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 find the probabilities for the committee selection, we’ll first determine the total number of ways to choose 4 students from a group of 22 (10 boys + 12 girls). **Total combinations:** The total number of ways to choose 4 members from 22 is given by the combination formula \( C(n, r) = \frac{n!}{r!(n - r)!} \). \[ C(22, 4) = \frac{22!}{4!(22 - 4)!} = \frac{22 \times 21 \times 20 \times 19}{4 \times 3 \times 2 \times 1} = 7315. \] **a. Probability that all 4 members will be girls:** The number of ways to choose 4 girls from 12 is: \[ C(12, 4) = \frac{12!}{4!(12 - 4)!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495. \] Thus, the probability \( P(\text{all girls}) \): \[ P(\text{all girls}) = \frac{C(12, 4)}{C(22, 4)} = \frac{495}{7315} \approx 0.0677. \] **b. Probability that all 4 members will be boys:** The number of ways to choose 4 boys from 10 is: \[ C(10, 4) = \frac{10!}{4!(10 - 4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210. \] Thus, the probability \( P(\text{all boys}) \): \[ P(\text{all boys}) = \frac{C(10, 4)}{C(22, 4)} = \frac{210}{7315} \approx 0.0287. \] **c. Probability that there will be at least 1 girl on the committee:** To find this probability, we can use the complement rule. This means we subtract the probability of the opposite event (no girls, meaning all boys) from 1. \[ P(\text{at least 1 girl}) = 1 - P(\text{all boys}) = 1 - \frac{210}{7315} \approx 1 - 0.0287 \approx 0.9713. \] So in summary: a. Probability of all girls ≈ 0.0677 b. Probability of all boys ≈ 0.0287 c. Probability of at least 1 girl ≈ 0.9713