/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 The computers of six faculty mem... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
a. \( \frac{1}{15} \); b. \( \frac{2}{5} \); c. \( \frac{14}{15} \); d. \( \frac{8}{15} \).

Step by step solution

01

Understanding the Problem

We have 6 computers: 2 laptops and 4 desktops. Two computers are to be randomly selected for setup. We need to calculate different probabilities based on which computers are selected.
02

Total Outcomes

The total number of ways to choose 2 computers out of 6 is given by the combination formula: \( \binom{6}{2} = 15 \). These are the 15 equally likely outcomes mentioned.
03

Probability of Both Laptops (a)

There are 2 laptop computers. The number of ways to choose 2 laptops out of these 2 is \( \binom{2}{2} = 1 \). Therefore, the probability that both computers selected are laptops is \( \frac{1}{15} \).
04

Probability of Both Desktops (b)

There are 4 desktops. The number of ways to choose 2 desktops out of these 4 is \( \binom{4}{2} = 6 \). So, the probability that both selected computers are desktops is \( \frac{6}{15} = \frac{2}{5} \).
05

Probability of At Least One Desktop (c)

The probability of at least one being a desktop is the complement of the event 'both are laptops'. Since 'both are laptops' has a probability of \( \frac{1}{15} \), the probability of at least one desktop is \( 1 - \frac{1}{15} = \frac{14}{15} \).
06

Probability of One of Each Type (d)

If one selected setup is a laptop and the other is a desktop, we first choose any 1 laptop out of 2 (\( \binom{2}{1} = 2 \)) and then choose any 1 desktop out of 4 (\( \binom{4}{1} = 4 \)). The total number of ways to choose one laptop and one desktop is \( 2 \times 4 = 8 \). The probability of this event is \( \frac{8}{15} \).

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Key Concepts

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

Combinatorics
Combinatorics is an area of mathematics that deals with counting, arranging, and combining objects. It's like solving puzzles where we figure out how different pieces fit together. In problems like ours, combinatorics helps us determine the different ways to select and pair computers when certain conditions are provided. For instance, when we need to set up two computers out of six, we use combinatorial methods to find the total number of possible selections. This involves using a special formula known as the "combination formula." This formula helps us understand different possible groupings without worrying about their order. Combinatorics is key to understanding probabilities in a situation where multiple outcomes are possible. Without it, calculating probabilities would be much harder, especially as the number of outcomes or items increases.
Random Selection
In our problem, we address a classic situation of random selection. This implies that each pair of computers from the six available has an equal chance of being selected for setup on a particular day. Random selection is grounded in the principle that all choices are equally likely, assuming no external bias or influence. This concept is crucial in probabilistic scenarios where fair outcomes are expected. Random selection ensures that every pair of computers, such as laptops and desktops, are considered without preference, which is fundamental to calculating the probabilities of different events. Understanding this helps ensure that the probabilities assigned to each event are valid and dependable.
Event Complement
The complement rule is a handy tool in probability, especially when calculating the chances of a scenario not happening proves simpler. The complement of an event is all the possible outcomes in the sample space that are not in the event itself. For example, in our exercise, we calculated the probability of selecting at least one desktop. Instead of directly figuring out this probability, we used the event complement approach by first calculating the probability of its opposite, which is selecting both laptops. By using the complement rule, we can then subtract this probability from 1 to find the probability of picking at least one desktop. This method is especially useful when direct calculation feels cumbersome or less intuitive.
Combination Formula
The combination formula, a crucial component in combinatorics, is used to determine the number of ways to choose a subset of items from a larger set, where the order of selection doesn't matter. It is expressed as \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \). In our exercise, we used the combination formula to calculate the number of ways to select 2 computers from a total of 6. By substituting \( n = 6 \) and \( r = 2 \) into the formula, we find that \( \binom{6}{2} = 15 \). This tells us there are 15 different ways to select any 2 computers out of the 6. We further applied this formula to determine the number of ways to select both laptops and both desktops. Remember, the factorial \( n! \) represents the product of all positive integers up to \( n \), which is a foundational concept in understanding combinations.

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Most popular questions from this chapter

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 \(P(B \mid A)\) ?

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