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

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Question Stats:

47% (04:53) correct
53% (01:48) wrong based on 108 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

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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.

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...