TITLE: If a student is absent twice, then what is the probability that the student will miss at least one test?
QUESTION [5 upvotes]: The probability that a teacher will give an unannounced test during
  any class meeting is 1/5 . If a student is absent twice, then the
  probability that the student will miss at least one test is ...?

Answer given : 9/25 
My attempt:

Let $G$: event that the student gives the test;
               $N$: event that the student does not give the test 
$P(G):1/5$
$P(N):4/5$
Then the sample space is : $(GG, NN, NG, GN)$ 
Required elements in the sample space is : $\{NN, NG, GN\}$ 
$P(NG)=P(GN)= 4/25$ and $P(NN)= 16/25$
Required probability = $P(NG)+P(GN)+P(NN)=24/25$

What am I doing wrong here?

REPLY [4 votes]: I'm not sure what you mean by the student "gives the test". So I will rephrase your variables as follows.
Let $G$ be the event that the teacher gives a test, and $N$ that the teacher does not give the test. Notice that the problem asks for the probability that at least one test is missed by a student. It is given that the student is absent on both days. Then we are interested in $GN\cup NG\cup GG$. Since the events are disjoint, we can add
$$P(GN\cup NG\cup GG) = P(GN)+P(NG)+P(GG) = \frac{1}{5}\cdot\frac{4}{5}+\frac{4}{5}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{5} = \frac{9}{25}$$
where I assume that test days are independent of one another. 
Notice that we can use the complement too:
$$P(GN\cup NG\cup GG) = 1-P(\text{Miss no test}) = 1-P(NN) = 1-\frac{4}{5}\cdot\frac{4}{5} = \frac{9}{25}.$$