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

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Director
Joined: 25 Aug 2007
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WE 1: 3.5 yrs IT
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18 Jun 2010, 04:25
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Difficulty:

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Question Stats:

50% (03:03) correct 50% (02:37) wrong based on 177 sessions

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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 $$\frac{1}{20}$$.
[Reveal] Spoiler: OA

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

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18 Jun 2010, 04:58
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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.

(2) $$\frac{c}{20}*\frac{c-1}{19}<\frac{1}{20}$$ --> $$c(c-1)<19$$ --> $$c<5$$ (4, 3, 2, 1, 0) --> $$s>15$$. Not sufficient.

(1)+(2) Not sufficient.

Hope it's clear.
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Director
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Re: Multiple trials - conditional probability and dependent even [#permalink]

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18 Jun 2010, 05:19
Thanks Bunuel. This solution is better.

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

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16 Jul 2010, 06:00
1
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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.

(2) $$\frac{c}{20}*\frac{c-1}{19}<\frac{1}{20}$$ --> $$c(c-1)<19$$ --> $$c<5$$ (4, 3, 2, 1, 0) --> $$s>15$$. Not sufficient.

(1)+(2) Not sufficient.

Hope it's clear.

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 $$s*(s-1)>19*5$$ ...

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

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16 Jul 2010, 06:56
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.

(2) $$\frac{c}{20}*\frac{c-1}{19}<\frac{1}{20}$$ --> $$c(c-1)<19$$ --> $$c<5$$ (4, 3, 2, 1, 0) --> $$s>15$$. Not sufficient.

(1)+(2) Not sufficient.

Hope it's clear.

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.

(2) $$\frac{c}{20}*\frac{c-1}{19}<\frac{1}{20}$$ --> $$c(c-1)<19$$ --> $$c<5$$ (4, 3, 2, 1, 0) --> $$s>15$$. Not sufficient.

(1)+(2) Not sufficient.

Hope it's clear.

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.

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

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29 Jul 2014, 17:26
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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.

(2) $$\frac{c}{20}*\frac{c-1}{19}<\frac{1}{20}$$ --> $$c(c-1)<19$$ --> $$c<5$$ (4, 3, 2, 1, 0) --> $$s>15$$. Not sufficient.

(1)+(2) Not sufficient.

Hope it's clear.

Thank Bunuel for exp
I stop at: a(a-1) > 19* 15, then I read the two options and try and error.
I got the answer E after more than 4 minutes

You solution could save a lot of time
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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).

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Re: For a trade show, two different cars are selected randomly   [#permalink] 14 Jun 2016, 00:20
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