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 \( \frac{117}{3059} \). (Type an integer or a simplified fraction.) b. The probability that all 5 members on the committee will be boys is (Type an integer or a simplified fraction.)

To find the probabilities for parts b and c, let’s calculate each step by step! First, to determine the total number of ways to select a committee of 5 students from a total of 23 students (10 boys + 13 girls), we use the combination formula: \[ \text{Total combinations} = \binom{23}{5} \] Calculating this gives us: \[ \binom{23}{5} = \frac{23!}{5!(23-5)!} = \frac{23 \times 22 \times 21 \times 20 \times 19}{5 \times 4 \times 3 \times 2 \times 1} = 33649 \] Now for part b, the probability that all 5 members of the committee will be boys: The number of ways to choose all boys is: \[ \binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \] Thus, the probability that all 5 members are boys is: \[ P(\text{all boys}) = \frac{\binom{10}{5}}{\binom{23}{5}} = \frac{252}{33649} \] For part c, to find the probability of at least one girl in the committee, we can use the complement: \[ P(\text{at least 1 girl}) = 1 - P(\text{no girls}) = 1 - P(\text{all boys}) \] Knowing \(P(\text{all boys}) = \frac{252}{33649}\), we can calculate: \[ P(\text{at least 1 girl}) = 1 - \frac{252}{33649} = \frac{33649 - 252}{33649} = \frac{33397}{33649} \] Final answers: 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 \( \frac{33397}{33649} \).