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

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Difficulty:

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Question Stats:

45% (03:00) correct
55% (01:44) wrong based on 76 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

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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.