The probability, of forming a 12 persons committee from 4 engineers, 2 doctors and 10 professors containing at least 3 engineers and at least 1 doctor, is

Solution

Maths sir · 4 steps

Split by exact counts of engineers and doctors; professors fill the rest so the committee has 12.

1. Total ways. There are 4 + 2 + 10 = 16 people. Any 12 of them can form the committee, so the total is C(16, 12) = C(16, 4) = 1820.
2. List the cases. Engineers can be 3 or 4, and doctors can be 1 or 2. That gives four cases, and in each one the professors make up the remaining places.

   Engineers	Doctors	Professors	Ways
   3	1	8	C(4, 3)C(2, 1)C(10, 8) = 4 × 2 × 45 = 360
   3	2	7	C(4, 3)C(2, 2)C(10, 7) = 4 × 1 × 120 = 480
   4	1	7	C(4, 4)C(2, 1)C(10, 7) = 1 × 2 × 120 = 240
   4	2	6	C(4, 4)C(2, 2)C(10, 6) = 1 × 1 × 210 = 210

3. Add the cases. Favourable ways = 360 + 480 + 240 + 210 = 1290.
4. So. The probability is 1290/1820 = 129/182, which is option (a).

The idea to keep. For "at least" conditions on several groups, list every allowed combination of exact counts and add them. Let the unrestricted group fill the remaining places.

Answer from the NTA final answer key. Solution by PrepWiser.