Question

The probability that a teacher will give an unannounced test during any class meeting is $\dfrac{1}{5}$. If a student is absent twice, the probability that he will miss at least one test is:
A.$\dfrac{7}{{25}}$
B.$\dfrac{9}{{25}}$
C.$\dfrac{{16}}{{25}}$
D.$\dfrac{{24}}{{25}}$

Answer 1

Hint: Here, we will first find the probability that no test was missed by the student. We will use the fact that the total probability of an experiment is 1. Then we will subtract the obtained probability from 1 to find the required probability.

Complete step-by-step answer:
Let $E$ be an event in which a teacher will give an unannounced test during any class meeting.
It is given that the probability that a teacher will give an unannounced test during any class meeting is $\dfrac{1}{5}$.
Therefore, $P\left( E \right) = \dfrac{1}{5}$
Now, if a student is absent twice, let us assume that on those two days, there was no unannounced test which took place.
Hence, the probability that no test was missed by the student $ = \dfrac{4}{5} \times \dfrac{4}{5} = \dfrac{{16}}{{25}}$
Now, we know that the total probability is equal to 1. Therefore,
The probability that the student will miss at least 1 test $ = 1 - \dfrac{{16}}{{25}}$
Taking LCM, we get
$ \Rightarrow $ The probability that the student will miss at least 1 test $ = \dfrac{{25 - 16}}{{25}}$
Subtracting the terms in the numerator, we get
$ \Rightarrow $ The probability that the student will miss at least 1 test $ = \dfrac{9}{{25}}$
Therefore, option B is the correct answer.

Note: An alternate way of solving this question is to solve it directly, i.e.
Since, we are required to find the probability that the student will miss at least 1 test, where it is given that:
The probability that a teacher will give an unannounced test during any class meeting is $\dfrac{1}{5}$
Hence, the probability that on the first day there would be a test but there would be no test on the second day,
 $P\left( {{E_1}} \right) = \dfrac{1}{5} \times \dfrac{4}{5} = \dfrac{4}{{25}}$
Similarly, it could be possible that the test would happen on the second day and on the first day there would be no test.
Hence, $P\left( {{E_2}} \right) = \dfrac{4}{5} \times \dfrac{1}{5} = \dfrac{4}{{25}}$
Also, it could happen that on both the days, the teacher would give an announced test, thus,
$P\left( {{E_3}} \right) = \dfrac{1}{5} \times \dfrac{1}{5} = \dfrac{1}{{25}}$
Now, these three events are mutually exclusive, thus,
The probability that the student will miss at least 1 test $ = P\left( {{E_1}} \right) + P\left( {{E_2}} \right) + P\left( {{E_3}} \right)$
Substituting the values in the above equation, we get
$ \Rightarrow $ The probability that the student will miss at least 1 test $ = \dfrac{4}{{25}} + \dfrac{4}{{25}} + \dfrac{1}{{25}} = \dfrac{9}{{25}}$
Therefore, this is the required answer.
Hence, option B is the correct answer.