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:

• A. $\frac{129}{182}$
• B. $\frac{103}{182}$
• C. $\frac{17}{26}$
• D. $\frac{19}{26}$

Hint:

The probability of an event is the ratio of the number of favorable outcomes to the total number of outcomes.

Answer 1

The Correct Option is A

1. Calculate the number of ways to form the committee: - 3 engineers + 1 doctor + 8 professors: \[ ^4C_3 \cdot ^2C_1 \cdot ^{10}C_8 = 360 \] - 3 engineers + 2 doctors + 7 professors: \[ ^4C_3 \cdot ^2C_2 \cdot ^{10}C_7 = 480 \] - 4 engineers + 1 doctor + 7 professors: \[ ^4C_4 \cdot ^2C_1 \cdot ^{10}C_7 = 240 \] - 4 engineers + 2 doctors + 6 professors: \[ ^4C_4 \cdot ^2C_2 \cdot ^{10}C_6 = 210 \]
2. Total number of favorable outcomes: \[ \text{Total} = 360 + 480 + 240 + 210 = 1290 \]
3. Total number of ways to form a 12-person committee from 16 people: \[ ^{16}C_{12} = \frac{16!}{12! \cdot 4!} = 1820 \]
4. Calculate the probability: \[ \text{Probability} = \frac{1290}{1820} = \frac{129}{182} \] Therefore, the correct answer is (1) $\frac{129}{182}$.

Answer 2

We have 4 engineers (E), 2 doctors (D), and 10 professors (P): total 16 people. We form a 12-person committee with at least 3 E and at least 1 D.

Concept Used:

Total committees: \( \binom{16}{12}=\binom{16}{4}=1820 \).

Favourable committees are counted by cases on numbers of engineers (3 or 4) and doctors (1 or 2), with professors filling the rest.

Step-by-Step Solution:

Case 1: \(e=3,\ d=1 \Rightarrow p=12-3-1=8\)

\[ \binom{4}{3}\binom{2}{1}\binom{10}{8}=4\cdot2\cdot45=360. \]

Case 2: \(e=3,\ d=2 \Rightarrow p=7\)

\[ \binom{4}{3}\binom{2}{2}\binom{10}{7}=4\cdot1\cdot120=480. \]

Case 3: \(e=4,\ d=1 \Rightarrow p=7\)

\[ \binom{4}{4}\binom{2}{1}\binom{10}{7}=1\cdot2\cdot120=240. \]

Case 4: \(e=4,\ d=2 \Rightarrow p=6\)

\[ \binom{4}{4}\binom{2}{2}\binom{10}{6}=1\cdot1\cdot210=210. \]

Favourable count: \(360+480+240+210=1290\).

Final Computation & Result

\[ \text{Probability}=\frac{1290}{\binom{16}{12}}=\frac{1290}{1820}=\frac{129}{182}. \]

Answer: \( \dfrac{129}{182} \).