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For a trade show, two different cars are selected randomly [#permalink]

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18 Jun 2010, 04:25

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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 \(\frac{1}{20}\).

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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]

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16 Jul 2010, 06:00

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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)?

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).
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Re: Multiple trials - conditional probability and dependent even [#permalink]

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18 Jul 2010, 08:11

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]

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29 Jul 2014, 17:26

Hello from the GMAT Club BumpBot!

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Re: For a trade show, two different cars are selected randomly [#permalink]

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07 Aug 2014, 22:28

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.

Re: For a trade show, two different cars are selected randomly [#permalink]

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14 Jun 2016, 00:20

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.
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