A committee of $m$ students is being formed by randomly selecting from a population of n undergraduate students in an engineering college (note: m < n). Assume that you are also included in this population of students. What is the probability that you will be among the $m$ selected students?
Since you're choosing $m$ amount of students to be in a committee from a population $n$, then the probability of any student being in the committee is $\left(\frac{m}n \right)$. But the probability of you being chosen is $\left(\frac{1}m \right)$. Since the students being selected are not replaced, I used the following equation:
$$
\frac{\left(\frac{1}m \right) \left(\frac{m-1}n \right)}{\left(\frac{m}n \right)}
$$
I feel like I am misunderstanding how permutations are used in this situation, so any insights on how to solve this question would be appreciated.
Best Answer
A person is fixed in the committee, it must select $m-1$ from $n-1$ people, that is $\dbinom{n-1}{m-1}$
The number of all cases, selecting $m$ people from $n$ is $\dbinom{n}{m}$
$P=\dfrac {\dbinom{n-1}{m-1}}{\dbinom{n}{m}}=\dfrac{m}{n}$