The probability that a teacher will give a surprise test during any class meeting is $\dfrac{3}{5}$. if a student is absent on two days, then the probability that he will miss at least one test is

Hint: First we’ll make the cases for how many days the student had missed the test and will find the probability for the individual case. Then adding the probability of all those cases will give the probability that he will miss at least one test i.e. the required answer.

Complete step by step Answer:

Given data: probability of test on a particular day$ = \dfrac{3}{5}$
Since the probability of all the cases is equal to 1
Probability of no test on a particular day$ = 1 - \dfrac{3}{5}$
$ = \dfrac{2}{5}$
Case I:
Let the student misses the test on both the days he was absent
Therefore, the probability of missing both tests$ = \dfrac{3}{5} \times \dfrac{3}{5}...............(i)$
Case II:
Let the student misses the test only on one day
Number of ways of selecting on which day he missed the test$ = {}^2{C_1}$
Now, the probability of missing test on any one day\[ = {}^2{C_1} \times \dfrac{3}{5} \times \dfrac{2}{5}\]
Therefore, the probability that he will miss at least one test will the sum of the probability of both the cases
Required probability\[ = \dfrac{3}{5} \times \dfrac{3}{5} + {}^2{C_1} \times \dfrac{3}{5} \times \dfrac{2}{5}\]
Using \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
\[ = \dfrac{9}{{25}} + \dfrac{{2!}}{{1!\left( {2 - 1} \right)!}} \times \dfrac{6}{{25}}\]
Since \[1! = 1\],
\[ = \dfrac{9}{{25}} + 2 \times \dfrac{6}{{25}}\]
Taking LCM and simplifying numerator, we get,
\[ = \dfrac{{9 + 12}}{{25}}\]
Adding the numerator, we get,
\[ = \dfrac{{21}}{{25}}\]
Hence, the probability that he will miss at least one test is \[\dfrac{{21}}{{25}}\].

Note: Remember for each individual case we multiply the probability as we’ve done in equation(i), do not do the addition here as the probability for the second day is dependent on that first day as if he misses the test on the first day then only he second day can be counted, so where remember always where we multiply or add the respective probabilities.