Q259 / 269 · NDA 2025 (Phase 1) · Mathematics
Consider the following for the two (02) items that follow:
A committee of 6 members is formed from a group of 7 gentlemen and 4 ladies.
What is the probability that the committee includes at least 2 ladies?
A committee of 6 members is formed from a group of 7 gentlemen and 4 ladies.
What is the probability that the committee includes at least 2 ladies?
- A$\dfrac{41}{66}$
- B$\dfrac{47}{66}$
- C$\dfrac{49}{66}$
- Correct answerD$\dfrac{53}{66}$
Solution
Total number of people is $7 + 4 = 11$.
Total number of ways to select a committee of $6$ members is $^{11}C_{6} = 462$.
The number of ways to form a committee with less than $2$ ladies (i.e., $0$ or $1$ lady) is $^{4}C_{0} \times ^{7}C_{6} + ^{4}C_{1} \times ^{7}C_{5}$.
$^{4}C_{0} \times ^{7}C_{6} = 1 \times 7 = 7$
$^{4}C_{1} \times ^{7}C_{5} = 4 \times 21 = 84$
Total number of ways with less than $2$ ladies is $7 + 84 = 91$.
Number of ways to form a committee with at least $2$ ladies is $462 - 91 = 371$.
Required probability is $\dfrac{371}{462} = \dfrac{53}{66}$.
Answer: $\dfrac{53}{66}$
Total number of ways to select a committee of $6$ members is $^{11}C_{6} = 462$.
The number of ways to form a committee with less than $2$ ladies (i.e., $0$ or $1$ lady) is $^{4}C_{0} \times ^{7}C_{6} + ^{4}C_{1} \times ^{7}C_{5}$.
$^{4}C_{0} \times ^{7}C_{6} = 1 \times 7 = 7$
$^{4}C_{1} \times ^{7}C_{5} = 4 \times 21 = 84$
Total number of ways with less than $2$ ladies is $7 + 84 = 91$.
Number of ways to form a committee with at least $2$ ladies is $462 - 91 = 371$.
Required probability is $\dfrac{371}{462} = \dfrac{53}{66}$.
Answer: $\dfrac{53}{66}$
Hence, the correct answer is option: 'D'.