1. Key Concepts and Formulas

• Probability: The probability of an event $A$ is the ratio of the number of favorable outcomes to the total number of possible outcomes. $P(A) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$
• Combinations: Used when the order of selection does not matter, such as forming a committee. The number of ways to choose $k$ items from a set of $n$ distinct items is given by: ${^n C_k} = \frac{n!}{k!(n-k)!}$
• Combinations Property: A useful property for simplifying calculations is ${^n C_k} = {^n C_{n-k}}$.

2. Step-by-Step Solution

Step 1: Understand the Problem and Identify Constraints

We need to form a committee of 12 persons from a total pool of 16 individuals, comprising:

• Engineers (E): 4 available
• Doctors (D): 2 available
• Professors (P): 10 available
• Total people available: $4 + 2 + 10 = 16$

The committee must satisfy two conditions:

• At least 3 engineers (i.e., $E \ge 3$)
• At least 1 doctor (i.e., $D \ge 1$)

Our goal is to find the probability that a randomly formed 12-person committee meets these criteria.

Step 2: Calculate the Total Number of Possible Committees (Denominator)

First, we determine the total number of ways to form a 12-person committee from the 16 available people without any restrictions. Since the order of selection does not matter, we use combinations.

$\text{Total possible outcomes} = {^{16} C_{12}}$ Using the property ${^n C_k} = {^n C_{n-k}}$, we can calculate this as ${^{16} C_{16-12}} = {^{16} C_4}$: ${^{16} C_4} = \frac{16 \times 15 \times 14 \times 13}{4 \times 3 \times 2 \times 1}$ ${^{16} C_4} = \frac{16}{4 \times 2} \times \frac{15}{3} \times 14 \times 13$ ${^{16} C_4} = 2 \times 5 \times 7 \times 13 = 10 \times 91 = 1820$ So, there are 1820 total possible ways to form a 12-person committee. This will be the denominator of our probability.

Step 3: Calculate the Number of Favorable Committees (Numerator)

Next, we identify the number of ways to form a 12-person committee that satisfies both conditions: at least 3 engineers and at least 1 doctor. Let $E$, $D$, and $P$ denote the number of engineers, doctors, and professors selected, respectively. The total committee size is 12: $E + D + P = 12$.

Based on the available personnel and conditions:

• Engineers (E): Must be $E \ge 3$. Since there are only 4 engineers available, $E$ can be 3 or 4.
• Doctors (D): Must be $D \ge 1$. Since there are only 2 doctors available, $D$ can be 1 or 2.
• Professors (P): The number of professors is determined by $P = 12 - E - D$. There are 10 professors available, so $0 \le P \le 10$.

We will consider all possible combinations of $(E, D)$ that satisfy the conditions and then determine the number of professors $(P)$ for each case, ensuring $P$ is within the available range.

• Case 3.1: 3 Engineers and 1 Doctor

  • Number of Engineers ($E=3$): Choose 3 from 4 $\implies {^4 C_3} = 4$ ways.
  • Number of Doctors ($D=1$): Choose 1 from 2 $\implies {^2 C_1} = 2$ ways.
  • Number of Professors ($P$): $P = 12 - 3 - 1 = 8$. (Valid, as $0 \le 8 \le 10$). Choose 8 from 10 $\implies {^{10} C_8} = {^{10} C_2} = \frac{10 \times 9}{2 \times 1} = 45$ ways.
  • Number of ways for Case 3.1: ${^4 C_3} \times {^2 C_1} \times {^{10} C_8} = 4 \times 2 \times 45 = 360$.

• Case 3.2: 3 Engineers and 2 Doctors

  • Number of Engineers ($E=3$): Choose 3 from 4 $\implies {^4 C_3} = 4$ ways.
  • Number of Doctors ($D=2$): Choose 2 from 2 $\implies {^2 C_2} = 1$ way.
  • Number of Professors ($P$): $P = 12 - 3 - 2 = 7$. (Valid, as $0 \le 7 \le 10$). Choose 7 from 10 $\implies {^{10} C_7} = {^{10} C_3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$ ways.
  • Number of ways for Case 3.2: ${^4 C_3} \times {^2 C_2} \times {^{10} C_7} = 4 \times 1 \times 120 = 480$.

• Case 3.3: 4 Engineers and 1 Doctor

  • Number of Engineers ($E=4$): Choose 4 from 4 $\implies {^4 C_4} = 1$ way.
  • Number of Doctors ($D=1$): Choose 1 from 2 $\implies {^2 C_1} = 2$ ways.
  • Number of Professors ($P$): $P = 12 - 4 - 1 = 7$. (Valid, as $0 \le 7 \le 10$). Choose 7 from 10 $\implies {^{10} C_7} = {^{10} C_3} = 120$ ways (from Case 3.2).
  • Number of ways for Case 3.3: ${^4 C_4} \times {^2 C_1} \times {^{10} C_7} = 1 \times 2 \times 120 = 240$.

• Case 3.4: 4 Engineers and 2 Doctors

  • Number of Engineers ($E=4$): Choose 4 from 4 $\implies {^4 C_4} = 1$ way.
  • Number of Doctors ($D=2$): Choose 2 from 2 $\implies {^2 C_2} = 1$ way.
  • Number of Professors ($P$): $P = 12 - 4 - 2 = 6$. (Valid, as $0 \le 6 \le 10$). Choose 6 from 10 $\implies {^{10} C_6} = {^{10} C_4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 10 \times 3 \times 7 = 210$ ways.
  • Number of ways for Case 3.4: ${^4 C_4} \times {^2 C_2} \times {^{10} C_6} = 1 \times 1 \times 210 = 210$.

The total number of favorable outcomes is the sum of the ways from these mutually exclusive cases: $\text{Total favorable outcomes} = 360 + 480 + 240 + 210 = 1290$

Step 4: Calculate the Final Probability

Finally, we calculate the probability by dividing the total number of favorable outcomes by the total number of possible outcomes: $P(\text{committee with conditions}) = \frac{\text{Total favorable outcomes}}{\text{Total possible outcomes}} = \frac{1290}{1820}$ Simplifying the fraction by dividing both numerator and denominator by 10: $P = \frac{129}{182}$ To check for further simplification, we find the prime factors: $129 = 3 \times 43$ $182 = 2 \times 7 \times 13$ Since there are no common prime factors, the fraction is in its simplest form.

3. Common Mistakes & Tips

• Exhaustive Case Analysis: When dealing with "at least" conditions for multiple categories, ensure you systematically list all valid combinations of those categories. Then, calculate the remaining members for the third category and verify they are within their allowed range.
• "At Least" vs. "Exactly": Remember that "at least 3 engineers" means 3 OR 4 engineers (since only 4 are available), not just exactly 3.
• Combinations vs. Permutations: For committee selection, the order of selection does not matter, so always use combinations ($^n C_k$), not permutations ($^n P_k$).
• Check Constraints for All Categories: After selecting members for the restricted categories, ensure that the number of members needed for the remaining category (professors, in this case) is both non-negative and does not exceed the available pool.

4. Summary

To determine the probability, we first calculated the total number of ways to form a 12-person committee from 16 individuals using combinations. Then, we systematically identified all possible compositions of the committee that satisfied the conditions of having at least 3 engineers and at least 1 doctor. This involved listing mutually exclusive cases for the number of engineers and doctors, calculating the corresponding number of professors, and summing up the combinations for each valid case. Finally, the probability was found by dividing the total number of favorable outcomes by the total possible outcomes and simplifying the resulting fraction.

The final answer is $\boxed{\frac{129}{182}}$ which corresponds to option (A).