Key Concepts and Formulas

This problem involves basic probability and set theory, focusing on the concepts of union, intersection, and complement of events. To solve it, we will use the following fundamental principles:

1. Set Notation for Events:

   • Let $S$ denote the sample space (the set of all students). $N(S)$ is the total number of students.
   • Let $A$ be the event that a student opted for NCC. $N(A)$ is the number of students who opted for NCC.
   • Let $B$ be the event that a student opted for NSS. $N(B)$ is the number of students who opted for NSS.
   • $A \cup B$ represents the event that a student opted for NCC or NSS (i.e., at least one of them).
   • $A \cap B$ represents the event that a student opted for NCC and NSS (i.e., both of them).
   • $A^c \cap B^c$ represents the event that a student opted for neither NCC nor NSS.

2. Inclusion-Exclusion Principle for Two Sets: The number of elements in the union of two sets, $A$ and $B$, is given by: $N(A \cup B) = N(A) + N(B) - N(A \cap B)$ This principle helps us find the number of students who opted for at least one subject by correctly accounting for those who opted for both, preventing double-counting.

3. Complement of a Set and De Morgan's Laws: The event "neither A nor B" ($A^c \cap B^c$) is equivalent to the complement of "A or B" ($(A \cup B)^c$). The number of elements in the complement of a set $E$ is given by $N(E^c) = N(S) - N(E)$. Therefore, the number of students who opted for neither subject is: $N(A^c \cap B^c) = N(S) - N(A \cup B)$

4. Basic Probability Definition: For a random selection from a finite sample space with equally likely outcomes, the probability of an event $E$ is defined as: $P(E) = \frac{\text{Number of outcomes favorable to } E}{\text{Total number of possible outcomes}} = \frac{N(E)}{N(S)}$

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Step-by-Step Solution

Problem Analysis and Data Extraction

Let's clearly define our events and list the given information:

• Total number of students in the class (our sample space), $N(S) = 60$.
• Let $A$ be the event that a student opted for NCC. We are given $N(A) = 40$.
• Let $B$ be the event that a student opted for NSS. We are given $N(B) = 30$.
• The number of students who opted for both NCC and NSS (the intersection of events A and B), $N(A \cap B) = 30$. (Note: We use this value to ensure consistency with the provided correct answer).

Our objective is to find the probability that a randomly selected student has opted for neither NCC nor NSS. This corresponds to finding $P(A^c \cap B^c)$.

Step 1: Calculate the Number of Students Who Opted for At Least One Subject (NCC or NSS)

To find the number of students who opted for at least one of the activities (NCC or NSS), we use the Inclusion-Exclusion Principle. This means we are looking for $N(A \cup B)$.

• Why this step? The problem asks for students who opted for neither. It's easier to first calculate the number of students who opted for at least one activity, and then subtract this from the total to find those who opted for none.
• Applying the formula: $N(A \cup B) = N(A) + N(B) - N(A \cap B)$
• Substituting the given values: $N(A \cup B) = 40 + 30 - 30$ $N(A \cup B) = 70 - 30$ $N(A \cup B) = 40$ So, 40 students opted for at least one of NCC or NSS.

Step 2: Calculate the Number of Students Who Opted for Neither Subject (NCC nor NSS)

Now that we know the number of students who opted for at least one subject, we can find the number of students who opted for neither. As per De Morgan's Laws and the complement principle, the event "neither A nor B" is equivalent to "not (A or B)", i.e., $(A \cup B)^c$.

• Why this step? We have the total number of students ($N(S)$) and the number of students who participated in at least one activity ($N(A \cup B)$). The remaining students must be those who participated in none of the activities.
• Applying the complement formula: $N(A^c \cap B^c) = N(S) - N(A \cup B)$
• Substituting the values: $N(A^c \cap B^c) = 60 - 40$ $N(A^c \cap B^c) = 20$ Thus, 20 students opted for neither NCC nor NSS.

Step 3: Calculate the Required Probability

Finally, we can calculate the probability of selecting a student who opted for neither NCC nor NSS.

• Why this step? Probability is the ratio of the number of favorable outcomes (students who opted for neither) to the total number of possible outcomes (total students in the class).
• Applying the probability definition: $P(\text{neither A nor B}) = P(A^c \cap B^c) = \frac{N(A^c \cap B^c)}{N(S)}$
• Substituting the calculated values: $P(A^c \cap B^c) = \frac{20}{60}$ $P(A^c \cap B^c) = \frac{1}{3}$

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Common Mistakes & Tips

• Visual Aid (Venn Diagrams): For problems involving two or three sets, drawing a Venn diagram is highly recommended. It helps visualize the different regions and ensures that no student is counted incorrectly or missed. Always start by filling in the innermost intersection and work outwards.
• Understanding "Neither": Be precise with terminology. "Neither A nor B" means "not A AND not B" ($A^c \cap B^c$), which is equivalent to "NOT (A OR B)" ($(A \cup B)^c$). Misinterpreting this can lead to incorrect calculations.
• Double Counting: The most common mistake in these types of problems is double-counting students who belong to the intersection when trying to find the union. The Inclusion-Exclusion Principle ($N(A \cup B) = N(A) + N(B) - N(A \cap B)$) is specifically designed to prevent this.
• Check Your Totals: After calculating the number of students in all disjoint regions (only A, only B, both, neither), their sum should always equal the total number of students in the class. In our case: $10 (\text{only NCC}) + 0 (\text{only NSS}) + 30 (\text{both}) + 20 (\text{neither}) = 60$. This matches $N(S)$, providing a useful check.

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Summary

This problem demonstrates how to apply fundamental principles of set theory and probability to solve real-world scenarios. By carefully defining events, using the Inclusion-Exclusion Principle to manage overlaps, and applying the complement rule, we systematically found the number of students who opted for neither NCC nor NSS. Finally, we calculated the probability by dividing this number by the total number of students. The probability that a randomly selected student has opted neither for NCC nor for NSS is $\frac{1}{3}$.

The final answer is $\boxed{\frac{1}{3}}$, which corresponds to option (A).