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: Understanding the Question:
This question is from the topic of Probability.
We are given the total number of students in a class and their distribution among five different school houses.
We need to calculate the probability of selecting a student who is not from house A, B, or C.

Step 2: Key Formula or Approach:
The probability of an event \(E\) is given by the formula:
\[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \]
Total possible outcomes equals the total number of students.
Favorable outcomes are the students who do not belong to house A, B, or C, which means they must belong to house D or E.

Step 3: Detailed Explanation:
Total number of students in the class = 23.
The students are distributed as follows:
- House A: 4 students
- House B: 8 students
- House C: 5 students
- House D: 2 students
- House E: Rest of the students.
Let us calculate the number of students in House E:
\[ \text{House E} = 23 - (\text{House A} + \text{House B} + \text{House C} + \text{House D}) \]
\[ \text{House E} = 23 - (4 + 8 + 5 + 2) \]
\[ \text{House E} = 23 - 19 = 4\text{ students} \]
We want the probability that the selected student is not from House A, B, and C.
These students must be from either House D or House E:
\[ \text{Favorable students} = \text{House D} + \text{House E} \]
\[ \text{Favorable students} = 2 + 4 = 6 \]
Now, compute the probability:
\[ P(\text{Not A, B, C}) = \frac{\text{Favorable students}}{\text{Total students}} \]
\[ P(\text{Not A, B, C}) = \frac{6}{23} \]

Step 4: Final Answer:
The probability that the selected student is not from house A, B and C is \(\frac{6}{23}\).