A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 students registered for the course.

(a) Compute the probability that two or fewer will withdraw.
(b) Compute the probability that exactly four will withdraw.
(c) Compute the probability that more than three will withdraw.
(d) Compute the expected number of withdrawals.

Solution

Let $X$ no. of students who withdraw without completing the introductory statistics course out of 20. Given that $n=20$, $p=0.20$.

Here $X\sim B(20, 0.20)$.

The probability mass function of $X$ is

$$ \begin{aligned} P(X=x) & = \binom{20}{x} (0.20)^x (0.80)^{20-x},\\ & \qquad x=0,1,2,\cdots, 20 \end{aligned} $$

(a) The probability that two or fewer will withdraw

$$ \begin{eqnarray*} P(X\leq 2) &=& P(X=0) +P(X=1) +P(X=2)\\ &=& \binom{20}{0}(0.20)^0(0.80)^{20}+\binom{20}{1}(0.20)^1(0.80)^{19} + \binom{20}{2}(0.20)^2(0.80)^{18}\\ &=& 0.0115+0.0576+0.1369\\ &=& 0.2060. \end{eqnarray*} $$

(b) The probability that exactly four will withdraw

$$ \begin{eqnarray*} P(X= 4) &=& \binom{20}{4}(0.20)^4(0.80)^{16}\\ &=& 0.2182. \end{eqnarray*} $$

(c) The probability that more than three will withdraw\

$$ \begin{eqnarray*} P(X> 3) &=& 1- P(X\leq 2)\\ &=& 1- 0.2060\\ &=& 0.7940. \end{eqnarray*} $$

The expected number of withdrawals $\mu = np = 20\times 0.20 = 4$.