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.

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:

For a trade show, two different cars are selected randomly [#permalink]
18 Jun 2010, 03:25

2

This post received KUDOS

4

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

44% (03:53) correct
56% (01:54) wrong based on 81 sessions

For a trade show, two different cars are selected randomly from a lot of 20 cars. If all the cars on the lot are either sedans or convertibles, is the probability that both cars selected will be sedans greater than \frac{3}{4}?

1) At least three-fourths of the cars are sedans. 2) The probability that both of the cars selected will be convertibles is less than \frac{1}{20}.

Re: Multiple trials - conditional probability and dependent even [#permalink]
18 Jun 2010, 03:58

1

This post received KUDOS

Expert's post

ykaiim wrote:

For a trade show, two different cars are selected randomly from a lot of 20 cars. If all the cars on the lot are either sedans or convertibles, is the probability that both cars selected will be sedans greater than \frac{3}{4}?

1) At least three-fourths of the cars are sedans. 2) The probability that both of the cars selected will be convertibles is less than \frac{1}{20}.

Let the # of sedans be s and the # of convertibles be c.

Given: s+c=20. Question: is \frac{s}{20}*\frac{s-1}{19}>\frac{3}{4}? --> Is s>17 (18, 19, 20)?

(1) s\geq{\frac{3}{4}*20} --> s\geq{15}. Not sufficient.

Re: Multiple trials - conditional probability and dependent even [#permalink]
16 Jul 2010, 05:00

1

This post received KUDOS

Bunuel wrote:

ykaiim wrote:

For a trade show, two different cars are selected randomly from a lot of 20 cars. If all the cars on the lot are either sedans or convertibles, is the probability that both cars selected will be sedans greater than \frac{3}{4}?

1) At least three-fourths of the cars are sedans. 2) The probability that both of the cars selected will be convertibles is less than \frac{1}{20}.

Let the # of sedans be s and the # of convertibles be c.

Given: s+c=20. Question: is \frac{s}{20}*\frac{s-1}{19}>\frac{3}{4}? --> Is s>17 (18, 19, 20)?

(1) s\geq{\frac{3}{4}*20} --> s\geq{15}. Not sufficient.

I know this is basics, but somehow I got stuck, can someone quickly explain how to solve the equation: \frac{s}{20}*\frac{s-1}{19}>\frac{3}{4}? --> Is s>17 (18, 19, 20)?

Re: Multiple trials - conditional probability and dependent even [#permalink]
16 Jul 2010, 05:56

Expert's post

AndreG wrote:

Bunuel wrote:

ykaiim wrote:

For a trade show, two different cars are selected randomly from a lot of 20 cars. If all the cars on the lot are either sedans or convertibles, is the probability that both cars selected will be sedans greater than \frac{3}{4}?

1) At least three-fourths of the cars are sedans. 2) The probability that both of the cars selected will be convertibles is less than \frac{1}{20}.

Let the # of sedans be s and the # of convertibles be c.

Given: s+c=20. Question: is \frac{s}{20}*\frac{s-1}{19}>\frac{3}{4}? --> Is s>17 (18, 19, 20)?

(1) s\geq{\frac{3}{4}*20} --> s\geq{15}. Not sufficient.

I know this is basics, but somehow I got stuck, can someone quickly explain how to solve the equation: \frac{s}{20}*\frac{s-1}{19}>\frac{3}{4}? --> Is s>17 (18, 19, 20)?

I get stuck at s*(s-1)>19*5 ...

First of all \frac{s}{20}*\frac{s-1}{19}>\frac{3}{4} --> s*(s-1)>19*15 --> we know s is an integer --> couple of substitutions gives s>17 (of course you can also solve quadratic inequality by in this case trial method works best). _________________

Re: Multiple trials - conditional probability and dependent even [#permalink]
18 Jul 2010, 07:11

AndreG wrote:

Bunuel wrote:

ykaiim wrote:

For a trade show, two different cars are selected randomly from a lot of 20 cars. If all the cars on the lot are either sedans or convertibles, is the probability that both cars selected will be sedans greater than \frac{3}{4}?

1) At least three-fourths of the cars are sedans. 2) The probability that both of the cars selected will be convertibles is less than \frac{1}{20}.

Let the # of sedans be s and the # of convertibles be c.

Given: s+c=20. Question: is \frac{s}{20}*\frac{s-1}{19}>\frac{3}{4}? --> Is s>17 (18, 19, 20)?

(1) s\geq{\frac{3}{4}*20} --> s\geq{15}. Not sufficient.

I know this is basics, but somehow I got stuck, can someone quickly explain how to solve the equation: \frac{s}{20}*\frac{s-1}{19}>\frac{3}{4}? --> Is s>17 (18, 19, 20)?

I get stuck at s*(s-1)>19*5 ...

great question to ask. I find myself getting stuck at places like this.

Re: For a trade show, two different cars are selected randomly [#permalink]
29 Jul 2014, 16:26

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: For a trade show, two different cars are selected randomly [#permalink]
07 Aug 2014, 21:28

Bunuel wrote:

ykaiim wrote:

For a trade show, two different cars are selected randomly from a lot of 20 cars. If all the cars on the lot are either sedans or convertibles, is the probability that both cars selected will be sedans greater than \frac{3}{4}?

1) At least three-fourths of the cars are sedans. 2) The probability that both of the cars selected will be convertibles is less than \frac{1}{20}.

Let the # of sedans be s and the # of convertibles be c.

Given: s+c=20. Question: is \frac{s}{20}*\frac{s-1}{19}>\frac{3}{4}? --> Is s>17 (18, 19, 20)?

(1) s\geq{\frac{3}{4}*20} --> s\geq{15}. Not sufficient.