yourshail123 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 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
