Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

1

This post received KUDOS

4

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

74% (02:37) correct
26% (01:22) wrong based on 98 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

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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).

Originally, I was supposed to have an in-person interview for Yale in New Haven, CT. However, as I mentioned in my last post about how to prepare for b-school interviews...

Hi Starlord, As far as i'm concerned, the assessments look at cognitive and emotional traits - they are not IQ tests or skills tests. The games are actually pulled from...

Interested in applying for an MBA? In the fourth and final part of our live QA series with guest expert Chioma Isiadinso, co-founder of consultancy Expartus and former admissions...

The 2015 Pan American Games and more than 6,000 athletes have come to Toronto. With them they have brought a great positive vibe and it made me think...