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:

Total number of persons available: 4 engineers (E), 2 doctors (D) and 10 professors (P).
Hence total persons = $$4+2+10 = 16$$.

We have to form a committee of 12 persons subject to the two constraints:
  • at least 3 engineers
  • at least 1 doctor.

Probability $$= \dfrac{\text{number of favourable committees}}{\text{total number of possible committees}}$$.

Total committees
Choosing any 12 out of 16 persons gives $$\binom{16}{12} = \binom{16}{4} = 1820$$ committees.

Favourable committees
Let the committee contain $$e$$ engineers, $$d$$ doctors and $$p$$ professors.
We need $$e \ge 3$$, $$d \ge 1$$ and $$e+d+p = 12$$ with the limits $$0 \le p \le 10$$.

Possible integer triples $$(e,d,p)$$ satisfying these conditions are:

• $$e = 3,\; d = 1,\; p = 8$$
• $$e = 3,\; d = 2,\; p = 7$$
• $$e = 4,\; d = 1,\; p = 7$$
• $$e = 4,\; d = 2,\; p = 6$$

For each case count the committees using $$\binom{n}{r}$$.

Case 1: $$(e,d,p) = (3,1,8)$$
$$\binom{4}{3}\binom{2}{1}\binom{10}{8} = 4 \times 2 \times 45 = 360$$

Case 2: $$(e,d,p) = (3,2,7)$$
$$\binom{4}{3}\binom{2}{2}\binom{10}{7} = 4 \times 1 \times 120 = 480$$

Case 3: $$(e,d,p) = (4,1,7)$$
$$\binom{4}{4}\binom{2}{1}\binom{10}{7} = 1 \times 2 \times 120 = 240$$

Case 4: $$(e,d,p) = (4,2,6)$$
$$\binom{4}{4}\binom{2}{2}\binom{10}{6} = 1 \times 1 \times 210 = 210$$

Add the favourable counts:
$$360 + 480 + 240 + 210 = 1290$$.

Probability
$$\text{P} = \dfrac{1290}{1820} = \dfrac{129}{182}$$ (dividing numerator and denominator by 10).

Thus the required probability is $$\dfrac{129}{182}$$, which matches Option A.