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Magazine 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
\(P\left( {{A_0}\mid P} \right) = 0.087\),
\({\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}\mid {\rm{P}}} \right){\rm{ = 0}}{\rm{.652}}\),
\({\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}\mid {\rm{P}}} \right){\rm{ = 0}}{\rm{.261}}\)
Step by step solution
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
\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{0}}}} \right){\rm{ = 0}}{\rm{.2}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = 0}}{\rm{.5}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.3}}\end{array}\)
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.
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
\({\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}\mid {\rm{B}}} \right){\rm{ = }}\frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}{\rm{\c{C}B}}} \right)}}{{{\rm{P(B)}}}}{\rm{ = }}\frac{{{\rm{P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{j}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{j}}}} \right)}}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{P}} \left( {{\rm{B}}\mid {{\rm{A}}_{\rm{i}}}} \right){\rm{*P}}\left( {{{\rm{A}}_{\rm{i}}}} \right)}}{\rm{,}}\;\;\;{\rm{j = 1,2, \ldots ,n}}\)
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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