Question

A mail-order computer software business has six telephone lines. Let \(x\) denote the number of lines in use at a specified time. The probability distribution of \(x\) is as follows: $$ \begin{array}{lrrrrrrr} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ p(x) & .10 & .15 & .20 & .25 & .20 & .06 & .04 \end{array} $$ Write each of the following events in terms of \(x\), and then calculate the probability of each one: 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

The probabilities for the events are: a. 0.70, b. 0.45, c. 0.55, d. 0.71, e. 0.45, f. 0.45.

Step-by-step solution

Creating a Probability Table

Initially, make a note that \(x\) can have values from 0 to 6 and the corresponding probabilities are 0.10, 0.15, 0.20, 0.25, 0.20, 0.06, 0.04, respectively. The sum of these probabilities should total to 1.

Calculating the probability for event a

Event a is defined as 'At most three lines are in use'. This means \(x\) can be 0, 1, 2, or 3. We calculate the probability by summing up the probabilities for each possible value of \(x\): \(p(x=0) + p(x=1) + p(x=2) + p(x=3) = 0.10 + 0.15 + 0.20 + 0.25 = 0.70\).

Calculating the probability for event b

Event b is defined as 'Fewer than three lines are in use'. This means \(x\) can be 0, 1, or 2. Therefore, we add the probabilities for these values: \(p(x=0) + p(x=1) + p(x=2) = 0.10 + 0.15 + 0.20 = 0.45\).

Calculating the probability for event c

Event c is 'At least three lines are in use'. This means \(x\) can be 3, 4, 5, or 6. So, the probability is \(p(x=3) + p(x=4) + p(x=5) + p(x=6) = 0.25 + 0.20 + 0.06 + 0.04 = 0.55\).

Calculating the probability for event d

Event d is 'Between two and five lines (inclusive) are in use'. This means \(x\) can be 2, 3, 4 or 5. Therefore the probability is \(p(x=2) + p(x=3) + p(x=4) + p(x=5) = 0.20 + 0.25 + 0.20 + 0.06 = 0.71\).

Calculating the probability for event e

Event e is 'Between two and four lines (inclusive) are not in use'. This means \(x\) can be 0, 1 or 2. So, the probability is \(p(x=0) + p(x=1) + p(x=2) = 0.10 + 0.15 + 0.20 = 0.45\).

Calculating the probability for event f

Event f is 'At least four lines are not in use'. This means \(x\) can be 0, 1, or 2. So, probability is \(p(x=0) + p(x=1) + p(x=2) = 0.10 + 0.15 + 0.20 = 0.45\).

Key concepts

Discrete Random Variables

Taking the first dive into probability, let's talk about discrete random variables. These are a cornerstone concept and relate to countable outcomes. Consider a call center with multiple phone lines; in this case, the number of lines in use at any given time is a prime example of a discrete random variable. It can take on finite or countable values, such as 0, 1, 2, all the way up to 6.

In essence, a discrete random variable, which we can denote as 'X', represents distinct outcomes of a random experiment, like flipping a coin, where 'X' can be 0 for tails and 1 for heads. Similarly, for our mail-order computer software business with six telephone lines, 'X' could be 0 when no lines are in use or 6 when all lines are engaged, with each state having a certain probability linked to it. Each possible state of 'X' is a fundamental piece of the puzzle in understanding the behaviour of systems influenced by chance.

Probability Mass Function

So, how do we quantify the likelihood of each outcome for our discrete random variable? Enter the probability mass function (PMF), symbolized as p(x). It is the function that gives us the probability that a discrete random variable is exactly equal to some value. For example, in our textbook problem, the PMF tells us the probability of 0 lines being in use is 0.10, 1 line is 0.15, and so on.

The PMF is interesting because it helps us create a complete picture of all possible outcomes and their probabilities. Mathematically, for a PMF to be valid, it must satisfy two conditions: the sum of all probabilities must equal 1, and the probability for each individual outcome must be between 0 and 1. This makes perfect sense because the probabilities of all possible outcomes must cover every scenario (hence totaling to 1), and no single outcome can be a certainty or an impossibility (hence the 0 to 1 range).

Cumulative Probability

Now that we have our PMF defined, we can explore cumulative probability. This entails looking at the probability of the event where our discrete random variable 'X' is less than or equal to a certain value. It's like summarizing the probabilities up to a point. This cumulative aspect is captured by the cumulative distribution function (CDF), which is obtained by summing up the probabilities from the start of the distribution up to a certain point 'x'.

In our exercise, when asked to find the probability that at most three lines are in use, we found it by adding up the probabilities up to and including when three lines are used—essentially, the CDF for 3. It is a powerful tool that helps students handle questions such as 'What's the probability of getting three or fewer calls?' without having to consider each possible outcome separately. The cumulative probability can answer 'at most', 'less than', 'greater than', or 'at least' types of questions, all vital in understanding the probability distribution of a discrete random variable.