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Consider X to be the number of lines in use for a mail order business.

Following table shows the pmf of random variableX:

a.

Consider X to be the probabilitythat at most three lines in use at a specified time.

The required probability is\(P\left( {X \le 3} \right)\). It can be computed as:

\(P\left( {X \le 3} \right) = P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X = 2} \right) + P\left( {X = 3} \right)\)

Substitute the values of\(P\left( {X = 0} \right) = 0.1\),\(P\left( {X = 1} \right) = 0.15\),\(P\left( {X = 2} \right) = 0.2\)and\(P\left( {X = 3} \right) = 0.25\)in the above formula,

\(\begin{array}{c}P\left( {X \le 3} \right) = 0.1 + 0.15 + 0.2 + 0.25\\ = 0.70\end{array}\)

Thus, the required probability is 0.70.

b.

Consider X to be the probability that fewer than three lines in use at a specified time.

The required probability is\(P\left( {X < 3} \right)\). It can be computed as:

\(P\left( {X < 3} \right) = P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X = 2} \right)\)

Substitute the values of\(P\left( {X = 0} \right) = 0.1\),\(P\left( {X = 1} \right) = 0.15\),and\(P\left( {X = 2} \right) = 0.2\)in the above formula,

\(\begin{array}{c}P\left( {X < 3} \right) = 0.1 + 0.15 + 0.2\\ = 0.45\end{array}\)

Thus, the required probability is 0.45.

c.

Consider X to be the probability that at least three lines in use at a specified time.

The required probability is\(P\left( {X \ge 3} \right)\). It can be computed as:

\(P\left( {X \ge 3} \right) = P\left( {X = 3} \right) + P\left( {X = 4} \right) + P\left( {X = 5} \right) + P\left( {X = 6} \right)\)

Substitute the values of\(P\left( {X = 3} \right) = 0.25\),\(P\left( {X = 4} \right) = 0.2\),\(P\left( {X = 5} \right) = 0.06\)and\(P\left( {X = 6} \right) = 0.04\)in the above formula,

\(\begin{array}{c}P\left( {X \ge 3} \right) = 0.25 + 0.2 + 0.06 + 0.04\\ = 0.55\end{array}\)

Thus, the required probability is 0.55.

d.

Consider X to be the probability that between two and five lines in use, inclusive

at a specified time.

The required probability is\(P\left( {2 \le X \le 5} \right)\). It can be computed as:

\(P\left( {2 \le X \le 5} \right) = P\left( {X = 2} \right) + P\left( {X = 3} \right) + P\left( {X = 4} \right) + P\left( {X = 5} \right)\)

Substitute the values of\(P\left( {X = 2} \right) = 0.2\),\(P\left( {X = 3} \right) = 0.25\),\(P\left( {X = 4} \right) = 0.2\), and\(P\left( {X = 5} \right) = 0.06\)in the above formula,

\(\begin{array}{c}P\left( {2 \le X \le 5} \right) = 0.2 + 0.25 + 0.2 + 0.06\\ = 0.71\end{array}\)

Thus, the required probability is 0.71.

e.

Consider X to be the probability that between two and four lines not in use, inclusive

at a specified time.

The required probability is\(P\left( {2 \le X \le 4} \right)\). It can be computed as:

\(P\left( {2 \le X \le 4} \right) = P\left( {X = 2} \right) + P\left( {X = 3} \right) + P\left( {X = 4} \right)\)

Substitute the values of\(P\left( {X = 2} \right) = 0.2\),\(P\left( {X = 3} \right) = 0.25\),\(P\left( {X = 4} \right) = 0.2\)in the above formula,

\(\begin{array}{c}P\left( {2 \le X \le 4} \right) = 0.2 + 0.25 + 0.2\\ = 0.65\end{array}\)

Thus, the required probability is 0.65.

f.

Consider X to be the probability that at least four lines not in use, inclusive

at a specified time.

The required probability is\(P\left( {X \le 2} \right)\). It can be computed as:

\(P\left( {X \le 2} \right) = P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X = 2} \right)\)

Substitute the values of\(P\left( {X = 0} \right) = 0.1\),\(P\left( {X = 1} \right) = 0.15\), and\(P\left( {X = 2} \right) = 0.2\)in the above formula,

\(\begin{array}{c}P\left( {X \le 2} \right) = 0.1 + 0.15 + 0.2\\ = 0.45\end{array}\)

Thus, the required probability is 0.45.