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Chapter 2: Q68E (page 85)

A friend who lives in Los Angeles makes frequent consulting trips to Washington, D.C.; \({\rm{50\% }}\)of the time she travels on airline\({\rm{\# 1}}\), \({\rm{30\% }}\) of the time on airline \({\rm{\# 2}}\), and the remaining \({\rm{20\% }}\) of the time on airline #3. For airline \({\rm{\# 1}}\), flights are late into D.C. \({\rm{30\% }}\) of the time and late into L.A. \({\rm{10\% }}\) of the time. For airline\({\rm{\# 3}}\), these percentages are \({\rm{25\% }}\) and \({\rm{20\% }}\), whereas for airline #3 the percentages are \({\rm{40\% }}\) and \({\rm{25\% }}\). If we learn that on a particular trip she arrived late at exactly one of the two destinations, what are the posterior probabilities of having flown on airlines \({\rm{\# 1}}\), \({\rm{\# 2}}\), and \({\rm{\# 3}}\)? Assume that the chance of a late arrival in L.A. is unaffected by what happens on the flight to D.C. (Hint: From the tip of each first-generation branch on a tree diagram, draw three second-generation branches labeled, respectively, \({\rm{2}}\) late, \({\rm{2}}\) late, and \({\rm{2}}\) late.)

Short Answer

Expert verified

The probabilities are

\(\begin{array}{l}P\left( {{A_1}\mid 1} \right) = 0.466\\P\left( {{A_2}\mid 1} \right) = 0.288\\P\left( {{A_3}\mid 1} \right) = 0.247\end{array}\)

Step by step solution

01

Definition

The updated chance of an event occurring after additional information is taken into account is known as posterior probability.

02

Finding probabilities

The solution is

The probability of being late at exactly one city is obtained by the following calculation

\({\rm{0}}{\rm{.5*0}}{\rm{.3*0}}{\rm{.9 + 0}}{\rm{.5*0}}{\rm{.7*0}}{\rm{.1 + 0}}{\rm{.3*0}}{\rm{.25*0}}{\rm{.8 + 0}}{\rm{.3*0}}{\rm{.75*0}}{\rm{.2 + 0}}{\rm{.2*0}}{\rm{.4*0}}{\rm{.75 + 0}}{\rm{.2*0}}{\rm{.6*0}}{\rm{.25}}\)

\({\rm{ = 0}}{\rm{.365}}\)

The probability she uses \({\rm{1}}\) is

\(\frac{{{\rm{0}}{\rm{.5*0}}{\rm{.3*0}}{\rm{.9 + 0}}{\rm{.5*0}}{\rm{.7*0}}{\rm{.1}}}}{{{\rm{0}}{\rm{.365}}}}\)

The probability she uses \({\rm{2}}\) is

\(\begin{array}{l}\frac{{{\rm{0}}{\rm{.3*0}}{\rm{.25*0}}{\rm{.8 + 0}}{\rm{.3*0}}{\rm{.75*0}}{\rm{.2}}}}{{{\rm{0}}{\rm{.365}}}}\\{\rm{ = 0}}{\rm{.288}}\end{array}\)

The probability she uses \({\rm{3}}\) is

\(\begin{array}{l}\frac{{{\rm{0}}{\rm{.2*0}}{\rm{.4*0}}{\rm{.75 + 0}}{\rm{.2*0}}{\rm{.6*0}}{\rm{.25}}}}{{{\rm{0}}{\rm{.365}}}}\\{\rm{ = 0}}{\rm{.247}}\end{array}\)

Therefore, the probabilities are

\(\begin{array}{l}P\left( {{A_1}\mid 1} \right) = 0.466\\P\left( {{A_2}\mid 1} \right) = 0.288\\P\left( {{A_3}\mid 1} \right) = 0.247\end{array}\)

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