91Ó°ÊÓ

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. \begin{tabular}{l|ccccccc} \(x\) & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline\(p(x)\) & \(.10\) & \(.15\) & \(.20\) & \(.25\) & \(.20\) & \(.06\) & \(.04\) \end{tabular} 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\\}

Step by step solution

01

Understanding the problem

We are given a probability mass function (pmf) for the number of telephone lines in use, represented by the variable \(X\), and need to calculate the probability of various events related to this distribution.

02

Calculate probability of event a

For the event \{at most three lines are in use\}, calculate \(P(X \leq 3)\). Sum the probabilities for \(x = 0, 1, 2, 3\): \[P(X \leq 3) = p(0) + p(1) + p(2) + p(3) = 0.10 + 0.15 + 0.20 + 0.25 = 0.70\]

03

Calculate probability of event b

For the event \{fewer than three lines are in use\}, calculate \(P(X < 3)\). Sum the probabilities for \(x = 0, 1, 2\): \[P(X < 3) = p(0) + p(1) + p(2) = 0.10 + 0.15 + 0.20 = 0.45\]

04

Calculate probability of event c

For the event \{at least three lines are in use\}, calculate \(P(X \geq 3)\). Use the complement rule by subtracting \(P(X < 3)\) from 1: \[P(X \geq 3) = 1 - P(X < 3) = 1 - 0.45 = 0.55\]

05

Calculate probability of event d

For the event \{between two and five lines, inclusive, are in use\}, calculate \(P(2 \leq X \leq 5)\). Sum the probabilities for \(x = 2, 3, 4, 5\): \[P(2 \leq X \leq 5) = p(2) + p(3) + p(4) + p(5) = 0.20 + 0.25 + 0.20 + 0.06 = 0.71\]

06

Calculate probability of event e

For the event \{between two and four lines, inclusive, are not in use\}, calculate \(P(X \leq 4)\). Consider not in use meaning \(x = 0, 1, 2, 3, 4\): \[P(X \leq 4) = p(0) + p(1) + p(2) + p(3) + p(4) = 0.10 + 0.15 + 0.20 + 0.25 + 0.20 = 0.90\]

07

Calculate probability of event f

For the event \{at least four lines are not in use\}, this is equivalent to \{two or fewer lines are in use\}. Calculate \(P(X \leq 2)\): \[P(X \leq 2) = p(0) + p(1) + p(2) = 0.10 + 0.15 + 0.20 = 0.45\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Discrete Random Variables

When dealing with discrete random variables, we focus on outcomes that can be counted, usually as whole numbers. This is different from continuous variables, which can take any value within a range. Here, the random variable \(X\), represents the number of telephone lines in use and can take values of 0, 1, 2, 3, 4, 5, or 6. Each of these values has an associated probability, given by the probability mass function (pmf).

• A discrete random variable can only take on specific values.
• Each outcome has a probability that is defined by the pmf.
• The sum of all probabilities for a discrete random variable equals 1.

For example, for \(x = 3\) lines in use, the pmf specifies that \(p(3) = 0.25\). This indicates a 25% chance that precisely three lines will be in use. Understanding discrete random variables is fundamental when calculating probabilities for different scenarios.

Complement Rule

The complement rule is a useful concept in probability. It states that the probability of an event occurring (event A) is equal to 1 minus the probability of it not occurring (event \(A^c\)). This is especially handy when calculating probabilities of events described as "at least" a certain number.

• The complement of an event \(A\) is termed \(A^c\), indicating that \(A\) does not happen.
• The rule is mathematically represented as \(P(A) = 1 - P(A^c)\).

For example, when calculating the probability that at least three lines are in use \(P(X \geq 3)\), the complement rule helps simplify the process. Instead of adding probabilities for \(x = 3, 4, 5, 6\), we calculate \(1 - P(X < 3)\). Since we already know \(P(X < 3) = 0.45\), we find that \(P(X \geq 3) = 1 - 0.45 = 0.55\). This concept can make certain probability calculations much more efficient.

Probability Calculations

Calculating probabilities involves understanding the scenario and correctly applying the right formulas or rules. Often, this involves summing up the probabilities of individual outcomes that make up an event.

• Find the probability of combined events by summing their individual probabilities.
• Use cumulative addition for probabilities across multiple outcomes.

For instance, to find the probability of 2 to 5 lines inclusive being in use, i.e., \(P(2 \leq X \leq 5)\), you sum the probabilities for each value of \(x\): \[P(2 \leq X \leq 5) = p(2) + p(3) + p(4) + p(5) = 0.20 + 0.25 + 0.20 + 0.06 = 0.71\]This summed value tells us there is a 71% chance that between 2 and 5 lines will be in use. Breaking down problems into smaller, manageable pieces helps simplify the calculations.

Cumulative Probability

Cumulative probability involves summing up probabilities from the start of the distribution up to a specific value. It provides insights into the likelihood of the random variable falling within a certain range, which is crucial in decision-making processes.

• Cumulative probability is noted as \(P(X \leq k)\) for discrete variables and is calculated by summing probabilities from 0 to \(k\).
• It is especially useful for events described with terms like "at most."

For the mailing scenario, to find the probability of at most three lines being in use, we identify it as a cumulative probability \(P(X \leq 3)\): \[P(X \leq 3) = p(0) + p(1) + p(2) + p(3) = 0.10 + 0.15 + 0.20 + 0.25 = 0.70\]This means there is a 70% chance that three or fewer lines are in use at the specified time. Cumulative probability helps narrow down ranges of interest efficiently.