JEE Mathematics 2025 Question with Solution

Given: There are $4$ engineers, $2$ doctors, and $10$ professors, so total people $=16$.

Find: The probability that a $12$-person committee contains at least $3$ engineers and at least $1$ doctor.

Total number of committees is

$$\binom{16}{12}=\binom{16}{4}=1820$$

Now count favourable committees by cases on the number of engineers and doctors.

Case 1: $3$ engineers, $1$ doctor, $8$ professors

$$\binom{4}{3}\binom{2}{1}\binom{10}{8}=4\cdot 2\cdot 45=360$$

Case 2: $3$ engineers, $2$ doctors, $7$ professors

$$\binom{4}{3}\binom{2}{2}\binom{10}{7}=4\cdot 1\cdot 120=480$$

Case 3: $4$ engineers, $1$ doctor, $7$ professors

$$\binom{4}{4}\binom{2}{1}\binom{10}{7}=1\cdot 2\cdot 120=240$$

Case 4: $4$ engineers, $2$ doctors, $6$ professors

$$\binom{4}{4}\binom{2}{2}\binom{10}{6}=1\cdot 1\cdot 210=210$$

So the total number of favourable committees is

$$360+480+240+210=1290$$

Therefore, the required probability is

$$\frac{1290}{1820}=\frac{129}{182}$$

Hence, the correct option is A.

The event requires at least $3$ engineers and at least $1$ doctor in a committee of $12$.

So the possible pairs $(e,d)$ are:

• $(3,1)$
• $(3,2)$
• $(4,1)$
• $(4,2)$

For each case, the remaining members are professors.

Thus,

$$\begin{aligned} N(F) &= \binom{4}{3}\binom{2}{1}\binom{10}{8} + \binom{4}{3}\binom{2}{2}\binom{10}{7} \\ &\quad + \binom{4}{4}\binom{2}{1}\binom{10}{7} + \binom{4}{4}\binom{2}{2}\binom{10}{6} \\ &= 360+480+240+210 \\ &= 1290 \end{aligned}$$

Also,

$$N(S)=\binom{16}{12}=1820$$

Therefore,

$$P(F)=\frac{N(F)}{N(S)}=\frac{1290}{1820}=\frac{129}{182}$$

Therefore, the required probability is $\frac{129}{182}$.