Math Problem Statement

A mail-order company 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)

0.11

0.15

0.20

0.25

0.19

0.07

0.03

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}

Solution

Let's go through each of the events step by step using the provided probability mass function (pmf) for XX, the number of lines in use. Let’s start calculating each probability:

Given PMF of XX:

x & p(x) \\ \hline 0 & 0.11 \\ 1 & 0.15 \\ 2 & 0.20 \\ 3 & 0.25 \\ 4 & 0.19 \\ 5 & 0.07 \\ 6 & 0.03 \\ \end{array}$$ ### (a) Probability that at most three lines are in use, $$P(X \leq 3)$$ This event includes $$X = 0, 1, 2,$$ and $$3$$: $$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$$ $$= 0.11 + 0.15 + 0.20 + 0.25 = 0.71$$ ### (b) Probability that fewer than three lines are in use, $$P(X < 3)$$ This event includes $$X = 0, 1,$$ and $$2$$: $$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$$ $$= 0.11 + 0.15 + 0.20 = 0.46$$ ### (c) Probability that at least three lines are in use, $$P(X \geq 3)$$ This event includes $$X = 3, 4, 5,$$ and $$6$$: $$P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$$ $$= 0.25 + 0.19 + 0.07 + 0.03 = 0.54$$ ### (d) Probability that between two and five lines, inclusive, are in use, $$P(2 \leq X \leq 5)$$ This event includes $$X = 2, 3, 4,$$ and $$5$$: $$P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$$ $$= 0.20 + 0.25 + 0.19 + 0.07 = 0.71$$ ### (e) Probability that between two and four lines, inclusive, are not in use, $$P(2 \leq (6 - X) \leq 4)$$ Here, we are interested in the event where the number of lines *not* in use (i.e., $$6 - X$$) is between two and four, inclusive. This corresponds to $$X$$ values of $$2, 3,$$ and $$4$$ (since $$6 - 2 = 4$$, $$6 - 3 = 3$$, and $$6 - 4 = 2$$): $$P(2 \leq (6 - X) \leq 4) = P(X = 2) + P(X = 3) + P(X = 4)$$ $$= 0.20 + 0.25 + 0.19 = 0.64$$ ### (f) Probability that at least four lines are not in use, $$P(6 - X \geq 4)$$ This event means that the number of lines in use, $$X$$, is at most 2 (since $$6 - X \geq 4$$ implies $$X \leq 2$$): $$P(6 - X \geq 4) = P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$$ $$= 0.11 + 0.15 + 0.20 = 0.46$$ ### Summary of Results: 1. **(a)** $$P(X \leq 3) = 0.71$$ 2. **(b)** $$P(X < 3) = 0.46$$ 3. **(c)** $$P(X \geq 3) = 0.54$$ 4. **(d)** $$P(2 \leq X \leq 5) = 0.71$$ 5. **(e)** $$P(2 \leq (6 - X) \leq 4) = 0.64$$ 6. **(f)** $$P(6 - X \geq 4) = 0.46$$ Would you like any further details on these calculations? --- ### Related Questions: 1. How would the probabilities change if there were seven lines instead of six? 2. What is the probability that exactly three lines are in use? 3. If the probability distribution were uniform, what would the probabilities be for each value of $$X$$? 4. How would we calculate the mean and variance of $$X$$? 5. What is the probability that an even number of lines is in use? **Tip:** When working with probabilities, always double-check if events are inclusive or exclusive, as this affects the calculations.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Discrete Random Variables
Probability Mass Function (PMF)

Formulas

P(X ≤ k) = P(X = 0) + P(X = 1) + ... + P(X = k)
P(X ≥ k) = P(X = k) + P(X = k+1) + ... + P(X = n)

Theorems

Additive Rule of Probability

Suitable Grade Level

Grades 11-12 or introductory college level