Understanding Unannounced Test Probability

This problem involves calculating the probability of a specific event occurring over multiple independent trials. We are given the probability that a teacher gives an unannounced test during any given class, and we need to find the probability that a student misses at least one test given they are absent twice.

Defining Probabilities

Let $P(T)$ be the probability that a teacher gives an unannounced test in a class. We are given: $P(T) = \frac{1}{5}$

The probability that a teacher does *not* give an unannounced test in a class is the complement of $P(T)$: $P(\text{No Test}) = 1 - P(T) = 1 - \frac{1}{5} = \frac{5}{5} - \frac{1}{5} = \frac{4}{5}$

Calculating the Probability of Missing At Least One Test

The student is absent twice. We want to find the probability that the student misses *at least one* test during these two absences. It's easier to calculate the probability of the complementary event, which is that the student misses *no* tests during their two absences. This means that on both occasions the student was absent, no test was given.

Assuming the occurrences of tests on different days are independent events:

• The probability of the student missing no test on the first absence is the probability that no test was given on that day: $P(\text{No Test on 1st Absence}) = \frac{4}{5}$.
• The probability of the student missing no test on the second absence is the probability that no test was given on that day: $P(\text{No Test on 2nd Absence}) = \frac{4}{5}$.

The probability of missing no tests across both absences is the product of the probabilities for each absence:

$$ P(\text{Misses No Tests in 2 Absences}) = P(\text{No Test on 1st Absence}) \times P(\text{No Test on 2nd Absence}) $$ $$ P(\text{Misses No Tests in 2 Absences}) = \frac{4}{5} \times \frac{4}{5} = \frac{16}{25} $$

The probability of missing *at least one* test is 1 minus the probability of missing *no* tests:

$$ P(\text{Misses At Least One Test}) = 1 - P(\text{Misses No Tests in 2 Absences}) $$ $$ P(\text{Misses At Least One Test}) = 1 - \frac{16}{25} $$ $$ P(\text{Misses At Least One Test}) = \frac{25}{25} - \frac{16}{25} = \frac{9}{25} $$

Conclusion

Therefore, the probability that a student misses at least one test when they are absent twice is $\frac{9}{25}$.