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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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New post 18 Jun 2010, 04:25
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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 \(\frac{1}{20}\).

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

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New post 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.

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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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

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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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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 23 Nov 2017, 16:07
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Hi All,

We're told that two different cars are selected randomly from a lot of 20 cars (and that all of the cars are either sedans or convertibles). We're asked if the probability that BOTH cars selected will be sedans is greater than 3/4. This is a YES/NO question. We can solve it by TESTing VALUES.

1) At least three-fourths of the cars are sedans.

IF... there are 15 sedans and 5 convertibles, then the probability of selecting 2 sedans is....
(15/20)(14/19) = (3/4)(14/19)
You don't have to actually calculate this value, since we're multiplying 3/4 by a positive fraction, the answer will be LESS then 3/4 and the answer to the question is NO.

IF... there are 19 sedans and 1 convertible, then the probability of selecting 2 sedans is....
(19/20)(18/19) = 18/20 = 90% and the answer to the question is YES.
Fact 1 is INSUFFICIENT

2) The probability that both of the cars selected will be convertibles is less than 1/20.

IF... there are 15 sedans and 5 convertibles, then the probability of selecting 2 convertibles is....
(5/20)(4/19) = 1/19.... This is NOT a match for the information in Fact 2 though, so there MUST BE FEWER than 5 convertibles.

IF... there are 16 sedans and 4 convertibles, then the probability of selecting 2 convertibles is....
(4/20)(3/19) = 3/95
The probability of selecting 2 sedans under these circumstances is...
(16/20)(15/19) = (15/20)(16/19) - since we're multiplying 3/4 by a positive fraction, the answer will be LESS then 3/4 and the answer to the question is NO.

IF... there are 19 sedans and 1 convertible, then the probability of selecting 2 sedans is....
(19/20)(18/19) = 18/20 = 90% and the answer to the question is YES.
Fact 2 is INSUFFICIENT

Combined, we know that there must be FEWER than 5 convertibles. From the work that we did in Fact 2 (above), we have proof that the answer could be NO (if there are 4 convertibles) and YES (if there is just 1 convertible).
Combined, INSUFFICIENT

Final Answer:

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Re: For a trade show, two different cars are selected randomly &nbs [#permalink] 23 Nov 2017, 16:07
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