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Chapter 3: Q13E (page 107)

A mail-order computer business has six telephone lines. Let X denote the number of lines in use at a specified time. Suppose the pmf of X is as given in the accompanying table.

X

0

1

2

3

4

5

6

p(x)

.10

.15

.20

.25

.20

.06

.04

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}

Short Answer

Expert verified

a. The probability that at most three lines are in use is 0.70.

b. The probability that fewer than three lines are in use is 0.45.

c. The probability that at least three lines are in use is 0.55.

d. The probability that between two and five lines, inclusive, are in use is 0.71.

e. The probability that between two and four lines, inclusive, are not in use is 0.65.

f. The probability that at least four lines are not in use is 0.45.

Step by step solution

01

Given information

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

Following table shows the pmf of random variableX:

X

0

1

2

3

4

5

6

P(X)

0.1

0.15

0.2

0.25

0.2

0.06

00.04

02

Compute the probability that at most three lines in use.

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.

03

Compute the probability that fewer than three lines in use.

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.

04

Compute the probability that at least three lines in use.

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.

05

Compute the probability that between two andfive lines in use, inclusive

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.

06

Compute the probability that between two and four lines not in use, inclusive

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.

07

Compute the probability that at least 4 lines are not in use

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.

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