To solve this problem, we need to use the combination formula which is C(n, r) = n! / [r!(n-r)!], where n is the total number of options, r is the number of options chosen, and "!" denotes a factorial.

First, we calculate the total number of ways to choose 14 students out of 31 (6 freshmen + 7 sophomores + 10 juniors + 8 seniors). This is C(31, 14).

Next, we calculate the number of ways to choose 2 freshmen out of 6, 3 sophomores out of 7, 4 juniors out of 10, and 5 seniors out of 8. These are C(6, 2), C(7, 3), C(10, 4), and C(8, 5) respectively.

The probability that the committee is made up of 2 freshmen, 3 sophomores, 4 juniors, and 5 seniors is the product of these four combinations divided by the total number of ways to choose 14 students.

So, the probability is [C(6, 2) * C(7, 3) * C(10, 4) * C(8, 5)] / C(31, 14).

Let's calculate it:

C(6, 2) = 6! / [2!(6-2)!] = 15
C(7, 3) = 7! / [3!(7-3)!] = 35
C(10, 4) = 10! / [4!(10-4)!] = 210
C(8, 5) = 8! / [5!(8-5)!] = 56
C(31, 14) = 31! / [14!(31-14)!] = 2035800

So, the probability is (15 * 35 * 210 * 56) / 2035800 = 0.000392 or 0.0392% when rounded to six decimal places.

Question

To solve this problem, we need to use the combination formula which is C(n, r) = n! / [r!(n-r)!], where n is the total number of options, r is the number of options chosen, and "!" denotes a factorial.

First, we calculate the total number of ways to choose 14 students out of 31 (6 freshmen + 7 sophomores + 10 juniors + 8 seniors). This is C(31, 14).

Next, we calculate the number of ways to choose 2 freshmen out of 6, 3 sophomores out of 7, 4 juniors out of 10, and 5 seniors out of 8. These are C(6, 2), C(7, 3), C(10, 4), and C(8, 5) respectively.

The probability that the committee is made up of 2 freshmen, 3 sophomores, 4 juniors, and 5 seniors is the product of these four combinations divided by the total number of ways to choose 14 students.

So, the probability is [C(6, 2) * C(7, 3) * C(10, 4) * C(8, 5)] / C(31, 14).

Let's calculate it:

C(6, 2) = 6! / [2!(6-2)!] = 15
C(7, 3) = 7! / [3!(7-3)!] = 35
C(10, 4) = 10! / [4!(10-4)!] = 210
C(8, 5) = 8! / [5!(8-5)!] = 56
C(31, 14) = 31! / [14!(31-14)!] = 2035800

So, the probability is (15 * 35 * 210 * 56) / 2035800 = 0.000392 or 0.0392% when rounded to six decimal places.

Knowee AI · Accepted Answer

To solve this problem, we need to use the combination formula which is C(n, r) = n! / [r!(n-r)!], where n is the total number of options, r is the number of options chosen, and "!" denotes a factorial.

First, we calculate the total number of ways to choose 14 students out of 31 (6 freshmen + 7 sophomores + 10 juniors + 8 seniors). This is C(31, 14).

Next, we calculate the number of ways to choose 2 freshmen out of 6, 3 sophomores out of 7, 4 juniors out of 10, and 5 seniors out of 8. These are C(6, 2), C(7, 3), C(10, 4), and C(8, 5) respectively.

The probability that the committee is made up of 2 freshmen, 3 sophomores, 4 juniors, and 5 seniors is the product of these four combinations divided by the total number of ways to choose 14 students.

So, the probability is [C(6, 2) * C(7, 3) * C(10, 4) * C(8, 5)] / C(31, 14).

Let's calculate it:

C(6, 2) = 6! / [2!(6-2)!] = 15
C(7, 3) = 7! / [3!(7-3)!] = 35
C(10, 4) = 10! / [4!(10-4)!] = 210
C(8, 5) = 8! / [5!(8-5)!] = 56
C(31, 14) = 31! / [14!(31-14)!] = 2035800

So, the probability is (15 * 35 * 210 * 56) / 2035800 = 0.000392 or 0.0392% when rounded to six decimal places.

6 freshmen, 7 sophomores, 10 juniors, and 8 seniors are eligible to be on a committee. If a committee of 14 students is chosen at random, what is the probability that it is made up of 2 freshmen, 3 sophomores, 4 juniors, and 5 seniors?Round your answer to 6 decimal places as needed.

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