I think we will have to consider different scenarios here. Let me give it a shot...

This a YES & NO problem.
Hence, we just need to find out whether the probability will be greater than or lesser than 3/4

Let the event that both cars selected are sedans be 'Event A'

Statement (1):
At least three-fourth are sedans
We will have to test Probability of event A for possible combinations
Lets start with 15 sedans and 5 convertibles
15s + 5c P(A) = 15/20 x 14/19 = 3/4 x 14/19 < 0.75
16s + 4c P(A) = 16/20 x 15/19 = 12/19 < 0.75
17s + 3c P(A) = 17/20 x 16/19 = 68/95 < 0.75
18s + 2c P(A) = 18/20 x 17/19 = 153/190 > 0.75
Thus, 19s + 4c will also have P(A) > 0.75

Hence from the available information (i.e. 20 cars of which at least 3/4th are sedans), we cannot conclusively determine whether probability of selecting two sedans is greater than or lesser than 0.75. INSUFFICIENT

(2) Let the event that both cars selected are convertibles = B
For 18s + 2c, P(B) = 2/20 x 1/19 = 1/190 < 1/20. We know, P(A)>0.75 from above.
For 17s + 3c, P(B) = 3/20 x 2/19 = 3/190 < 1/20. We know, P(A)<0.75 from above.

Hence, with the information in statement (2), we cannot conclusively say whether probability of selecting two sedans is greater or lesser than 0.75. INSUFFICIENT

Statement (1) and (2) together do not help in any way.

The answer should be (E) Both statements are insufficient.

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*Post edited on 28/09/10 to correct the interpretation and the answer.

the problem above too hard to me, so only small attempt
we have 20 cars, x-sedans and (20-x) conver, and we need to discover is prob of picking two sedans >3/4
is (x/20)*(x-1)/19>3/4, after little transformimg is
x^2-x>285, or x(x-1)>285. here dont try to solve quadratic equation it is unsolvable with integer roots ( i checked). let us see, what is the least value of x ( assuming that x is integer) allows the expression to be right
x=17, 17*16=272 less that 285 does`t fit

x=18, 18*17=306 >285, here i rephrased the problem, is x-number of sedans>17, where x is integer. to the st

(1) the number of sedans >=3/4*20=15
insufficient as it can be 15, 16 ,17,18
(2) here again too many work, the number of convertibles =(20-x)
((20-x)/20)*(20-x-1)/19<1/20, transforming,
(20-x)*(19-x)<19
again do not try to solve the quaratic equation, it is unsolvable with integers. start to insert
x=19.0<19 yes
x=18, 2*1<19. yes
x=17, 6<19 yes from here we can see that the value of x is not fixed within needed range, st 2 insyff
both also insuff
my pick for E

IMO E

My approach:

Statement 1:
Sedan's are atleast 3/4 of total , so sedan can be 15 , 16 , 17 , 18 , 19

for 15 Sedan probability => 15 * 14 / ( 20 * 19) = .6....
for 19 SEDAN probability => 19 * 18 / ( 20 * 19) = .9.....

Hence Statement 1 doesn't help us in determining whether the probability will always be > or < 3/4

Statement 2
Probability of both the cars be convertibles < 1/20
If N be the number of convertibles
n ( n-1) / ( 20 * 19) < 1/20
=> n(n-1) < 19
hence n = 2 sedan = 18 Prob = 18 * 17 / ( 20 * 19) = .8...... > 3/4
n = 3 sedan = 17 Prob = 17 * 16 / ( 20 * 19) = .71....... < 3/4
n = 4 ...........................

Hence Statement doesn't help us in determining whether the probability will always be > or < 3/4

Joining Both Statement 1 and Statement 2 We still have the option of Sedan = 18 0r 17. hence this can also not help to determine whether the probability is > 3/4 or < 3/4