The probability, of forming a 12 persons committee from 4...

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 $\frac{19}{26}$

Correct answer

A. $\frac{129}{182}$

Step-by-step solution

\begin{aligned}& 3 \text{ engineering }+1 \text{ doctor }+8 \text{ Prof } \rightarrow{ }^4 \mathrm{C}_3 \cdot{ }^2 \mathrm{C}_1 \cdot{ }^{10} \mathrm{C}_8 \\ & =360 \\ & 3 \text{ engineering }+2 \text{ doctors }+7 \text{ Prof } \rightarrow{ }^4 \mathrm{C}_3 \cdot{ }^2 \mathrm{C}_2 \cdot{ }^{10} \mathrm{C}_7 \\ & =480 \\ & 4 \text{ engineering }+1 \text{ doctor }+7 \text{ Prof } \rightarrow{ }^4 \mathrm{C}_4 \cdot{ }^2 \mathrm{C}_1 \cdot{ }^{10} \mathrm{C}_7 \\ & =240 \\ & 4 \text{ engineering }+2 \text{ doctors }+6 \text{ Prof } \rightarrow{ }^4 \mathrm{C}_4 \cdot{ }^2 \mathrm{C}_2 \cdot{ }^{10} \mathrm{C}_6 \\ & =210\end{aligned}

Total

=1290

Req. probability

=\frac{1290}{{ }^{16} \mathrm{C}_{12}}=\frac{1290}{1820}=\frac{129}{182}

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