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In a shipment of 20 cars, 3 are found to be defective. If [#permalink]
20 Nov 2006, 08:28

1

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A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

75% (02:36) correct
25% (01:27) wrong based on 114 sessions

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?

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.

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)

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).

Re: In a shipment of 20 cars, 3 are found to be defective. If [#permalink]
25 Nov 2012, 14:46

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 .. ???

Re: In a shipment of 20 cars, 3 are found to be defective. If [#permalink]
26 Nov 2012, 02:12

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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).

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