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

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kidderek
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In a shipment of 20 cars, 3 are found to be defective. If [#permalink] In a shipment of 20 cars, 3 are found to be defective. If   Updated on: 26 Nov 2012, 02:57
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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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C

Originally posted by kidderek on 20 Nov 2006, 09:28.
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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Re: In a shipment of 20 cars, 3 are found to be defective. If [#permalink] In a shipment of 20 cars, 3 are found to be defective. If   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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[#permalink] In a shipment of 20 cars, 3 are found to be defective. If   20 Nov 2006, 09:37
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Out of 3 defective cars – select 1 => 3 possible ways
Out of 17 good cars – select 3 => 17 C 3 possible ways

Probability = (3 * 17 C 3) / (20 C 4) = 8 /19
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kidderek
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[#permalink] In a shipment of 20 cars, 3 are found to be defective. If   20 Nov 2006, 09:52
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anindyat wrote:
Out of 3 defective cars – select 1 => 3 possible ways
Out of 17 good cars – 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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[#permalink] In a shipment of 20 cars, 3 are found to be defective. If   20 Nov 2006, 11:01
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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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[#permalink] In a shipment of 20 cars, 3 are found to be defective. If   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
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C [#permalink] In a shipment of 20 cars, 3 are found to be defective. If   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
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[#permalink] In a shipment of 20 cars, 3 are found to be defective. If   21 Nov 2006, 01:29
kidderek wrote:
anindyat wrote:
Out of 3 defective cars – select 1 => 3 possible ways
Out of 17 good cars – 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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[#permalink] In a shipment of 20 cars, 3 are found to be defective. If   21 Nov 2006, 16:21
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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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[#permalink] In a shipment of 20 cars, 3 are found to be defective. If   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
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Re: In a shipment of 20 cars, 3 are found to be defective. If [#permalink] In a shipment of 20 cars, 3 are found to be defective. If   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] In a shipment of 20 cars, 3 are found to be defective. If   18 Apr 2020, 10:33
Bunuel wrote:
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.


I have a doubt here: the question assumes that you select the cars sequentially, i.e. one after the other. But what if I decided to select the 4 cars at the same time? Could you use the binomial distribution to model the problem?
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Re: In a shipment of 20 cars, 3 are found to be defective. If [#permalink] In a shipment of 20 cars, 3 are found to be defective. If   18 Apr 2020, 10:51
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louis3092 wrote:
Bunuel wrote:
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.


I have a doubt here: the question assumes that you select the cars sequentially, i.e. one after the other. But what if I decided to select the 4 cars at the same time? Could you use the binomial distribution to model the problem?


Mathematically the probability of picking simultaneously, or picking them one at a time (without replacement) is the same.
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Re: In a shipment of 20 cars, 3 are found to be defective. If [#permalink] In a shipment of 20 cars, 3 are found to be defective. If   19 Apr 2020, 11:35
kidderek wrote:
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


defective cars = 3 and not defective = 17
for 4 cars chosen at random
3/20 * 17/19*16/18 * 15/17 = 2/19
and these 4 cars can be arranged in 4!/3! ways ; 4 ways
4*2/19 ; 8/19
OPTION C
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Re: In a shipment of 20 cars, 3 are found to be defective. If [#permalink] In a shipment of 20 cars, 3 are found to be defective. If   13 Sep 2022, 23:09
P(d) = 3/20; p(nd) = 17/20

Keeping exactly 1 so, choosing 4 would mean

3/20 * 17/19 * 16/18 * 15/17 = 2/19

2/19 * No of arrangements

2/19 * 4!/3! * 1!

=8/19

or 2/19 + 2/19 + 2/19 + 2/19 = 8/19
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Re: In a shipment of 20 cars, 3 are found to be defective. If [#permalink] In a shipment of 20 cars, 3 are found to be defective. If   14 Sep 2022, 20:44
Think of probability as probability = no. of desire outcomes/no. of total outcomes

total outcomes: choosing 4 out of 20 cars
= \(\frac{20!}{16!*4!}= \frac{20*19*18*17}{4*3*2*1} = 19*17*5*3\)

desired outcome: exactly 1 of 4 is defective

choose 1 defective out of 3 = \(\frac{3!}{2!*1!} = 3 \)

choose 3 cars out of remaining 17 cars = \(\frac{17!}{14!*3!} = \frac{17*16*15}{3*2*1} = 17*8*5\)

total ways to choose 1 defective out of 4 = \(17*8*5*3\)

\(probability = \frac{17*8*5*3}{19*17*5*3} = \frac{8}{9}\)
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Re: In a shipment of 20 cars, 3 are found to be defective. If [#permalink] In a shipment of 20 cars, 3 are found to be defective. If   05 Sep 2023, 11:56
Can this be solved via the binomial formula of (nPr).P^r(1-p)^n-r?Bunuel
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Re: In a shipment of 20 cars, 3 are found to be defective. If [#permalink]
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