Correct Answer: A) $\frac{43}{63}$ To determine the probability, we first calculate the total number of possible committees and then the number of committees satisfying both given conditions. Step 1: Calculate the Total Possible Committees. Total members = 5 gentlemen + 4 ladies = 9. Number of ways to form a committee of 5 members from 9 is $\binom{9}{5}$. $\binom{9}{5} = \frac{9!}{5!(9-5)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 9 \times 2 \times 7 = 126$. Step 2: Calculate Favorable Committees (satisfying both conditions). We address Condition (ii) first by considering two cases: Case 1: Mr. A is NOT selected for the committee. If Mr. A is not selected, Condition (ii) is vacuously satisfied. We need to select 5 members from the remaining 4 gentlemen (excluding Mr. A) and 4 ladies (including Ms. B). Total members available = $4$ gentlemen + $4$ ladies = $8$. Number of ways to select 5 members from these 8 is $\binom{8}{5} = \frac{8!}{5!3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$. Now, from these 56 committees, we must ensure Condition (i) is met: at least 2 gentlemen and at least 1 lady. Committees that violate Condition (i) within this case: * 1 gentleman and 4 ladies: $\binom{4}{1} \binom{4}{4} = 4 \times 1 = 4$. * 0 gentlemen and 5 ladies: Not possible as there are only 4 ladies. So, committees satisfying both conditions in Case 1 = $56 - 4 = 52$. Case 2: Mr. A IS selected for the committee. According to Condition (ii), if Mr. A is selected, Ms. B must also be selected. So, Mr. A and Ms. B are already in the committee. We need to select 3 more members. The remaining pool is 4 gentlemen (excluding Mr. A) and 3 ladies (excluding Ms. B). Total 7 members. Number of ways to select 3 additional members from these 7 is $\binom{7}{3} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$. Now, from these 35 committees (which already include Mr. A and Ms. B), we must ensure Condition (i) is met: at least 2 gentlemen and at least 1 lady. * Since Ms. B is already in, the 'at least 1 lady' condition is satisfied. * Since Mr. A is already in, we need 'at least 1 more gentleman' among the 3 additional members chosen to meet the 'at least 2 gentlemen' condition. Committees that violate the 'at least 2 gentlemen' condition within this case (i.e., committees with Mr. A, Ms. B, and less than 1 more gentleman among the 3 additional members): * 0 gentlemen and 3 ladies among the 3 additional members: $\binom{4}{0} \binom{3}{3} = 1 \times 1 = 1$. This committee (Mr. A, Ms. B, and 3 other ladies) would have only 1 gentleman (Mr. A), violating Condition (i). So, committees satisfying both conditions in Case 2 = $35 - 1 = 34$. Step 3: Calculate the Probability. Total favorable committees = $52$ (from Case 1) + $34$ (from Case 2) = $86$. Total possible committees = $126$. Probability = $\frac{ ext{Favorable Committees}}{ ext{Total Possible Committees}} = \frac{86}{126}$. Simplifying the fraction by dividing both numerator and denominator by 2: $\frac{43}{63}$. Thus, the probability that the committee satisfies both conditions is $\frac{43}{63}$.