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# In a shipment of 20 cars, 3 are found to be defective. If

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SVP
Joined: 24 Aug 2006
Posts: 2130
In a shipment of 20 cars, 3 are found to be defective. If [#permalink]

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20 Nov 2006, 09:28
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35% (medium)

Question Stats:

76% (02:27) correct 24% (01:19) wrong based on 165 sessions

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In a shipment of 20 cars, 3 are found to be defective. If four cars are selected at random, what is the probability that exactly one of the four will be defective?

(A) 170/1615
(B) 3/20
(C) 8/19
(D) 3/5
(E) 4/5
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Nov 2012, 02:57, edited 1 time in total.
Renamed the topic and edited the question.
Senior Manager
Joined: 08 Jun 2006
Posts: 335
Location: Washington DC

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20 Nov 2006, 09:37
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Out of 3 defective cars â€“ select 1 => 3 possible ways
Out of 17 good cars â€“ select 3 => 17 C 3 possible ways

Probability = (3 * 17 C 3) / (20 C 4) = 8 /19
SVP
Joined: 24 Aug 2006
Posts: 2130

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20 Nov 2006, 09:52
anindyat wrote:
Out of 3 defective cars â€“ select 1 => 3 possible ways
Out of 17 good cars â€“ select 3 => 17 C 3 possible ways

Probability = (3 * 17 C 3) / (20 C 4) = 8 /19

Different yet very effective. My approach is more convoluted

I did 4C1 (3/20 * 17/19 * 16/18 * 15/17) = 48,960/116,280

I was reducing the fraction and got stuck at 136/323. Couldn't reduce it for the life of me. 8/19 looked closest so I divided the numerator and denominator by 8/19 and got 17 as the divisor.
Intern
Joined: 08 Jul 2006
Posts: 11

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20 Nov 2006, 11:01
D : Defective car
G : OK car

The posibilities are:

GGGD
GGDG
GDGG
DGGG

The probability for each possibility is:

17/20*16/19*15/18*3/17
17/20*16/19*3/18*15/17
17/20*3/19*16/18*15/17
3/20*17/19*16/18*15/17

The answer for this problem is the sum of the probabilities of all possibilities

4*(17*16*15*3)/(20*19*18*17) = 8/19
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

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20 Nov 2006, 18:02
D - defective
N - non-defective

DNNN --> P = (3/20)(17/19)(16/18)(15/17) = 2/19
NDNN --> P = (17/20)(3/19)(16/18)(15/17) = 2/19
NNDN --> 2/19
NNND --> 2/19

Total = 4 * 2/19 = 8/19

Ans C
Senior Manager
Joined: 01 Sep 2006
Posts: 301
Location: Phoenix, AZ, USA

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21 Nov 2006, 00:27
In a shipment of 20 cars, 3 are found to be defective. If four cars are selected at random, what is the probability that exactly one of the four will be defective?

(A) 170/1615
(B) 3/20
(C) 8/19
(D) 3/5
(E) 4/5

chances of 1st to be defective and rest not def
(3/20)*(17/19)*(16/18)*(15/17)

Prob for 2nd def rest ok is same ..

so 4*((3/20)*(17/19)*(16/18)*(15/17))= 8/19
Director
Joined: 18 Jul 2006
Posts: 524

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21 Nov 2006, 01:29
kidderek wrote:
anindyat wrote:
Out of 3 defective cars â€“ select 1 => 3 possible ways
Out of 17 good cars â€“ select 3 => 17 C 3 possible ways

Probability = (3 * 17 C 3) / (20 C 4) = 8 /19

Different yet very effective. My approach is more convoluted

I did 4C1 (3/20 * 17/19 * 16/18 * 15/17) = 48,960/116,280

I was reducing the fraction and got stuck at 136/323. Couldn't reduce it for the life of me. 8/19 looked closest so I divided the numerator and denominator by 8/19 and got 17 as the divisor.

4*(3/20*17/19*16/18*15/17)
4*(3/18*16/20*15/19)
4*(4/6*3/19)
4*(2/19) (its easy to do on paper as you can mark out).
Manager
Joined: 13 Sep 2006
Posts: 212

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21 Nov 2006, 16:21
I'm having trouble w/ the numerator on this one..

I know that the denominator is 20c4....

but what are the numbe of combinations of having one defect?

GGGD
GGDG
GDGG
DGGG

wouldn't each defect (out of the 3) have the same grouping above? So the total would be 12?

Thanks for anyone who can set me straight.
Senior Manager
Joined: 19 Jul 2006
Posts: 358

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22 Nov 2006, 07:56
amorica wrote:
I'm having trouble w/ the numerator on this one..

I know that the denominator is 20c4....

but what are the numbe of combinations of having one defect?

GGGD
GGDG
GDGG
DGGG

wouldn't each defect (out of the 3) have the same grouping above? So the total would be 12?

Thanks for anyone who can set me straight.

i think u r confusing with permutation .... the sequence of defect is not important here.

numerator = ( 1 out of three defective) and ( 3 out of 17 good ones)
= 3C1 * 17C3
Intern
Joined: 29 Aug 2012
Posts: 26
Schools: Babson '14
GMAT Date: 02-28-2013
Re: In a shipment of 20 cars, 3 are found to be defective. If [#permalink]

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25 Nov 2012, 15:46
D = defective, F = functional

I'm having trouble in comprehending why is it in 4 different ways, why is the order important as we are just selecting randomly 4 cars ... so DFFF, FDFF, FFDF, FFFD should accordingly be just 1 case .. ???
Math Expert
Joined: 02 Sep 2009
Posts: 39745
Re: In a shipment of 20 cars, 3 are found to be defective. If [#permalink]

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26 Nov 2012, 03:12
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himanshuhpr wrote:
D = defective, F = functional

I'm having trouble in comprehending why is it in 4 different ways, why is the order important as we are just selecting randomly 4 cars ... so DFFF, FDFF, FFDF, FFFD should accordingly be just 1 case .. ???

In a shipment of 20 cars, 3 are found to be defective. If four cars are selected at random, what is the probability that exactly one of the four will be defective?

(A) 170/1615
(B) 3/20
(C) 8/19
(D) 3/5
(E) 4/5

APPROACH #1:

$$P=\frac{C^1_3*C^3_{17}}{C^4_{20}}=\frac{8}{19}$$, where $$C^1_3$$ is number of ways to select 1 defective car out of 3, $$C^3_{17}$$ is number of ways to select 3 not defective car out of 17, and $$C^4_{20}$$ is total number of ways to select 4 cars out of 20.

APPROACH #2:

We need the probability of DFFF: $$P(DFFF)=\frac{4!}{3!}*\frac{3}{20}*\frac{17}{19}*\frac{16}{18}*\frac{15}{17}$$. We are multiplying by $$\frac{4!}{3!}=4$$, since DFFF case can occur in 4 ways: DFFF (the first car is defective and the second, third and fourth are not), FDFF, FFDF, FFFD (basically number of permutations of 4 letter DFFF out of which 3 N's are identical).

Hope it's clear.
_________________
Re: In a shipment of 20 cars, 3 are found to be defective. If   [#permalink] 26 Nov 2012, 03:12
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