For a trade show, two different cars are selected randomly

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).

great question to ask. I find myself getting stuck at places like this.

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

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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Rich

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