From a group of 10 boys and 13 girls, a committee of 5 students is chosen at random. a. What is the probability that all 5 members on the committee will be girls? b. What is the probability that all 5 members of the committee will be boys? c. What is the probability that there will be at least 1 girl on the committee? a. The probability that all 5 members on the committee will be girls is (Type an integer or a simplified fraction.)

To find the required probabilities, we can use combinations. a. The total number of ways to choose 5 girls from 13 is calculated as \( \binom{13}{5} \). The total number of ways to choose any 5 students from the group of 23 (10 boys + 13 girls) is \( \binom{23}{5} \). Therefore, the probability that all 5 members are girls is: \[ P(\text{all girls}) = \frac{\binom{13}{5}}{\binom{23}{5}} = \frac{1287}{33649} \approx 0.0383. \] b. Similarly, for all boys, the total number of ways to choose 5 boys from 10 is \( \binom{10}{5} \). Using the same total number of ways to choose any 5 students, the probability that all members are boys is: \[ P(\text{all boys}) = \frac{\binom{10}{5}}{\binom{23}{5}} = \frac{252}{33649} \approx 0.0075. \] c. To find the probability that there is at least 1 girl on the committee, we can calculate it as 1 minus the probability that all are boys or girls. \[ P(\text{at least 1 girl}) = 1 - (P(\text{all boys}) + P(\text{all girls})) = 1 - \left(\frac{252}{33649} + \frac{1287}{33649}\right) = 1 - \frac{1539}{33649} = \frac{32110}{33649} \approx 0.9543. \] Thus, the final answers are: a. The probability that all 5 members on the committee will be girls is \( \frac{1287}{33649} \). b. The probability that all 5 members of the committee will be boys is \( \frac{252}{33649} \). c. The probability that there will be at least 1 girl on the committee is \( \frac{32110}{33649} \).