DS : Car Selection

[yourshail123]

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 3/4?
A) At least three-fourths of the cars are sedans.
B) The probability that both of the cars selected will be convertibles is less than 1/20.

[anuprajan5]

The answer is B

Statement 1- At least three-fourths of the cars are sedans.
This says that the least number of sedans is 15 and convertibles is 5. Calculating this, the probability of selecting both sedans is 21/38 (<3/4). But it could be 19 sedans and 1 convertible. probability is greater than 3/4. Insufficient.

Statement 2 - This says that the probability of selecting both sedans is greater than 19/20 (always greater than 3/4) Sufficient.

[jkaustubh]

The answer is B

Statement 1- At least three-fourths of the cars are sedans.
This says that the least number of sedans is 15 and convertibles is 5. Calculating this, the probability of selecting both sedans is 21/38 (<3/4). But it could be 19 sedans and 1 convertible. probability is greater than 3/4. Insufficient.

Statement 2 - This says that the probability of selecting both sedans is greater than 19/20 (always greater than 3/4) Sufficient.

but what in case one car is a sedan and the another one is convertible

according to me the answer is E

[soni_pallavi]

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 3/4?
A) At least three-fourths of the cars are sedans.
B) The probability that both of the cars selected will be convertibles is less than 1/20.

The answer is E

Statement 1

Sedans = S > = 3/4*20 > = 15

Now when S = 15 then the Probability of choosing 2 Sedans = 15C2 / 20C2 = 21/38 which is less than 3/4

Now when S = 18 then the Probability of choosing 2 Sedans = 18C2 / 20C2 is greater than 3/4

INSUFFICIENT

Statement 2

Lets take the number of Convertibles as X ; the statement says that

XC2/20C2 < 1/20

XC2 we can write as X(x-1)(x-2)!/2!(x-2)!= x(x-1)/2

20C2 = 190

The equation becomes x(x-1)/2 < 190/20 = x(x-1) < 19 = x^2 - x < 19

This equation is only satisfied for x=0,1,2,3 and 4 leaving the number of Sedans as 20,19,18,17,16

From this also we will not get a definite answer

INSUFFICIENT

Combining

No of Sedans = 16,17,18,19 and this again doesnt give us a definite

INSUFFICIENT

answer Hence E