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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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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
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Last edited by Bunuel on 26 Nov 2012, 02:57, edited 1 time in total.
Renamed the topic and edited the question.



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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



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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.



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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



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D  defective
N  nondefective
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



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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
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



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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).



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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.



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Posts: 358

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



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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 .. ???



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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. Answer: C. 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). Answer: C. Hope it's clear.
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Re: In a shipment of 20 cars, 3 are found to be defective. If
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