A committee of three people is selected at random from a set consisting of five teachers, six parents of students, and three alumni. a. What is the probability of the event that the committee consists of all teachers? b. What is the probability of the event that the committee has no teachers? a. The probability the committee consists of all teachers is (Type an integer or a simplified fraction.)

To find the probability for part (a), we first find the total number of ways to choose 3 people from the entire set of 14 (5 teachers + 6 parents + 3 alumni). This can be calculated using the combination formula: \[ \text{Total ways} = \binom{14}{3} = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = 364 \] Next, we find the number of ways to choose 3 teachers from the 5 teachers: \[ \text{Ways to choose all teachers} = \binom{5}{3} = \frac{5 \times 4}{2 \times 1} = 10 \] Now, the probability that the committee consists of all teachers is: \[ P(\text{all teachers}) = \frac{\text{Ways to choose all teachers}}{\text{Total ways}} = \frac{10}{364} = \frac{5}{182} \] For part (b), to find the probability that the committee has no teachers, we count the total number of ways to choose 3 people from the 9 remaining (6 parents + 3 alumni): \[ \text{Ways to choose no teachers} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \] Thus, the probability that the committee has no teachers is: \[ P(\text{no teachers}) = \frac{\text{Ways to choose no teachers}}{\text{Total ways}} = \frac{84}{364} = \frac{21}{91} \] So the answers are: a. \(\frac{5}{182}\) b. \(\frac{21}{91}\)