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

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Director
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18 Jun 2010, 03:25
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Question Stats:

49% (02:59) correct 51% (02:29) wrong based on 199 sessions

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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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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.

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

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18 Jun 2010, 04:19
Thanks Bunuel. This solution is better.

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

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16 Jul 2010, 05:00
1
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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.

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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16 Jul 2010, 05:56
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.

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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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.

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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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.

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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23 Nov 2017, 15:07
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Expert's post
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

[Reveal] Spoiler:
E

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Re: For a trade show, two different cars are selected randomly   [#permalink] 23 Nov 2017, 15:07
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