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 house A, B and C is :

• A. \(\frac{4}{23}\)
• B. \(\frac{6}{23}\)
• C. \(\frac{8}{23}\)
• D. \(\frac{17}{23}\)

Hint:

An alternative method uses the complementary probability:
Total students from A, B, C = \(4 + 8 + 5 = 17\).
The probability of selecting a student from A, B, C is \(\frac{17}{23}\).
Therefore, the probability of NOT selecting from A, B, C is:
\[ 1 - \frac{17}{23} = \frac{6}{23} \]
This avoids calculating the number of students in House E.

Answer 1

The Correct Option is B

Step 1: Set up the complement idea.
Instead of counting students from D and E directly, we can just find the chance of picking from A, B or C, and subtract that from 1, since every student belongs to exactly one house.
Step 2: Find the combined probability of A, B and C.
\[ P(A \text{ or } B \text{ or } C) = \frac{4}{23} + \frac{8}{23} + \frac{5}{23} = \frac{17}{23} \]
Step 3: Subtract from 1 to get the answer. \[ P(\text{not } A, B, C) = 1 - \frac{17}{23} = \frac{23 - 17}{23} = \frac{6}{23} \] \[ \boxed{\dfrac{6}{23}} \]