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For a trade show, two different cars are selected randomly

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

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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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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.

Answer: E.

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

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New post 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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New post 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.

(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.

Answer: E.

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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New post 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.

Answer: E.

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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New post 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.

Answer: E.

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

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New post 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.

Answer: E.

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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New post 14 Jun 2016, 00:20
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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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