Consider the following for the two (02) items that follow: A...

Total number of people = $7 + 4 = 11$

Total number of ways to select a committee of $6$ members = $^{11}C_{6}$

To form a committee of $6$ members with exactly $3$ gentlemen, the remaining $3$ members must be ladies.

Number of ways to select exactly $3$ gentlemen and $3$ ladies = $^{7}C_{3} \times ^{4}C_{3}$

Required probability = $\dfrac{^{7}C_{3} \times ^{4}C_{3}}{^{11}C_{6}}$

$^{11}C_{6} = \dfrac{11 \times 10 \times 9 \times 8 \times 7 \times 6}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 462$

$^{7}C_{3} = \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

$^{4}C_{3} = 4$

Required probability = $\dfrac{35 \times 4}{462} = \dfrac{140}{462} = \dfrac{10}{33}$

Answer: $\dfrac{10}{33}$