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 probabilities, we will first determine the total number of ways to choose a committee of 5 members from the group of 23 students (10 boys and 13 girls). The total combination of choosing 5 students out of 23 can be calculated using the combination formula: \[ \text{Total combinations} = \binom{n}{r} = \frac{n!}{r!(n-r)!} \] Here, \( n = 23 \) and \( r = 5 \). **Total combinations:** \[ \binom{23}{5} = \frac{23!}{5!(23-5)!} = \frac{23 \times 22 \times 21 \times 20 \times 19}{5 \times 4 \times 3 \times 2 \times 1} = 33649 \] Now, let's answer the questions: a. **Probability that all 5 members will be girls:** The number of ways to choose 5 girls from 13 is: \[ \binom{13}{5} = \frac{13!}{5!(13-5)!} = \frac{13 \times 12 \times 11 \times 10 \times 9}{5 \times 4 \times 3 \times 2 \times 1} = 1287 \] The probability is: \[ P(\text{all girls}) = \frac{\binom{13}{5}}{\binom{23}{5}} = \frac{1287}{33649} \] b. **Probability that all 5 members will be boys:** The number of ways to choose 5 boys from 10 is: \[ \binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \] The probability is: \[ P(\text{all boys}) = \frac{\binom{10}{5}}{\binom{23}{5}} = \frac{252}{33649} \] c. **Probability that there will be at least 1 girl on the committee:** We can find this by taking the complement of the probability that there are no girls (i.e., all boys): \[ P(\text{at least 1 girl}) = 1 - P(\text{all boys}) \] Using our previous result: \[ P(\text{at least 1 girl}) = 1 - \frac{252}{33649} = \frac{33649 - 252}{33649} = \frac{33397}{33649} \] In conclusion: - 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 \(\frac{33397}{33649}\).