The probability, of forming a 12 persons committee from 4 engineers, 2 doctors and 10 professors containing…
- $\frac{129}{182}$
- $\frac{103}{182}$
- $\frac{17}{26}$
- $\frac{19}{26}$
Solution
& 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 $=\frac{1290}{{ }^{16} \mathrm{C}_{12}}=\frac{1290}{1820}=\frac{129}{182}$ .
Asked in: JEE Main 2025 (04 Apr Shift 1)