From a class of 15 students (7 from Maharashtra, 5 from Karnataka, 3 from Goa), four students are selected at random. What is the probability that at least one selected student is from Karnataka?

Introduction / Context:This is a classical “at least one” selection problem from disjoint groups. The cleanest approach is to compute the complement (no one from Karnataka) and subtract from 1.

Given Data / Assumptions:

• Total students = 15 (7 Maharashtra, 5 Karnataka, 3 Goa).
• We choose 4 students uniformly at random without replacement.
• Event of interest: at least one from Karnataka among the selected four.

Concept / Approach:Use complement: P(≥1 from Karnataka) = 1 − P(none from Karnataka). “None from Karnataka” means all four are from the remaining 10 students (Maharashtra + Goa).

Step-by-Step Solution:Total ways = C(15, 4) = 1365.Ways with none from Karnataka = C(10, 4) = 210.Therefore, probability = 1 − 210/1365 = 1155/1365 = 11/13.

Verification / Alternative check:Arithmetic check: 1365 − 210 = 1155; divide numerator and denominator by 105 to get 11/13.

Why Other Options Are Wrong:12/13 overstates; 10/15 is not a valid hypergeometric probability; 1/15 is the complement of something else, not this event; 9/13 undercounts.

Common Pitfalls:Trying to sum “exactly k” cases directly; the complement method is faster and less error-prone here.

Final Answer:11/13