A class has 23 students: 4 from house A, 8 from B, 5 from C, 2 from D and the rest from E. One student is selected at random to be the monitor. The probability that the selected student is not from A, B and C is
\(\dfrac{4}{23}\)
\(\dfrac{6}{23}\)
\(\dfrac{8}{23}\)
\(\dfrac{17}{23}\)
Step 1: Total students in the class
Total = 23 students.
Step 2: Students from each house
Step 3: Find number of students in House E
Add students from A, B, C, D: \(4 + 8 + 5 + 2 = 19\).
So, House E = Total – (A + B + C + D) = \(23 - 19 = 4\).
Step 4: Which houses are not allowed?
The student should NOT be from A, B, C. That means the student must be from D or E only.
Step 5: Favourable outcomes
Students in D = 2
Students in E = 4
Total favourable = \(2 + 4 = 6\).
Step 6: Probability formula
Probability = (Favourable outcomes) ÷ (Total outcomes)
Probability = \(\dfrac{6}{23}\).
Final Answer: Option B (\(\dfrac{6}{23}\)).