If a student is absent twice, then what is the probability that the student will miss at least one test?

Question

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?

Answer 1

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: $$P(GN\cup NG\cup GG) = 1-P(\text{Miss no test}) = 1-P(NN) = 1-\frac{4}{5}\cdot\frac{4}{5} = \frac{9}{25}.$$