X/4 - 5 ≤ 0 --> X ≤ 20 so we need Pr(x ≤ 20)

1 tells us there are 8 integers in set T, which is obviously insufficient.
2 tells us the larger set of integers, between 1 and 25 inclusive that contains set T but we still don't know the intersection of the two sets. e.g. T could be {1,2,3,4,5,6,7,8} for which the probability would be 1 or it could be {17,18,19,20,21,22,23,24} for which the probability would be 1/2. So the answer is E.

OK, let's consider the following:
If we consider (1) and (2):
Using 1, we can construct a 'finite' list of unique 8-integer sets:
[1,2,3,4,5,6,7,8]
[1,2,3,4,5,6,7,9]
....
[18,19,20,21,22,23,24,25]

Let's assume that the number of those sets is N. The probability of selecting a random 'set' is 1/N

Now for each set, we can find the probability of x<=20. Ex:
[1,2,3,4,5,6,7,8,] --> P1(x<=20) = 0
...
[18,19,20,21,22,23,24,25] --> PN(x<=20) = 3/8

Using (1) and (2) can't we calculate P(x<=20) for all sets?
P1 x 1/N
+
P2 x 1/N
+
...
+
PN * 1/N
?

The way I read the question, it isn't about the probability of a randomly selected X being &#8804; 20 in a randomly selected set of 8 numbers between 1 and 25. Only X is selected at random but set T is A specific set and is already defined.