From a group of 10 boys and 12 girls, a committee of 4 students is chosen at random. a. What is the probability that all 4 members on the committee will be girls? b. What is the probability that all 4 members of the committee will be boys? c. What is the probability that there will be at least 1 girl on the committee?

To answer your questions, let's calculate the probabilities step by step. a. First, find the total number of ways to choose 4 students from 22 (10 boys + 12 girls): \[ \text{Total ways} = \binom{22}{4} = \frac{22!}{4!(22-4)!} = 7315 \] Next, calculate the number of ways to choose 4 girls from 12: \[ \text{Ways to choose 4 girls} = \binom{12}{4} = \frac{12!}{4!(12-4)!} = 495 \] Now, the probability that all 4 members on the committee will be girls: \[ P(\text{all girls}) = \frac{\binom{12}{4}}{\binom{22}{4}} = \frac{495}{7315} \approx 0.0677 \] b. Similarly, calculate the number of ways to choose 4 boys from 10: \[ \text{Ways to choose 4 boys} = \binom{10}{4} = \frac{10!}{4!(10-4)!} = 210 \] Now, the probability that all 4 members of the committee will be boys: \[ P(\text{all boys}) = \frac{\binom{10}{4}}{\binom{22}{4}} = \frac{210}{7315} \approx 0.0287 \] c. To find the probability that there will be at least 1 girl, we can use the complement rule. The complement of "at least 1 girl" is "no girls", which means all boys in this case: \[ P(\text{at least 1 girl}) = 1 - P(\text{all boys}) \] Substituting the probability of all boys: \[ P(\text{at least 1 girl}) = 1 - \frac{210}{7315} = \frac{7315 - 210}{7315} = \frac{7105}{7315} \approx 0.9713 \] So, summarizing: - a. Probability that all members are girls: \( \approx 0.0677 \) - b. Probability that all members are boys: \( \approx 0.0287 \) - c. Probability that there is at least 1 girl: \( \approx 0.9713 \)