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The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop machines and the other four have chosen desktop machines. Suppose that only two of the setups can be done on a particular day, and the two computers to be set up are randomly selected from the six (implying 15 equally likely outcomes; if the computers are numbered \(1,2, \ldots, 6\), then one outcome consists of computers 1 and 2 , another consists of computers 1 and 3 , and so on). a. What is the probability that both selected setups are for laptop computers? b. What is the probability that both selected setups are desktop machines? c. What is the probability that at least one selected setup is for a desktop computer? d. What is the probability that at least one computer of each type is chosen for setup?

Step by step solution

01

Identify Total Possible Outcomes

The computers are numbered from 1 to 6, with two laptops and four desktops. The selection of any two computers from these six results in \( \binom{6}{2} = 15 \) possible outcomes. This confirms the information given in the problem.

02

Calculate Probability Both Laptops are Selected

We need both selected computers to be laptops. The laptops can be selected in \( \binom{2}{2} = 1 \) way (since both must be chosen). To find the probability, divide this by the total number of outcomes: \[ P(\text{both laptops}) = \frac{1}{15} \]

03

Calculate Probability Both Desktops are Selected

There are \( \binom{4}{2} = 6 \) ways to select any two desktops from four. The probability is the number of ways to choose two desktops divided by the total outcomes: \[ P(\text{both desktops}) = \frac{6}{15} = \frac{2}{5} \]

04

Calculate Probability of At Least One Desktop

At least one desktop means not selecting two laptops. From Step 2, the probability of selecting both laptops is \( \frac{1}{15} \). Thus, the probability of at least one desktop is the complement: \[ P(\text{at least one desktop}) = 1 - \frac{1}{15} = \frac{14}{15} \]

05

Calculate Probability of One of Each Type

We want one laptop and one desktop. There are two ways to select one laptop from two and four ways to select one desktop from four, leading to \( 2 \times 4 = 8 \) mixed setups. Thus, the probability is: \[ P(\text{one of each}) = \frac{8}{15} \]

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Key Concepts

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

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arrangement, and combination of objects. The basic idea behind combinatorics is to enumerate possible outcomes without actually listing them. This is particularly useful in probability calculations. In our problem, the combination formula is used to determine the number of ways we can select a subset from a larger set.

The formula for combinations, also known as "n choose k," is given by:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

The notation \( \binom{6}{2} \) represents the number of ways to select 2 computers from 6 total computers. This results in 15 possible combinations of selecting computer pairs, as calculated using the combination formula.

In our example, combinatorics helps calculate the distinct ways two setups can be selected from among the laptops and desktops.

Complementary Probability

Complementary probability revolves around the idea of calculating the probability of an event by subtracting the probability of its complement from 1. This approach is often easier than calculating the desired probability directly when the complement is simpler to determine.

In the exercise, we used complementary probability to find one of the probabilities: the chance that at least one desktop is selected. Instead of directly counting every favorable outcome involving desktops, we calculated the probability of the complement—selecting only laptops—and subtracted it from 1.

Thus, knowing \( P(\text{both laptops}) = \frac{1}{15} \), we got:

• \( P(\text{at least one desktop}) = 1 - \frac{1}{15} = \frac{14}{15} \)

Complementary probability can simplify complex problems by focusing on the simpler complement event.

Probability Calculations

Probability calculations are essential to determining the likelihood of specific outcomes in any random event. To calculate the probability, you divide the number of favorable outcomes by the total possible outcomes. This provides a measure of how likely it is for an event to occur.

For instance, to find the probability that both setups involve desktops, we counted the successful arrangements of desktops (6 ways) and divided by the total arrangements (15). This method gives:

• \( P(\text{both desktops}) = \frac{6}{15} = \frac{2}{5} \)

Probability calculations allow us to assess different potential scenarios and prepare for outcomes based on their likelihood. It's fundamental in making informed decisions in uncertain situations.

Sample Space

The sample space is a fundamental concept in probability, representing the set of all possible outcomes in a certain experiment. Defining the sample space is the initial and crucial step in solving probability problems.

In this exercise, the sample space is made up of all the pairs of computers that could potentially be set up out of the six available. There are 15 such pairs, each equally likely, as identified by \( \binom{6}{2} = 15 \). This includes every possible combination of selecting two computers.

Understanding the sample space lets you know the denominators in probability calculations, ensuring that you're looking at the event outcomes relative to the entire set of possibilities. This equips you with a complete view of the problem context.