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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] New post 18 Jun 2010, 03:25
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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}.
[Reveal] Spoiler: OA

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Re: Multiple trials - conditional probability and dependent even [#permalink] New post 18 Jun 2010, 03:58
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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.

Answer: E.

Hope it's clear.
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Re: Multiple trials - conditional probability and dependent even [#permalink] New post 18 Jun 2010, 04:19
Thanks Bunuel. This solution is better.
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Re: Multiple trials - conditional probability and dependent even [#permalink] New post 16 Jul 2010, 05:00
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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] New post 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.

(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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

(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] New post 29 Jul 2014, 16:26
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Re: For a trade show, two different cars are selected randomly [#permalink] New post 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.

(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] 07 Aug 2014, 21:28
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