A school has five houses A, B, C, D, and E. A class has \[23\] students, \[\;4\] from house A, $8$ from house B,\[\;5\] from house C, \[\;2\] from house D and rest from house E. A single student is selected at random to be the class monitor. The probability that the selected student is not from A, B, and C is?

Hint: First we take \[n\left( S \right)\] be the total number of students,
Also, \[n(X)\] be the total number of students from D and E.
We can find the probability of the selected students from D and E.
That is \[P(X)\]is equal to the total number of students from D and E.
Here in this question we using probability formula in general,
\[P(A) = \dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of favorable outcomes}}}}\]

Complete step-by-step answer:
It is given that the total number of student is 23 and five houses are denoted by A, B, C, D, E
Total number of students
\[n\left( S \right) = 23\]
Number of \[\;4\] student from house A
Number of \[\;8\] student from house B
Number of \[\;5\] student from house C
Number of \[\;2\] student from house D
The selected student is not from A, B and C means students from D and E
So the number of students in houses A, B, and C
\[ = 4 + 8 + 5 = 17\]
Number of students from D and E
\[ = 23 - 17 = 6\]
\[n(X) = 6\]
The probability that the selected student is not from A, B, and C.
\[P(X) = \dfrac{{{\text{Number of students from D and E}}}}{{{\text{Total number of students}}}}\]
\[P(X) = \dfrac{{n(X)}}{{n(S)}}\]
\[P(X) = \dfrac{6}{{23}}\]

Therefore, the probability that the selected students is not from A, B, and C \[ = \dfrac{6}{{23}}\]

Note: Finding the selected students is not from A, B and C. For that subtract the number of students in houses A, B and C from the total number of students.