Question

A committee consists of 4 B. Tech, 3 M. Tech, and 2 Ph.D. students. Find the probability of removing 2 students from the same category and the third one from any other category.

Hint:

Use combinations to calculate the probability for selections of groups.

Answer 1

Correct Answer: 0.654

Total students = 9. Ways to choose 2 students from same category (B.Tech, M.Tech, or PhD): \[ \binom{4}{2} = 6, \binom{3}{2} = 3, \binom{2}{2} = 1 \] Ways to choose the third student from a different category: \[ \text{For B.Tech}: 3 (\text{M.Tech} + \text{PhD}) \text{and so on for others.} \] Total favorable outcomes = 6 \times 3 = 18 (B. Tech chosen and other categories). Total ways to select 3 students = \(\binom{9}{3} = 84\). Thus, probability: \[ P = \frac{18}{84} = 0.6547 \approx 0.655 \]