a group of 6 seniors, 5 juniors, and 4 sophomores run for student council. the council has 6 members. assume…

a group of 6 seniors, 5 juniors, and 4 sophomores run for student council. the council has 6 members. assume that each student has an equal chance of being elected to student council. determine each probability and express your answers as fractions in lowest terms. sample problem what is the probability that the students elect 2 seniors, 2 juniors, and 2 sophomores? $\frac{{6choose2}\times{5choose2}\times{4choose2}}{{15choose6}}=\frac{15\times10\times6}{5005}=\frac{900}{5005}=\frac{180}{1001}$ the probability of choosing 2 seniors, 2 juniors, and 2 sophomores is $\frac{180}{1001}$. what is the probability that the students elect 1 senior, 4 juniors, and 1 sophomore? enter the answer in the space provided. use numbers instead of words.
Answer
Explanation:
Step1: Calculate total number of students
Let's assume there are 6 seniors, 5 juniors and 4 sophomores. So the total number of students is $6 + 5+4=15$. The council has 6 members. The total number of ways to choose 6 - member council from 15 students is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 15$ and $r = 6$. So, ${15}C{6}=\frac{15!}{6!(15 - 6)!}=\frac{15!}{6!9!}=\frac{15\times14\times13\times12\times11\times10}{6\times5\times4\times3\times2\times1}=5005$.
Step2: Calculate number of favorable - cases
We want to choose 1 senior out of 6, 4 juniors out of 5 and 1 sophomore out of 4. The number of ways to choose 1 senior out of 6 is ${6}C{1}=\frac{6!}{1!(6 - 1)!}=\frac{6!}{1!5!}=6$. The number of ways to choose 4 juniors out of 5 is ${5}C{4}=\frac{5!}{4!(5 - 4)!}=\frac{5!}{4!1!}=5$. The number of ways to choose 1 sophomore out of 4 is ${4}C{1}=\frac{4!}{1!(4 - 1)!}=\frac{4!}{1!3!}=4$. The number of favorable cases is the product of these combinations: ${6}C{1}\times_{5}C_{4}\times_{4}C_{1}=6\times5\times4 = 120$.
Step3: Calculate the probability
The probability $P$ is the number of favorable cases divided by the total number of cases. So $P=\frac{{6}C{1}\times_{5}C_{4}\times_{4}C_{1}}{{15}C{6}}=\frac{120}{5005}=\frac{24}{1001}$.
Answer:
$\frac{24}{1001}$