The grade appeal process at a university requires that a jury be structured by selecting six individuals randomly from a pool of nine students and seven faculty. (a) What is the probability of selecting a jury of all students? (b) What is the probability of selecting a jury of all faculty? (c) What is the probability of selecting a jury of four students and two faculty? Question content area bottom Part 1 (a) What is the probability of selecting a jury of all students? 0.01049 (Round to five decimal places as needed.) Part 2 (b) What is the probability of selecting a jury of all faculty? 0.00087 (Round to five decimal places as needed.) Part 3 (c) What is the probability of selecting a jury of four students and two faculty? enter your response here (Round to five decimal places as needed.)
To solve the problem, we will use combinations to calculate the probabilities for each part.
First, we need to calculate the total number of ways to select 6 individuals from the pool of 9 students and 7 faculty (total of 16 individuals).
[ \text{Total combinations} = \binom{16}{6} ]
Calculating this:
[ \binom{16}{6} = \frac{16!}{6!(16-6)!} = \frac{16!}{6! \cdot 10!} = \frac{16 \times 15 \times 14 \times 13 \times 12 \times 11}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 8008 ]
To find the probability of selecting a jury of all students (6 students), we calculate the number of ways to choose 6 students from 9.
[ \text{Combinations of all students} = \binom{9}{6} ]
Calculating this:
[ \binom{9}{6} = \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 ]
Now, the probability of selecting a jury of all students is:
[ P(\text{all students}) = \frac{\text{Combinations of all students}}{\text{Total combinations}} = \frac{84}{8008} \approx 0.01049 ]
To find the probability of selecting a jury of all faculty (6 faculty), we calculate the number of ways to choose 6 faculty from 7.
[ \text{Combinations of all faculty} = \binom{7}{6} ]
Calculating this:
[ \binom{7}{6} = 7 ]
Now, the probability of selecting a jury of all faculty is:
[ P(\text{all faculty}) = \frac{\text{Combinations of all faculty}}{\text{Total combinations}} = \frac{7}{8008} \approx 0.00087 ]
To find the probability of selecting a jury of 4 students and 2 faculty, we calculate the combinations for both groups.
[ \text{Combinations of 4 students} = \binom{9}{4} ]
Calculating this:
[ \binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 ]
[ \text{Combinations of 2 faculty} = \binom{7}{2} ]
Calculating this:
[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 ]
Now, the total combinations for selecting 4 students and 2 faculty is:
[ \text{Total combinations for 4 students and 2 faculty} = \binom{9}{4} \times \binom{7}{2} = 126 \times 21 = 2646 ]
Now, the probability of selecting a jury of 4 students and 2 faculty is:
[ P(4 \text{ students and } 2 \text{ faculty}) = \frac{\text{Total combinations for 4 students and 2 faculty}}{\text{Total combinations}} = \frac{2646}{8008} \approx 0.33049 ]