A class consists of 8 men and 7 ladies Two students were selected randomly to represent the class in a debate. Find the probability that the two students selected are: (a) both ladies; (b) both men; (c) of the same sex; (d) of different sexes.

To solve this problem, we will first determine the total number of students in the class and then calculate the probabilities for each scenario. ### Known Conditions: - Number of men (M) = 8 - Number of ladies (L) = 7 - Total number of students (T) = M + L = 8 + 7 = 15 ### Total Ways to Select 2 Students: The total number of ways to select 2 students from 15 can be calculated using the combination formula: \[ C(n, r) = \frac{n!}{r!(n-r)!} \] Where \( n \) is the total number of students and \( r \) is the number of students to choose. So, the total ways to select 2 students from 15 is: \[ C(15, 2) = \frac{15!}{2!(15-2)!} = \frac{15 \times 14}{2 \times 1} = 105 \] ### (a) Probability that both students selected are ladies: The number of ways to select 2 ladies from 7 is: \[ C(7, 2) = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 \] Thus, the probability \( P(A) \) that both selected are ladies is: \[ P(A) = \frac{C(7, 2)}{C(15, 2)} = \frac{21}{105} = \frac{1}{5} \] ### (b) Probability that both students selected are men: The number of ways to select 2 men from 8 is: \[ C(8, 2) = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28 \] Thus, the probability \( P(B) \) that both selected are men is: \[ P(B) = \frac{C(8, 2)}{C(15, 2)} = \frac{28}{105} = \frac{4}{15} \] ### (c) Probability that both students selected are of the same sex: The total ways to select 2 students of the same sex is the sum of the ways to select 2 ladies and 2 men: \[ P(C) = P(A) + P(B) = \frac{21 + 28}{105} = \frac{49}{105} = \frac{7}{15} \] ### (d) Probability that both students selected are of different sexes: The probability that the two students are of different sexes can be calculated as: \[ P(D) = 1 - P(C) = 1 - \frac{7}{15} = \frac{8}{15} \] ### Summary of Results: - (a) Probability both are ladies: \( \frac{1}{5} \) - (b) Probability both are men: \( \frac{4}{15} \) - (c) Probability both are of the same sex: \( \frac{7}{15} \) - (d) Probability both are of different sexes: \( \frac{8}{15} \)