The probability, of forming a 12 persons committee from 4 engineers, 2 doctors and 10 professors con

Q: Why can't we just use the complement method? It seems like it would be faster.

Ah, the complement method! It is a powerful tool, but be careful. The complement of 'at least 3 engineers and at least 1 doctor' involves the union of 'fewer than 3 engineers' OR '0 doctors'. Calculating the union of these sets often leads to double-counting errors (the Principle of Inclusion-Exclusion). In this specific problem, direct case-by-case summation is actually cleaner, more intuitive, and significantly less prone to logical traps.

Q: Why do we multiply the combinations within each case?

Great question! This is the Fundamental Counting Principle. We multiply because these are independent choices within a single committee formation. You choose the engineers AND the doctors AND the professors. The 'AND' is your mathematical signal to multiply the possibilities.

Q: Why do we add the results of the four cases at the end?

Because these cases are mutually exclusive. You cannot have a committee that is simultaneously 'Case 1' and 'Case 2'. Since these scenarios cannot happen at the same time, the total number of ways is simply the sum of the individual possibilities. Think of it as branching paths in a tree diagram; you add the leaves to get the total.

Question

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:

Options:

(A) $\frac{129}{182}$

(B) $\frac{103}{182}$

(C) $\frac{17}{26}$

(D) $\frac{19}{26}$

Sigma Percentile · Accepted Answer

### The Art of Counting: Mastering the Committee Problem

Welcome, future engineer. Today, we are not just solving a probability problem; we are stepping into the shoes of a committee organizer. Imagine you are standing in a room with sixteen brilliant minds: four engineers, two doctors, and ten professors.

Your mission is to form a committee of twelve. But there is a catch—a set of constraints that makes this more than just a simple selection. This is where the beauty of combinatorics comes alive.

### Phase 1: The Grand Sample Space

Before we dive into the constraints, we must understand the 'universe' of possibilities. We have a total pool of $16$ people, and we need to choose $12$. Mathematically, this is represented by the combination formula $\binom{16}{12}$.

Recall the symmetry property: $\binom{n}{r} = \binom{n}{n-r}$. This means choosing $12$ people to be on the committee is exactly the same as choosing $4$ people to be left out. Thus, $\binom{16}{12} = \binom{16}{4}$.

Calculating $\binom{16}{4}$ is much friendlier:

$$
\binom{16}{4} = \frac{16 	imes 15 	imes 14 	imes 13}{4 	imes 3 	imes 2 	imes 1} = 1820
$$

So, there are $1820$ possible ways to form any committee of twelve. This is our denominator, our total sample space $n(S)$.

### Phase 2: Decoding the Constraints

Now, let's look at the rules. We need at least $3$ engineers and at least $1$ doctor. The word 'at least' is the heartbeat of this problem, indicating that we have flexibility that must be managed carefully.

We have the following constraints:
- Engineers ($E$): We can have $3$ or $4$ (since there are only $4$ available).
- Doctors ($D$): We can have $1$ or $2$ (since there are only $2$ available).

Because these conditions are independent, we must break this down into mutually exclusive cases. If we try to do it all at once, we will double-count. Let's be systematic.

### Phase 3: The Case-by-Case Detective Work

Let's build our committee, case by case. Remember, the total committee size must always be $12$.

**Case 1: The Baseline ($3E, 1D$)**
We take $3$ engineers from $4$, and $1$ doctor from $2$. That is $4$ people. We need $8$ more to reach $12$, which must be professors chosen from $10$.

$$
	ext{Ways} = \binom{4}{3} 	imes \binom{2}{1} 	imes \binom{10}{8} = 4 	imes 2 	imes 45 = 360
$$

**Case 2: More Doctors ($3E, 2D$)**
We keep $3$ engineers, but now we take both doctors. That is $5$ people. We need $7$ more professors.

$$
	ext{Ways} = \binom{4}{3} 	imes \binom{2}{2} 	imes \binom{10}{7} = 4 	imes 1 	imes 120 = 480
$$

**Case 3: More Engineers ($4E, 1D$)**
We take all $4$ engineers and $1$ doctor. That is $5$ people. We need $7$ more professors.

$$
	ext{Ways} = \binom{4}{4} 	imes \binom{2}{1} 	imes \binom{10}{7} = 1 	imes 2 	imes 120 = 240
$$

**Case 4: The Maximum ($4E, 2D$)**
We take all $4$ engineers and both doctors. That is $6$ people. We need $6$ more professors.

$$
	ext{Ways} = \binom{4}{4} 	imes \binom{2}{2} 	imes \binom{10}{6} = 1 	imes 1 	imes 210 = 210
$$

### Phase 4: The Synthesis

We have analyzed every possible way to satisfy the constraints. Since these cases are mutually exclusive, we simply add them up to find the total number of favorable outcomes, $n(A)$:

$$
n(A) = 360 + 480 + 240 + 210 = 1290
$$

Finally, the probability $P(A)$ is the ratio of our favorable outcomes to the total sample space:

$$
P(A) = \frac{n(A)}{n(S)} = \frac{1290}{1820}
$$

By canceling the zeros, we arrive at our elegant final answer: **$\frac{129}{182}$**.

Animated Solution for Mathematics - Probability: 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:

Select Answer:

Visualized Solution

Defining the Sample Space Setup

Calculating Total Outcomes $n (S)$

Analyzing Constraints

Case 1: $3 E, 1 D, 8 P$

Case 2: $3 E, 2 D, 7 P$

Case 3: $4 E, 1 D, 7 P$

Case 4: $4 E, 2 D, 6 P$

Summing Favorable Outcomes $n (A)$

Final Probability Calculation

The Sigma Insight: Classical Definition of Probability

The Art of Counting

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

The probability that two randomly selected subsets of the set ${1, 2, 3, 4, 5}$ have exactly two elements in their intersection, is:

If 10 different balls has to be placed in 4 distinct boxes at random, then the probability that two of these boxes contain exactly 2 and 3 balls is :

Balls are drawn one-by-one without replacement from a box containing 2 black, 4 white and 3 red balls till all the balls are drawn. Find the probability that the balls drawn are in the order 2 black, 4 white and 3 red.

Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is

In a class of 60 students, 40 opted for NCC, 30 opted for NSS and 20 opted for both NCC and NSS. If one of these students is selected at random, then the probability that the student selected has opted neither for NCC nor for NSS is :

If 10 different balls are to be placed in 4 distinct boxes at random, then the probability that two of these boxes contain exactly 2 and 3 balls is :

$A$ is a set containing $n$ elements. A subset $P$ of $A$ is chosen at random. The set $A$ is reconstructed by replacing the elements of $P$ . A subset $Q$ of $A$ is again chosen at random. Find the probability that $P$ and $Q$ have no common elements.

If three letters can be posted to any one of the 5 different addresses, then the probability that the three letters are posted to exactly two addresses is:

If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is