Question

Consider a committee of 12 members is formed randomly out of 4 Engineers, 2 Doctors and 10 Professors. Find the probability that the committee has exactly 3 Engineers and 1 Doctor.

(1) \(\frac{17}{91}\)

(2) \(\frac{18}{91}\)

(3) \(\frac{15}{91}\)

(4) \(\frac{18}{71}\) 

Answer 1

**Answer:**

Total number of people = 4 Engineers + 2 Doctors + 10 Professors = 16

We need to form a committee of 12 members such that:
- 3 are Engineers
- 1 is a Doctor
- Remaining 8 are Professors

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Total number of ways to choose 12 members from 16:

\(\[ \binom{16}{12} \]\)

Favourable ways:
- Choosing 3 Engineers out of 4:  
\(\[ \binom{4}{3} \]\)
- Choosing 1 Doctor out of 2:  
\(\[ \binom{2}{1} \]\)
- Choosing 8 Professors out of 10:  
\(\[ \binom{10}{8} \]\)

So, total favourable ways:

\(\[ \binom{4}{3} \cdot \binom{2}{1} \cdot \binom{10}{8} \]\)

Required probability:

\(\[ \text{Probability} = \frac{\binom{4}{3} \cdot \binom{2}{1} \cdot \binom{10}{8}}{\binom{16}{12}} \]\)

Now calculating values:

\(\[ \binom{4}{3} = 4,\quad \binom{2}{1} = 2,\quad \binom{10}{8} = 45,\quad \binom{16}{12} = 1820 \]\)

\(\[ \text{Probability} = \frac{4 \cdot 2 \cdot 45}{1820} = \frac{360}{1820} = \frac{18}{91} \]\)

\(\[ \boxed{\text{Final Answer: } \frac{18}{91}} \] \)

Answer 2

Correct option is: (2) \(\frac{18}{91}\)  

\(P(3 E, 1 D)=\frac{{ }^{4} C_{3} \cdot{ }^{2} C_{1} \cdot 10{ }_{C_{8}}}{16_{C_{12}}}\)   

\(=\frac{4 \times 2 \times \frac{10 \times 9}{2}}{\frac{16 \times 15 \times 14 \times 13}{24}}=\frac{360 \times 24}{13 \times 7 \times 30 \times 16}\)      

\(=\frac{18}{91}\)