Independent odds, am I (+ friend) seeing this wrong or is there a mistake in the practice exam? I found this exercise in a practice exam:

Any student has a 90% chance of entering a University. Two students
  are applying. Assuming each student’s results are independent, what is
  the probability that at least one of them will be successful in
  entering the National University?
A. $0.50$
B. $0.65$
C. $0.88$
D. $0.90$
E. $0.96$

I think the answer is something different than the answers above, namely $0.99$.
$0.01 = (0.1 \times 0.1)$ is the chance of neither, so $1 - 0.01$ must be $0.99$ right? But it's not part of the possible answers.
Other way: $(0.9 \times 0.9) + (0.9 \times 0.1) + (0.1 \times 0.9) = 0.99$
Am I missing something here?

That's the standard way of doing it:
$P(\text{At Least One}) = 1 - P(\text{Both Fail}) = 1 - 0.1 \times 0.1 = 1 - 0.01 = 0.99$
And the not so standard way:
$P(1\text{ in}) + P(2\text{ in Without }1) = 0.9 + (0.9 \times 0.1) = .99$.
Or 
$$\begin{align}
P(1\text{ in Without }2) + P(2\text{in Without }1) + P(1\text{ in And }2\text{ in}) 
&= 0.9 \times 0.1 + 0.9 \times 0.1 + 0.9 \times 0.9 \\
&= 0.09 + 0.09 + 0.81 = .99
\end{align}$$
No matter how we cut it, you are right. They are wrong.
(It's probably just a typo.)
