# Solved:In a large university, 15 of the students are female. If a random sample of twenty students

#### ByDr. Raju Chaudhari

Aug 23, 2020

In a large university, 15% of the students are female. If a random sample of twenty students is selected.

a. What is the probability that the sample contains exactly four female students?
b. What is the probability that the sample will contain no female students?
c. What is the probability that the sample will contain more than two female students?
d. What is the probability that the sample will contain fewer than five female students?
e. What is the expected number of female students?

#### Solution

Let $X$ denote the number of female students and $n=20$ students are selected at random. Let $p$ be the probability that the selected student is female. Given that $p=0.15$.

Here $X\sim B(20, 0.15)$.

The probability mass function of $X$ is

$$P(X=x) = \binom{20}{x} (0.15)^x (1-0.15)^{20-x}, \; x=0,1,\cdots, 20.$$

a. The probability that the sample contains exactly 4 females is

\begin{aligned} P(X = 4) &= \binom{20}{4}(0.15)^4(1-0.15)^{16}\\ & = 0.1821 \end{aligned}

b. The probability that the sample contains no females is

\begin{aligned} P(X = 0) &= \binom{20}{0}(0.15)^0(1-0.15)^{20}\\ & = 0.0388 \end{aligned}

c. The probability that $X$ is more than $2$ is

\begin{aligned} P(X > 2) & =1-P(X\leq 2)\\ &= 1-\sum_{x=0}^{2} P(x)\\ &= 1- P(X=0) + P(X=1) +P(X=2)\\ & =1-\sum_{x=0}^{2}\binom{20}{x}(0.15)^x(1-0.15)^{20-x}\\ & = 1-\bigg(0.0388+0.1368+0.2293\bigg) \\ & = 0.5951 \end{aligned}

d. The probability that fewer than 5 females is

\begin{aligned} P(X < 5) & =\sum_{x=0}^{4} P(x)\\ & =\sum_{x=0}^{4}\binom{20}{x}(0.15)^x(1-0.15)^{20-x}\\ & = \bigg(0.0388+0.1368+0.2293+0.2428+0.1821\bigg) \\ & = 0.8298 \end{aligned}

The expected number of female students is $n*p = 20*0.15 = 3$.