1. In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that (i) The student opted for NCC or NSS. (ii) The student has opted neither NCC nor NSS. (iii) The student has opted NSS but not NCC.

To solve the problem, we'll use the principle of inclusion-exclusion and basic probability concepts. Let: - \( n(A) \) = Number of students who opted for NCC = 30 - \( n(B) \) = Number of students who opted for NSS = 32 - \( n(A \cap B) \) = Number of students who opted for both NCC and NSS = 24 (i) To find the number of students who opted for either NCC or NSS, we use: \[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \] Substituting the values: \[ n(A \cup B) = 30 + 32 - 24 = 38 \] The probability that a student opted for either NCC or NSS is: \[ P(A \cup B) = \frac{n(A \cup B)}{60} = \frac{38}{60} = \frac{19}{30} \] (ii) The number of students who opted for neither NCC nor NSS can be found by subtracting those who opted for either NCC or NSS from the total number of students: \[ n(\text{neither}) = 60 - n(A \cup B) = 60 - 38 = 22 \] The probability that a student has opted for neither NCC nor NSS is: \[ P(\text{neither}) = \frac{22}{60} = \frac{11}{30} \] (iii) The number of students who opted for NSS but not NCC can be calculated as: \[ n(B \text{ only}) = n(B) - n(A \cap B) = 32 - 24 = 8 \] The probability that a student has opted for NSS but not NCC is: \[ P(B \text{ only}) = \frac{n(B \text{ only})}{60} = \frac{8}{60} = \frac{2}{15} \] So the final answers are: (i) \( \frac{19}{30} \) (ii) \( \frac{11}{30} \) (iii) \( \frac{2}{15} \)