Question

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?

Answer 1

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.