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Chapter 2: Problem 23

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?

Short Answer

Expert verified
a: \( \frac{1}{15} \), b: \( \frac{6}{15} \), c: \( \frac{14}{15} \), d: \( \frac{8}{15} \).

Step by step solution

01

Identify the Total Number of Outcomes

There are 6 computers, and we need to select 2 setups for the day. The number of ways to select 2 computers out of 6 can be realized by using combinations: \( \binom{6}{2} \). Calculate this:\[ \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 \]Thus, there are 15 equally likely outcomes.
02

Calculate Probability of Selecting Both Laptops

Two of the six computers are laptops. We need to select both of these laptops. The number of ways to select 2 laptops from 2 laptops is given by \( \binom{2}{2} = 1 \). Therefore, the probability is:\[ P(\text{both laptops}) = \frac{1}{15} \]
03

Calculate Probability of Selecting Both Desktops

Four of the six computers are desktops. We need to select 2 desktops. The number of ways to select 2 desktops from 4 desktops is given by \( \binom{4}{2} \):\[ \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 \]Therefore, the probability is:\[ P(\text{both desktops}) = \frac{6}{15} \]
04

Calculate Probability of At least One Desktop

The probability of at least one desktop is the complement of selecting both laptops. We already found the probability of selecting both laptops in Step 2, which is \( \frac{1}{15} \). Therefore:\[ P(\text{at least one desktop}) = 1 - P(\text{both laptops}) = 1 - \frac{1}{15} = \frac{14}{15} \]
05

Calculate Probability of Selecting One Laptop and One Desktop

The probability that one of each type is selected can also be approached by finding the complement of having both selected as laptop or both as desktop. So subtract both scenarios from the total probability:\[ P(\text{one laptop, one desktop}) = 1 - P(\text{both laptops}) - P(\text{both desktops}) \]\[ P(\text{one laptop, one desktop}) = 1 - \frac{1}{15} - \frac{6}{15} = \frac{8}{15} \]

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Key Concepts

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

Combinatorics
Combinatorics is the branch of mathematics dealing with combinations, permutations, and counting. When you come across a scenario where you need to choose or arrange objects, you鈥檙e using combinatorial techniques. In our problem, we need to select 2 out of 6 computers. Here, the order doesn鈥檛 matter; it's all about which computers are chosen, not the sequence. Thus, we use combinations, not permutations.

The mathematical symbol for combinations is "C" or it can be represented using the binomial coefficient. The formula for calculating combinations is \( \binom{n}{k} \), where \( n \) is the total number of items to choose from, and \( k \) is the number of items to choose. Using this formula helps us determine the number of potential pairings. In our case, \( \binom{6}{2} \) gives us 15 different ways to choose 2 computers from 6, confirming that each possible pairing is an equally likely outcome.
  • Total Outcomes: Calculated using the formula for combinations.
  • Independent of Order: Combinations concern selection, not arrangement.
Conditional Probability
Conditional probability is the likelihood of an event occurring given that another event has already occurred. It refines the sample space to consider only relevant scenarios. In our exercise, each probability is figured based on certain conditions.

For the probability of both computers being laptops, we consider only the pairings that involve laptops. Since there are 2 laptops among the 6 computers, our focus narrows to selecting both laptops from possible combinations. This specific condition drastically limits the number of favorable outcomes.
  • Laptop Setup: Just 1 outcome supports this condition: both laptops.
  • Desktop Setup: A larger number of favorable configurations exist with 4 desktops.
Conditional probability allows us to zoom in on certain subsets of the whole, providing precise calculations for each distinct scenario. It's about understanding the probability under specific pre-set conditions.
Complementary Probability
Complementary probability involves identifying the probability of the complement of an event 鈥� i.e., all the outcomes that do not involve the event occurring. This is calculated by subtracting the probability of the desired event from 1 (the certainty of all outcomes).

In figuring out the chance of at least one desktop being chosen, we looked at its complement: selecting both as laptops. We know dynamically that the universe of probabilities must total 1, so the complement aids in efficiently arriving at certain probabilities.

Similarly, finding the probability of selecting one laptop and one desktop can be visualized as first removing the pairings containing only laptops or only desktops. Using complementary probability streamlines the process, making intricate calculations more manageable by focusing on what's "left over."
  • At Least One of Type: Subtraction simplifies finding probabilities of non-exact scenarios.
  • Efficient Calculation: Complement helps swiftly deduce probabilities indirectly.

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Most popular questions from this chapter

Use Venn diagrams to verify the following two relationships for any events \(A\) and \(B\) (these are called De Morgan's laws): a. \((A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}\) b. \((A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}\)

Suppose an individual is randomly selected from the population of all adult males living in the United States. Let \(A\) be the event that the selected individual is over \(6 \mathrm{ft}\) in height, and let \(B\) be the event that the selected individual is a professional basketball player. Which do you think is larger, \(P(A \mid B)\) or \(P(B \mid A)\) ? Why?

As of April 2006 , roughly 50 million .com web domain names were registered (e.g., yahoo.com). a. How many domain names consisting of just two letters in sequence can be formed? How many domain names of length two are there if digits as well as letters are permitted as characters? [Note: A character length of three or more is now mandated.] b. How many domain names are there consisting of three letters in sequence? How many of this length are there if either letters or digits are permitted? [Note: All are currently taken.] c. Answer the questions posed in (b) for four-character sequences. d. As of April \(2006,97,786\) of the four-character sequences using either letters or digits had not yet been claimed. If a four-character name is randomly selected, what is the probability that it is already owned?

A certain shop repairs both audio and video components. Let \(A\) denote the event that the next component brought in for repair is an audio component, and let \(B\) be the event that the next component is a compact disc player (so the event \(B\) is contained in \(A\) ). Suppose that \(P(A)=.6\) and \(P(B)=.05\). What is \(P(B \mid A) ?\)

An electronics store is offering a special price on a complete set of components (receiver, compact disc player, speakers, turntable). A purchaser is offered a choice of manufacturer for each component: Receiver: Kenwood, Onkyo, Pioneer, Sony, Sherwood Compact disc player: Onkyo, Pioneer, Sony, Technics Speakers: Boston, Infinity, Polk Turntable: Onkyo, Sony, Teac, Technics A switchboard display in the store allows a customer to hook together any selection of components (consisting of one of each type). Use the product rules to answer the following questions: a. In how many ways can one component of each type be selected? b. In how many ways can components be selected if both the receiver and the compact disc player are to be Sony? c. In how many ways can components be selected if none is to be Sony? d. In how many ways can a selection be made if at least one Sony component is to be included? e. If someone flips switches on the selection in a completely random fashion, what is the probability that the system selected contains at least one Sony component? Exactly one Sony component?
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