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Chapter 2: Q65E (page 84)

A large operator of timeshare complexes requires anyone interested in making a purchase to first visit the site of interest. Historical data indicates that \({\rm{20\% }}\) of all potential purchasers select a day visit, \({\rm{50\% }}\) choose a one-night visit, and \({\rm{30\% }}\)opt for a two-night visit. In addition, \({\rm{10\% }}\) of day visitors ultimately make a purchase, \({\rm{30\% }}\) of one-night visitors buy a unit, and \({\rm{20\% }}\) of those visiting for two nights decide to buy. Suppose a visitor is randomly selected and is found to have made a purchase. How likely is it that this person made a day visit? A one-night visit? A two-night visit?

Short Answer

Expert verified

\(P\left( {{A_0}\mid P} \right) = 0.087\),

\({\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}\mid {\rm{P}}} \right){\rm{ = 0}}{\rm{.652}}\),

\({\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}\mid {\rm{P}}} \right){\rm{ = 0}}{\rm{.261}}\)

Step by step solution

01

Determine first and second layer of event

The Multiplication Rule

\({\rm{P(A\c{C}B) = P(A}}\mid {\rm{B)*P(B)}}\)

First layer of event are

\(\begin{array}{l}{{\rm{A}}_{\rm{0}}}{\rm{ = \{ day visit \} }}\\{{\rm{A}}_{\rm{1}}}{\rm{ = \{ 1 night visit \} }}\\{{\rm{A}}_{\rm{2}}}{\rm{ = \{ 2 night visit \} }}\end{array}\)

and the second layer events are

\({\rm{P = \{ madeapurchase\} }}\)

\({{\rm{P}}^{\rm{\cent}}}{\rm{ = \{ didnotmakeapurchase\} }}{\rm{.}}\)

The adequate probabilities for the first layer are

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{0}}}} \right){\rm{ = 0}}{\rm{.2}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = 0}}{\rm{.5}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.3}}\end{array}\)

02

Determine conditional probabilities

The adequate conditional probabilities are

\(\begin{array}{l}{\rm{P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{0}}}} \right){\rm{ = 0}}{\rm{.1}}\\{\rm{P}}\left( {{{\rm{P}}^{\rm{\cent}}}\mid {{\rm{A}}_{\rm{0}}}} \right){\rm{ = 1 - P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{0}}}} \right){\rm{ = 0}}{\rm{.9}}\\{\rm{P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{1}}}} \right){\rm{ = 0}}{\rm{.3}}\\{\rm{P}}\left( {{{\rm{P}}^{\rm{\cent}}}\mid {{\rm{A}}_{\rm{1}}}} \right){\rm{ = 1 - P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{0}}}} \right){\rm{ = 0}}{\rm{.7}}\\{\rm{P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.2}}\\{\rm{P}}\left( {{{\rm{P}}^{\rm{\cent}}}\mid {{\rm{A}}_{\rm{2}}}} \right){\rm{ = 1 - P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{0}}}} \right){\rm{ = 0}}{\rm{.8}}\end{array}\)

The tree diagram also contains the probabilities of the intersections. We use the multiplication rule to calculate every of the four intersections displayed in the tree diagram

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{0}}}{\rm{\c{C}P}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{0}}}} \right){\rm{*P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{0}}}} \right){\rm{ = 0}}{\rm{.2*0}}{\rm{.1 = 0}}{\rm{.02}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{0}}}{\rm{\c{C}}}{{\rm{P}}^{\rm{\cent}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{0}}}} \right){\rm{*P}}\left( {{{\rm{P}}^{\rm{\cent}}}\mid {{\rm{A}}_{\rm{0}}}} \right){\rm{ = 0}}{\rm{.2*0}}{\rm{.9 = 0}}{\rm{.18}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{\c{C}P}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{*P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{1}}}} \right){\rm{ = 0}}{\rm{.5*0}}{\rm{.3 = 0}}{\rm{.15}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{\c{C}}}{{\rm{P}}^{\rm{\cent}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{*P}}\left( {{{\rm{P}}^{\rm{\cent}}}\mid {{\rm{A}}_{\rm{1}}}} \right){\rm{ = 0}}{\rm{.5*0}}{\rm{.7 = 0}}{\rm{.35}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}{\rm{\c{C}P}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{*P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.3*0}}{\rm{.2 = 0}}{\rm{.06}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}{\rm{\c{C}}}{{\rm{P}}^{\rm{\cent}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{*P}}\left( {{{\rm{P}}^{\rm{\cent}}}\mid {{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.3*0}}{\rm{.8 = 0}}{\rm{.24}}\end{array}\)

From the tree diagram given below, we will calculate everything easier.

03

Determine probability that the person made a day visit

First: A probability that the person made a day visit is

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{0}}}\mid {\rm{P}}} \right){\rm{ }}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{0}}}{\rm{\c{C}P}}} \right)}}{{{\rm{P}}\left( {{{\rm{A}}_{\rm{0}}}} \right){\rm{ \times P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{0}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ \times P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{1}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ \times P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{2}}}} \right)}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.02}}}}{{{\rm{0}}{\rm{.02 + 0}}{\rm{.15 + 0}}{\rm{.06}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.02}}}}{{{\rm{0}}{\rm{.23}}}}\\{\rm{ = 0}}{\rm{.087}}\end{array}\)

(1): here we use the Bayes' Theorem mentioned below.

Bayes" Theorem

For \({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{A}}_{\rm{n}}}\)mutually exclusive events, and \({\rm{\`E }}_{{\rm{i = 1}}}^{\rm{n}}{{\rm{A}}_{\rm{i}}}{\rm{ = \Omega }}\) the prior probabilities are\(\left. {{\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}} \right){\rm{,i = 1,2, \ldots ,n}}} \right)\). If \({\rm{B}}\) is event for which\({\rm{P(B) > 0}}\), than the posterior probability of \({{\rm{A}}_{\rm{i}}}\)given that \({\rm{B}}\) has occurred

\({\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}\mid {\rm{B}}} \right){\rm{ = }}\frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}{\rm{\c{C}B}}} \right)}}{{{\rm{P(B)}}}}{\rm{ = }}\frac{{{\rm{P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{j}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{j}}}} \right)}}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{P}} \left( {{\rm{B}}\mid {{\rm{A}}_{\rm{i}}}} \right){\rm{*P}}\left( {{{\rm{A}}_{\rm{i}}}} \right)}}{\rm{,}}\;\;\;{\rm{j = 1,2, \ldots ,n}}\)

04

Determine probability that the person made a one-night visit

Second: A probability that the person made a one-night visit is

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}\mid {\rm{P}}} \right)\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{\c{C}P}}} \right)}}{{{\rm{P}}\left( {{{\rm{A}}_{\rm{0}}}} \right){\rm{ \times P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{0}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ \times P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{1}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ \times P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{2}}}} \right)}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.15}}}}{{{\rm{0}}{\rm{.02 + 0}}{\rm{.15 + 0}}{\rm{.06}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.15}}}}{{{\rm{0}}{\rm{.23}}}}\\{\rm{ = 0}}{\rm{.652}}\end{array}\)

(1): here we use the Bayes' Theorem mentioned above.

Third: A probability that the person made a two-night visit is

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}\mid {\rm{P}}} \right)\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}{\rm{\c{C}P}}} \right)}}{{{\rm{P}}\left( {{{\rm{A}}_{\rm{0}}}} \right){\rm{ \times P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{0}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ \times P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{1}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ \times P}}\left( {{\rm{P}}\mid {{\rm{A}}_{\rm{2}}}} \right)}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.06}}}}{{{\rm{0}}{\rm{.02 + 0}}{\rm{.15 + 0}}{\rm{.06}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.06}}}}{{{\rm{0}}{\rm{.23}}}}\\{\rm{ = 0}}{\rm{.261}}\end{array}\)

(1): here we use the Bayes' Theorem mentioned above.

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Most popular questions from this chapter

Three components are connected to form a system as shown in the accompanying diagram. Because the components in the 2–3 subsystem are connected in parallel, that subsystem will function if at least one of the two individual components functions. For the entire system to function, component 1 must function and so must the 2–3 subsystem.

The experiment consists of determining the condition of each component (S(success) for a functioning componentand F (failure) for a non-functioning component).

a. Which outcomes are contained in the event Athat exactly two out of the three components function?

b. Which outcomes are contained in the event Bthat at least two of the components function?

c. Which outcomes are contained in the event Cthat the system functions?

d. List outcomes in C’, A \( \cup \)C, A \( \cap \)C, B \( \cup \)C, and B \( \cap \)C.

A quality control inspector is examining newly produced items for faults. The inspector searches an item for faults in a series of independent fixations, each of a fixed duration. Given that a flaw is actually present, let p denote the probability that the flaw is detected during any one fixation (this model is discussed in “Human Performance in Sampling Inspection,” Human Factors, \({\rm{1979: 99--105)}}{\rm{.}}\)

a. Assuming that an item has a flaw, what is the probability that it is detected by the end of the second fixation (once a flaw has been detected, the sequence of fixations terminates)?

b. Give an expression for the probability that a flaw will be detected by the end of the nth fixation.

c. If when a flaw has not been detected in three fixations, the item is passed, what is the probability that a flawed item will pass inspection?

d. Suppose \({\rm{10\% }}\) of all items contain a flaw (P(randomly chosen item is flawed) . \({\rm{1}}\)). With the assumption of part (c), what is the probability that a randomly chosen item will pass inspection (it will automatically pass if it is not flawed, but could also pass if it is flawed)?

e. Given that an item has passed inspection (no flaws in three fixations), what is the probability that it is actually flawed? Calculate for \({\rm{p = 5}}\).

A box in a supply room contains \({\rm{15}}\) compact fluorescent lightbulbs, of which \({\rm{5}}\) are rated \({\rm{13}}\)-watt, \({\rm{6}}\)are rated \({\rm{18}}\)-watt, and \({\rm{4}}\) are rated \({\rm{23}}\)-watt. Suppose that three of these bulbs are randomly selected.

a. What is the probability that exactly two of the selected bulbs are rated \({\rm{23}}\)-watt?

b. What is the probability that all three of the bulbs have the same rating?

c. What is the probability that one bulb of each type is selected?

d. If bulbs are selected one by one until a \({\rm{23}}\)-watt bulb is obtained, what is the probability that it is necessary to examine at least 6 bulbs?

The composer Beethoven wrote \({\rm{9}}\) symphonies, \({\rm{9}}\) piano concertos (music for piano and orchestra), and \({\rm{32}}\) piano sonatas (music for solo piano).

a. How many ways are there to play first a Beethoven symphony and then a Beethoven piano concerto?

b. The manager of a radio station decides that on each successive evening (\({\rm{7}}\) days per week), a Beethoven symphony will be played followed by a Beethoven piano concerto followed by a Beethoven piano sonata. For how many years could this policy be continued before exactly the same program would have to be repeated?

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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