Question

A mail-order company business has six telephone lines.

Let X denote the number of lines in use at a specified time.

Suppose the p. m .f of $X$ is as given in the accompanying table.

$x$	0	1	2	3	4	5	6
$\mathrm{P}\left(x\right)$	0.10	0.15	0.20	0.25	0.20	0.05	0.05

Calculate the probability of each of the following events.

(a) {at most three lines are in use}
(b) {fewer than three lines are in use}
(c) {at least three lines are in use}
(d) {between two and five lines, inclusive, are in use}
(e) {between two and four lines, inclusive, are not in use}
(f) {at least four lines are not in use}

Answer 1

To find Probability from the pmf X:

Step-1: (a) Find the probability in the event of at most three telephone lines are in use:

Here we need to find the probability for $X\le 3$.

$\begin{array}{rcl}P\left(X\le 3\right)& =& P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)\\ & =& 0.10+0.15+0.20+0.25\\ & =& 0.70\end{array}$

Hence the probability in the event of at most three telephone are in use is $0.70$

Step-2: (b) Probability of fewer than three telephone lines are in use.

Here we need to find the probability for $X<3$.

$\begin{array}{rcl}P\left(X<3\right)& =& P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)\\ & =& 0.10+0.15+0.20\\ & =& 0.45\end{array}$

Hence the probability in the event of fewer than three telephone lines are in use is $0.45$.

Step-3: (c) Probability of at least three telephone lines are in use.

Here we need to find the probability for $X\ge 3$.

$\begin{array}{rcl}P\left(X\ge 3\right)& =& P\left(X=3\right)+P\left(X=4\right)+P\left(X=5\right)+P\left(X=6\right)\\ & =& 0.25+0.20+0.05+0.05\\ & =& 0.55\end{array}$

Hence the probability in the event of at least three telephone lines are in use is $0.55$.

Step-4: (d) Probability between two and five lines, inclusive, are in use.

Here we need to find the probability for $2\le X\le 5$.

$\begin{array}{rcl}P\left(2\le X\le 5\right)& =& P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)+P\left(X=5\right)\\ & =& 0.20+0.25+0.20+0.05\\ & =& 0.70\end{array}$

Hence the probability between two and five lines, inclusive, are in use is $0.70$.

Step 5: (e) Probability between two and four lines, inclusive, are not in use.

Here we need to find the probability for $X=2orX=3orX=4$.

$\begin{array}{rcl}P\left(X=2\right)orP\left(X=3\right)orP\left(X=4\right)& =& P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)\\ & =& 0.20+0.25+0.20\\ & =& 0.65\end{array}$

Hence Probability between two and four lines, inclusive, are not in use $0.65$.

Step 6: (f) Probability at least four lines are not in use.

Here we need to find the probability for $X=0orX=1orX=2$.

$\begin{array}{rcl}P\left(X=0\right)orP\left(X=1\right)orP\left(X=2\right)& =& P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)\\ & =& 0.10+0.15+0.20\\ & =& 0.45\end{array}$

Hence, the probability at least four lines are not in use $0.45$.