11
P (of neither defective) = 1 - P (of all cases of defective) = 1 - 5/11 = 11/11 - 5/11 = 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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08 Jan 2017, 00:09
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.

Bunuel What if Probability (Neither pen will be defective) = 1-Probability (defective both time) = 1- (3/12)*(2/11) = 21/22 ?

I get the different yield.
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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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08 Jan 2017, 05:00
ziyuenlau 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.

Bunuel What if Probability (Neither pen will be defective) = 1-Probability (defective both time) = 1- (3/12)*(2/11) = 21/22 ?

I get the different yield.

P(Neither pen will be defective) = 1 - (P(both pens are defective) + P(one of the pens is defective)) = 1 - (3/12*2/11 + 2*3/12*9/11) = 6/11.

Hope it's clear.
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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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08 Jan 2017, 05:15
A shipment of 8 TV sets contains 2 black and white sets and 6 color sets. If 2 TV sets are to be chosen at random from this shipment, what is the probability that at least 1 of the 2 sets chosen will be a black and white set?

A) 1/7
B) 1/4
C) 5/14
D) 11/28
E) 13/26

PROBABILITY APPROACH:

P(at leas one)=1-P(none)=1-6/8*5/7=13/28.

Bunuel I am confused with this approach.
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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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08 Jan 2017, 06:35
ziyuenlau wrote:
A shipment of 8 TV sets contains 2 black and white sets and 6 color sets. If 2 TV sets are to be chosen at random from this shipment, what is the probability that at least 1 of the 2 sets chosen will be a black and white set?

A) 1/7
B) 1/4
C) 5/14
D) 11/28
E) 13/26

PROBABILITY APPROACH:

P(at leas one)=1-P(none)=1-6/8*5/7=13/28.

Bunuel I am confused with this approach.

This question is discussed here: a-shipment-of-8-television-sets-contains-2-black-and-white-sets-and-89561.html
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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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25 Mar 2017, 08:51
We can solve it using 2 methods: combinatorics and simple probability

1) Combinatorics
2C9/2C11=6/11
2) probability; 1st pen non-defective: 9/12 AND 2nd non defective 8/11
9/12*11/12=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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13 Apr 2017, 10:58
How do you know when to use the (9/12) * (8/11) and when to use (9/12) * (9/12)? It doesn't explicitly say that the customer picks one pen and then another pen afterward. Couldn't the customer have taken both pens at the same time?
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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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13 Apr 2017, 23:53
elhho wrote:
How do you know when to use the (9/12) * (8/11) and when to use (9/12) * (9/12)? It doesn't explicitly say that the customer picks one pen and then another pen afterward. Couldn't the customer have taken both pens at the same time?

1. If the drawing is with replacement it's explicitly mentioned. If it's not mentioned, then it's without replacement.

2. Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same.
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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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14 Apr 2017, 04:34
Bunuel wrote:
elhho wrote:
How do you know when to use the (9/12) * (8/11) and when to use (9/12) * (9/12)? It doesn't explicitly say that the customer picks one pen and then another pen afterward. Couldn't the customer have taken both pens at the same time?

1. If the drawing is with replacement it's explicitly mentioned. If it's not mentioned, then it's without replacement.

2. Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same.

Ohh...that makes sense

I definitely overlooked that. So, replacement = same numerator & denominator; no replacement = subtract one from numerator + denominator. Thanks!!
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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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19 Apr 2017, 15:15
1
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

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.

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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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13 Dec 2017, 16:01
Top Contributor
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

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.

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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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25 Feb 2018, 04:32
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.

YAY !

Can anyone competent in probability questiones explain why my methos is not correct

So we have total 12 pens out of which 3 are defective and 9 are non defective

defective = D
non defective = N

ok so i pick one pen and second in succession -> 1/12 * 2/11 = 2/132 DD (i get both defective)

next attempt: 1/12 * 2/11 = 2/132 --- > ND

one more: 1/12 * 2/11 = 2/132 ----> NN

another attempt: 1/12 * 2/11 = 2/132 ---> DN

so i have 1/3 probality why ist it correct ?
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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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21 Mar 2019, 11:48
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.
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

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

May I know why the answer is not (B) 2/9?
Re: In a box of 12 pens, a total of 3 are defective. If a customer buys 2   [#permalink] 21 Mar 2019, 11:48

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