Q41 / 75 · JEE Main 2025 (4 Apr Shift 1) · Mathematics
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:
- Correct answerA$\frac{129}{182}$
- B$\frac{103}{182}$
- C$\frac{17}{26}$
- D$\frac{19}{26}$
Solution
$\begin{aligned}<br/>& 3 \text { engineering }+1 \text { doctor }+8 \text { Prof } \rightarrow{ }^4 \mathrm{C}_3 \cdot{ }^2 \mathrm{C}_1 \cdot{ }^{10} \mathrm{C}_8 \\
& =360 \\
& 3 \text { engineering }+2 \text { doctors }+7 \text { Prof } \rightarrow{ }^4 \mathrm{C}_3 \cdot{ }^2 \mathrm{C}_2 \cdot{ }^{10} \mathrm{C}_7 \\
& =480 \\
& 4 \text { engineering }+1 \text { doctor }+7 \text { Prof } \rightarrow{ }^4 \mathrm{C}_4 \cdot{ }^2 \mathrm{C}_1 \cdot{ }^{10} \mathrm{C}_7 \\
& =240 \\
& 4 \text { engineering }+2 \text { doctors }+6 \text { Prof } \rightarrow{ }^4 \mathrm{C}_4 \cdot{ }^2 \mathrm{C}_2 \cdot{ }^{10} \mathrm{C}_6 \\
& =210<br/>\end{aligned}$
Total $=1290$
Req. probability $=\frac{1290}{{ }^{16} \mathrm{C}_{12}}=\frac{1290}{1820}=\frac{129}{182}$
Total $=1290$
Req. probability $=\frac{1290}{{ }^{16} \mathrm{C}_{12}}=\frac{1290}{1820}=\frac{129}{182}$
Hence, the correct answer is option: 'A'.