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Chapter 2: Q32E (page 73)

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:

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?

Short Answer

Expert verified
  1. \({\rm{240}}\) ways
  2. \({\rm{12}}\) ways
  3. \({\rm{108}}\) ways
  4. \({\rm{132}}\) ways
  5. At least one sony component

\({\rm{P( }}\)at least one sony component\({\rm{) = }}\frac{{{\rm{11}}}}{{{\rm{20}}}}{\rm{ = 0}}{\rm{.55 = 55\% }}\)

Exactly one sony cokmponent

\({\rm{P(}}\)exactly one sony component\({\rm{) = }}\frac{{{\rm{33}}}}{{{\rm{80}}}}{\rm{ = 0}}{\rm{.4125 = 41}}{\rm{.25\% }}\)

Step by step solution

01

Definition of product rule and also types of component

The product rule of probability refers to the occurrence of two or more independent occurrences at the same time. This is the sum of the likelihood of each of these occurrences occurring independently.

02

Determining the ways in which one can component of each type be selected 

Receiver, CD player, speakers, and turntable are the four sorts of components. With the exception of the receiver (which has five manufacturers) and speakers, all have four possible manufacturers (which have \({\rm{3}}\) possible manufacturers).

Receiver: \({\rm{5}}\) options

There are four methods to use a compact disc player.

There are three options for speakers.

\({\rm{4}}\)different ways to turn the turntable

The basic counting principle is as follows: If one event may happen in \({\rm{m}}\) ways and another can happen in \({\rm{n}}\)ways, the total number of ways the two events can happen in order is \({\rm{m*n}}\)

Make use of the basic counting principle

\({\rm{5*4*3*4 = 240}}\)

03

Determining the ways in which can components be selected if both the receiver and the compact disc player are to be Sony 

Counting principle: If one event may happen in\({\rm{m}}\)ways and another can happen in\({\rm{n}}\)ways, the number of ways the two events can happen in order is\({\rm{m*n}}\)

There is just one method to use the receiver (Sony), and there is only one way to use the compact disc player (Sony) (Sony)

\({\rm{1*1*3*4 = 12}}\)

04

Determining the ways in which components be selected if none is to be Sony?

The counting principle is as follows: If one event may happen in \({\rm{m}}\) ways and another can happen in \({\rm{n}}\) ways, the total number of ways the two events can happen in order is \({\rm{m*n}}\)

We are unable to choose Sony. The receiver only has four options (rather than five), the CD player only has three options (rather than four), and the turntable only has three options (instead of\({\rm{4}}\)).

\({\rm{4*3*3*3 = 108}}\)

05

Determining the ways in which selection be made if at least one Sony component is to be included

We know there are a total of 240 potential methods from section (a).

We know there are 108 methods to do it without using a Sony component because of section (c).

The total number of potential ways to choose at least one Sony component is then reduced by the number of conceivable ways to choose no Sony component:

\({\rm{240 - 108 = 132}}\)

06

Determining the probability that the system selected contains at least one Sony component

If we choose a Sony component, the other components can't have Sony components in them. Then, using the basic counting concept, we get:

Additional from the Sony receiver, there are no other Sony components: \({\rm{1*3*3*3 = 27}}\)

There are no additional Sony components save the compact disc player: \({\rm{4*1*3*3 = 36}}\)

There are no additional Sony components save the turntable: \({\rm{4*3*3*1 = 36}}\)

The number of positive outcomes divided by the total number of potential outcomes equals the probability:

\(\begin{aligned}{{\rm{P(\;At least one Sony component\;) = }}\frac{{{\rm{\# \;of favorable outcomes\;}}}}{{{\rm{\# \;of possible outcomes\;}}}}{\rm{ = }}\frac{{{\rm{132}}}}{{{\rm{240}}}}{\rm{ = }}\frac{{{\rm{11}}}}{{{\rm{20}}}}{\rm{ = 0}}{\rm{.55 = 55\% }}}\\{{\rm{P(\;Exactly one Sony component\;) = }}\frac{{{\rm{\# \;of favorable outcomes\;}}}}{{{\rm{\# \;of possible outcomes\;}}}}{\rm{ = }}\frac{{{\rm{27 + 36 + 36}}}}{{{\rm{240}}}}{\rm{ = }}\frac{{{\rm{99}}}}{{{\rm{240}}}}{\rm{ = }}\frac{{{\rm{33}}}}{{{\rm{80}}}}{\rm{ = 0}}{\rm{.4125 = 41}}{\rm{.25\% }}}\end{aligned}\)

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Most popular questions from this chapter

A system consists of two identical pumps, \(\# 1\) and \(\# 2\). If one pump fails, the system will still operate. However, because of the added strain, the remaining pump is now more likely to fail than was originally the case. That is, \(r\; = P\left( {\# 2\;fails|\# 1 fails} \right) > P\left( {\# 2 fails} \right) = q\). If at least one pump fails by the end of the pump design life in \(7\% \) of all systems and both pumps fail during that period in only 1%, what is the probability that pump \(\# 1\)will fail during the pump design life?

A college library has five copies of a certain text onreserve. Two copies (1 and 2) are first printings, and the other three (3, 4, and 5) are second printings. A student examines these books in random order, stopping only when a second printing has been selected. One possible outcome is 5, and another is 213.

a. List the outcomes in S.

b. Let Adenote the event that exactly one book must be examined. What outcomes are in A?

c. Let Bbe the event that book 5 is the one selected. What outcomes are in B?

d. Let Cbe the event that book 1 is not examined. What outcomes are in C?

Show that for any three events \({\rm{A,B}}\), and \({\rm{C}}\) with \({\rm{P(C) > 0}}\),\({\rm{P(A}} \cup {\rm{B}}\mid {\rm{C) = P(A}}\mid {\rm{C) + P(B}}\mid {\rm{C) - P(A}} \cap {\rm{B}}\mid {\rm{C)}}\).

A computer consulting firm presently has bids out on three projects. Let \({A_i}\) = {awarded project i}, for i=1, 2, 3, and suppose that\(P\left( {{A_1}} \right) = .22,\;P\left( {{A_2}} \right) = .25,\;P\left( {{A_3}} \right) = .28,\;P\left( {{A_1} \cap {A_2}} \right) = .11,\;P\left( {{A_1} \cap {A_3}} \right) = .05,\)\(P\left( {{A_2} \cap {A_3}} \right) = .07\),\(P\left( {{A_1} \cap {A_2} \cap {A_3}} \right) = .01\). Express in words each of the following events, and compute the probability of each event:

a. {\({A_1} \cup {A_2}\)}

b. \(A_1' \cap A_2'\)(Hint: \(\left( {{A_1} \cup {A_2}} \right)' = A_1' \cap A_2'\) )

c.\({A_1} \cup {A_2} \cup {A_3}\)

d. \(A_1' \cap A_2' \cap A_3'\)

e. \(A_1' \cap A_2' \cap {A_3}\)

f. \(\left( {A_1' \cap A_2'} \right) \cup {A_3}\)

Reconsider the system defect situation described in Exercise.

a. Given that the system has a type \(1\) defect, what is the probability that it has a type \({\bf{2}}\) defect?

b. Given that the system has a type \(1\) defect, what is the probability that it has all three types of defects?

c. Given that the system has at least one type of defect, what is the probability that it has exactly one type of defect?

d. Given that the system has both of the first two types of defects, what is the probability that it does not have the third type of defect?
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