5. A class consists of 8 men and 7 ladien, Two students were selected randomlyle 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 need to calculate the probabilities based on the total number of possible selections and the specific desired outcomes. The total number of students in the class is \(8 + 7 = 15\). For part (a), the probability of selecting both ladies requires calculating the number of ways to choose 2 ladies from 7 out of the total ways to choose 2 students from 15: \[ P(\text{both ladies}) = \frac{\binom{7}{2}}{\binom{15}{2}} = \frac{21}{105} = \frac{1}{5}. \] For part (b), we find the probability of selecting both men from the 8 available: \[ P(\text{both men}) = \frac{\binom{8}{2}}{\binom{15}{2}} = \frac{28}{105} = \frac{4}{15}. \] For part (c), we need the probability of selecting two students of the same sex, which includes both ladies and both men: \[ P(\text{same sex}) = P(\text{both ladies}) + P(\text{both men}) = \frac{1}{5} + \frac{4}{15}. \] Finding a common denominator (15), this gives us: \[ P(\text{same sex}) = \frac{3}{15} + \frac{4}{15} = \frac{7}{15}. \] Finally, for part (d), we calculate the probability of selecting students of different sexes. This can be done using the total probability and subtracting the probability of selecting students of the same sex: \[ P(\text{different sexes}) = 1 - P(\text{same sex}) = 1 - \frac{7}{15} = \frac{8}{15}. \] To summarize: (a) Probability both ladies: \(\frac{1}{5}\) (b) Probability both men: \(\frac{4}{15}\) (c) Probability same sex: \(\frac{7}{15}\) (d) Probability different sexes: \(\frac{8}{15}\)