# Solved: An academic department has 11 faculty members, 6 of whom are tenured. Five faculty members are to be randomly selected to serve on a committee

#### ByDr. Raju Chaudhari

Feb 26, 2021

An academic department has 11 faculty members, 6 of whom are tenured. Five faculty members are to be randomly selected to serve on a committee. Let X be a random variable for the number of tenured members on the committee.

(a) Specify the distribution for X, and give the parameters of the distribution.
(b) What is the probability that the tenured members on the committee will constitute a majority?
(c) Compute the mean and variance for the number of tenured members on the committee.

#### Solution

Let X denote the number of tenured members on the committee out of 11 faculty members.

The probability that a member is tenured is $p=6/11= 0.5455$.

(a) The random variable $X$ follows Binomial distribution with $n=5$ and $p=0.5455$. That is $X\sim B(5,0.5455)$.

The probability mass function of $X$ is
 \begin{aligned} P(X=x) &= \binom{5}{x} (0.5455)^x (1-0.5455)^{5-x},\\ & \quad \; x=0,1,\cdots, 5 \end{aligned}

(b) The probability that the tenured members on the committee will constitute a majority is $P(X=5)$.

Thus the required probability is

 \begin{aligned} P(X=5) &= \binom{5}{5} (0.5455)^{5} (1-0.5455)^{5-5}\\ &= 0.0483 \end{aligned}

(c) The mean number of tenured members on the committee is
 \begin{aligned} E(X) &= n*p\\ &= 5*0.5455\\ &= 2.7275 \end{aligned}

The variance of the number of tenured members on the committee is

 \begin{aligned} V(X)&=n*p*(1-p)\\ &= 5*0.5455*(1- 0.5455)\\ &= 1.2396 \end{aligned}