Answer :
Answer:
a) There is a 1.82% probability of selecting a jury of all students.
b) There is a 0.02% probability of selecting a jury of all faculty.
c) There is a 36.4% probability of selecting a jury of three students and three faculty.
Step-by-step explanation:
6 individuals are going to be selected. There are no replacements.So:
(a) What is the probability of selecting a jury of all students?
Initially, there are 9 students and 15 people in the pool.
So, the probability that the first person selected to the jury is a student is [tex]\frac{9}{15}[/tex].
Now, there are 8 students and 14 people in the pool. So, the probability that the second person selected to the pool is a student is [tex]\frac{8}{14}[/tex].
We do this for the other four positions of the jury, and we end up with the following probability.
[tex]P = \frac{9}{15}*\frac{8}{14}*\frac{7}{13}*\frac{6}{12}*\frac{5}{11}*\frac{4}{10} = 0.0182[/tex]
There is a 1.82% probability of selecting a jury of all students.
(b) What is the probability of selecting a jury of all faculty?
The same logic as a) above.
Initially, there are 6 faculty and 15 people in the pool.
So, the probability that the first person selected to the jury is a faculty is [tex]\frac{6}{15}[/tex].
Now, there are 5 faculty and 14 people in the pool. So, the probability that the second person selected to the pool is a faculty is [tex]\frac{5}{14}[/tex].
We do this for the other four positions of the jury, and we end up with the following probability.
[tex]P = \frac{6}{15}*\frac{5}{14}*\frac{4}{13}*\frac{3}{12}*\frac{2}{11}*\frac{1}{10} = 0.0002[/tex]
There is a 0.02% probability of selecting a jury of all faculty.
(c) What is the probability of selecting a jury of three students and three faculty?
I am going to initially find the probability of finding the following sequence:
J-J-J-F-F-F.
Initially, there are 9 students, 6 faculty, that is 15 people in the pool.
So, the probability that the first person selected to the jury is a student is [tex]\frac{9}{15}[/tex].
Now, there are 8 students and 14 people in the pool. So, the probability that the second person selected to the pool is a student is [tex]\frac{8}{14}[/tex].
Now, there are 7 students and 13 people in the pool. So, the probability that the third person selected to the pool is a student is [tex]\frac{7}{13}[/tex].
Now, there are 6 faculty and 12 people in the pool. So, the probability that the third person selected to the pool is a faculty is [tex]\frac{6}{12}[/tex].
Now, there are 5 faculty and 11 people in the pool. So, the probability that the third person selected to the pool is a faculty is [tex]\frac{5}{11}[/tex].
Now, there are 4 faculty and 10 people in the pool. So, the probability that the third person selected to the pool is a faculty is [tex]\frac{4}{10}[/tex].
So, the probability of getting just this sequence is
[tex]P = \frac{9}{15}*\frac{8}{14}*\frac{7}{13}*\frac{6}{12}*\frac{5}{11}*\frac{4}{10} = 0.0182[/tex]
However, any combination with 3 students and 3 faculty is valid, no matter the order. So, we need to multiply P by the combination of 6, with 3 by 3. So
[tex]P = 0.0182*C^{6}_{3,3} = 0.0182*\frac{6!}{(6-3)!3!} = 0.0182*\frac{6!}{3!}{3!} = 0.364[/tex]
There is a 36.4% probability of selecting a jury of three students and three faculty.