Mathematics, 23.10.2019 05:30 valencial0917

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 six 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 three students and three faculty?

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Answer from: nani122

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 $\frac{9}{15}$ .

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 $\frac{8}{14}$ .

We do this for the other four positions of the jury, and we end up with the following probability.

$P = \frac{9}{15}*\frac{8}{14}*\frac{7}{13}*\frac{6}{12}*\frac{5}{11}*\frac{4}{10} = 0.0182$

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 $\frac{6}{15}$ .

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 $\frac{5}{14}$ .

We do this for the other four positions of the jury, and we end up with the following probability.

$P = \frac{6}{15}*\frac{5}{14}*\frac{4}{13}*\frac{3}{12}*\frac{2}{11}*\frac{1}{10} = 0.0002$

There is a 0.02% probability of selecting a jury of all 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 $\frac{9}{15}$ .

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 $\frac{8}{14}$ .

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 $\frac{7}{13}$ .

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 $\frac{6}{12}$ .

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 $\frac{5}{11}$ .

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 $\frac{4}{10}$ .

So, the probability of getting just this sequence is

$P = \frac{9}{15}*\frac{8}{14}*\frac{7}{13}*\frac{6}{12}*\frac{5}{11}*\frac{4}{10} = 0.0182$

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

$P = 0.0182*C^{6}_{3,3} = 0.0182*\frac{6!}{(6-3)!3!} = 0.0182*\frac{6!}{3!}{3!} = 0.364$

There is a 36.4% probability of selecting a jury of three students and three faculty.

Answer from: Quest

step-by-step explanation:

smple

Answer from: Quest

The final answer is 3x3+6x2+8x+24+47x−2mark me brainliest

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