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Chapter 2: Q66E (page 85)

Consider the following information about travellers on vacation (based partly on a recent Travelocity poll): \({\rm{40\% }}\) check work email, \({\rm{30\% }}\) use a cell phone to stay connected to work, \({\rm{25\% }}\) bring a laptop with them, \({\rm{23\% }}\) both check work email and use a cell phone to stay connected, and \({\rm{51\% }}\) neither check work email nor use a cell phone to stay connected nor bring a laptop. In addition, \({\rm{88}}\) out of every \({\rm{100}}\) who bring a laptop also check work email, and \({\rm{70}}\) out of every \({\rm{100}}\) who use a cell phone to stay connected also bring a laptop. a. What is the probability that a randomly selected traveller who checks work email also uses a cell phone to stay connected? b. What is the probability that someone who brings a laptop on vacation also uses a cell phone to stay connected? c. If the randomly selected traveller checked work email and brought a laptop, what is the probability that he/ she uses a cell phone to stay connected?

Short Answer

Expert verified

a. \({\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{E}}}} \right){\rm{ = 0}}{\rm{.575}}\),

b. \({\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{L}}}} \right){\rm{ = 0}}{\rm{.84}}\),

c. \({\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ = 0}}{\rm{.91}}\)

Step by step solution

01

Determine denote events and given probabilities

Denote events

\(\begin{array}{l}{{\rm{A}}_{\rm{E}}}{\rm{ = \{ check work mail \} }}\\{{\rm{A}}_{\rm{C}}}{\rm{ = \{ use cell phone to stay connected to work \} ;}}\\{{\rm{A}}_{\rm{L}}}{\rm{ = \{ bring a laptop \} }}\end{array}\)

we are given the following probabilities

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}} \right)\rm &=& 0{\rm{.4}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}} \right) \rm &=& 0{\rm{.3}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{L}}}} \right) \rm &=& 0{\rm{.25}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{C}}}} \right) \rm &=& 0{\rm{.23}}\\{\rm{P}}\left( {{\rm{A}}_{\rm{E}}^{\rm{垄}}{\rm{脟}A}_{\rm{C}}^{\rm{垄}}{\rm{脟}A}_{\rm{L}}^{\rm{垄}}} \right)\rm &=& 0{\rm{.51}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}\mid {{\rm{A}}_{\rm{L}}}} \right)\rm &=& 0{\rm{.88}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{L}}}\mid {{\rm{A}}_{\rm{C}}}} \right)\rm &=& 0{\rm{.7}}\end{array}\)

02

Determine the value of \({\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{E}}}} \right)\)

(a):

We need to find the probability that a randomly selected traveler who checks work email also uses a cell phone to stay connected, which is a conditional probability of \({{\rm{A}}_{\rm{C}}}\)given that the event \({{\rm{A}}_{\rm{E}}}\)has occurred. Therefore,

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{E}}}} \right)\rm &=& \frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{E}}}} \right)}}{{{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}} \right)}}\\ \rm &=& \frac{{{\rm{0}}{\rm{.23}}}}{{{\rm{0}}{\rm{.4}}}}\\ \rm &=& 0{\rm{.575}}\end{array}\)

where we used the definition of conditional probability.

Conditional probability of \({\rm{A}}\) given that the event \({\rm{B}}\) has occurred, for which \({\rm{P}}\left( {\rm{B}} \right){\rm{ > 0,}}\)is

\({\rm{P(A}}\mid {\rm{B) = }}\frac{{{\rm{P(A{脟}B)}}}}{{{\rm{P(B)}}}}\)

for any two events \({\rm{A}}\)and\({\rm{B}}\).

03

Determine the value of \({\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{L}}}} \right)\)

(b):

The probability that someone who brings a laptop on vacation also uses a cell phone to stay connected is conditional probability of \({{\rm{A}}_{\rm{C}}}\)given that the event \({{\rm{A}}_{\rm{L}}}\)has occurred. We have

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{L}}}} \right)\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{L}}}{\rm{脟}}{{\rm{A}}_{\rm{C}}}} \right)}}{{{\rm{P}}\left( {{{\rm{A}}_{\rm{L}}}} \right)}}\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} \frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{L}}}\mid {{\rm{A}}_{\rm{C}}}} \right){\rm{*P}}\left( {{{\rm{A}}_{\rm{C}}}} \right)}}{{{\rm{P}}\left( {{{\rm{A}}_{\rm{L}}}} \right)}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.7*0}}{\rm{.3}}}}{{{\rm{0}}{\rm{.25}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.21}}}}{{{\rm{0}}{\rm{.25}}}}\\{\rm{ = 0}}{\rm{.84}}\end{array}\)

(1): here we use the definition of conditional probability, (2): here we use the multiplication rule given below.

The Multiplication Rule

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

04

Determine the value of \({\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{É}}{{\rm{A}}_{\rm{C}}}{\rm{É}}{{\rm{A}}_{\rm{L}}}} \right)\)

(c):

The probability we are asked to calculate is conditional probability of \({{\rm{A}}_{\rm{C}}}\) given that the event \({{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}\) occurred. We have the following

\({\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ = }}\frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right)}}{{{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right)}}\)

Using the proposition given below, the following is true

\({\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脡}}{{\rm{A}}_{\rm{C}}}{\rm{脡}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{E}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{C}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{L}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{C}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right)\)

or equally

\({\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脡}}{{\rm{A}}_{\rm{C}}}{\rm{脡}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{E}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{C}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{L}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{C}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right)\)

We are missing

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脡}}{{\rm{A}}_{\rm{C}}}{\rm{脡}}{{\rm{A}}_{\rm{L}}}} \right)\rm &=& 1 - P\left( {{{\left( {{{\rm{A}}_{\rm{E}}}{\rm{脡}}{{\rm{A}}_{\rm{C}}}{\rm{脡}}{{\rm{A}}_{\rm{L}}}} \right)}^{\rm{垄}}}} \right)\\ \rm &=& 1 - P\left( {{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right)\\ \rm &=& 1 - 0{\rm{.51}}\\ \rm &=& 0{\rm{.49}}\end{array}\)

05

Determine the value of \({\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{E}}}{\rm{Ç}}{{\rm{A}}_{\rm{L}}}} \right)\)

Now we can substitute the values

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脡}}{{\rm{A}}_{\rm{C}}}{\rm{脡}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{E}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{C}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{L}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{C}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{0}}{\rm{.49 - 0}}{\rm{.4 - 0}}{\rm{.3 - 0}}{\rm{.25 + 0}}{\rm{.22 + 0}}{\rm{.23 + 0}}{\rm{.21}}\\{\rm{ = 0}}{\rm{.2}}{\rm{.}}\end{array}\)

(1): here we calculated \({\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{l}}}} \right)\)as

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{L}}}} \right){\rm{*P}}\left( {{{\rm{A}}_{\rm{E}}}\mid {{\rm{A}}_{\rm{L}}}} \right)\\{\rm{ = 0}}{\rm{.25*0}}{\rm{.88}}\\{\rm{ = 0}}{\rm{.22,}}\end{array}\)

where the other probabilities were calculated previously.

Proposition: For every three events \({\rm{A,B}}\)and \({\rm{C}}\)the following is true

\({\rm{P(A脡 B脡 C) = P(A) + P(B) + P(C) - P(A脟B) - P(A脟C) - P(B脟C) + P(A脟B脟C)}}\)

Finally, we have

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}\mid {{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right){\rm{ = }}\frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{C}}}{\rm{脟}}{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right)}}{{{\rm{P}}\left( {{{\rm{A}}_{\rm{E}}}{\rm{脟}}{{\rm{A}}_{\rm{L}}}} \right)}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.2}}}}{{{\rm{0}}{\rm{.22}}}}\\{\rm{ = 0}}{\rm{.91}}\end{array}\)

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Most popular questions from this chapter

Consider randomly selecting a single individual and having that person test drive \({\rm{3}}\) different vehicles. Define events \({{\rm{A}}_{\rm{1}}}\), \({{\rm{A}}_{\rm{2}}}\), and \({{\rm{A}}_{\rm{3}}}\) by

\({{\rm{A}}_{\rm{1}}}\)=likes vehicle #\({\rm{1}}\)\({{\rm{A}}_{\rm{2}}}\)= likes vehicle #\({\rm{2}}\)\({{\rm{A}}_{\rm{3}}}\)=likes vehicle #\({\rm{3}}\)Suppose that\({\rm{ = }}{\rm{.65,}}\)\({\rm{P(}}{{\rm{A}}_3}{\rm{)}}\)\({\rm{ = }}{\rm{.70,}}\)\({\rm{P(}}{{\rm{A}}_{\rm{1}}} \cup {{\rm{A}}_{\rm{2}}}{\rm{) = }}{\rm{.80,P(}}{{\rm{A}}_{\rm{2}}} \cap {{\rm{A}}_{\rm{3}}}{\rm{) = 40,}}\)and\({\rm{P(}}{{\rm{A}}_{\rm{1}}} \cup {{\rm{A}}_{\rm{2}}} \cup {{\rm{A}}_{\rm{3}}}{\rm{) = }}{\rm{.88}}{\rm{.}}\)

a. What is the probability that the individual likes both vehicle #\({\rm{1}}\)and vehicle #\({\rm{2}}\)?

b. Determine and interpret\({\rm{p}}\)(\({{\rm{A}}_{\rm{2}}}\)|\({{\rm{A}}_{\rm{3}}}\)).

c. Are \({{\rm{A}}_{\rm{2}}}\)and \({{\rm{A}}_{\rm{3}}}\)independent events? Answer in two different ways.

d. If you learn that the individual did not like vehicle #\({\rm{1}}\), what now is the probability that he/she liked at least one of the other two vehicles?

Three components are connected to form a system as shown in the accompanying diagram. Because the components in the 2鈥�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鈥�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鈥�, A \( \cup \)C, A \( \cap \)C, B \( \cup \)C, and B \( \cap \)C.

At a certain gas station, \({\rm{40\% }}\)of the customers use regular gas \(\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{,35\% }}\) use plus gas\(\left( {{{\rm{A}}_{\rm{2}}}} \right)\), and \({\rm{25\% }}\) use premium\(\left( {{{\rm{A}}_{\rm{3}}}} \right)\). Of those customers using regular gas, only \({\rm{30\% }}\) fill their tanks (event \({\rm{B}}\) ). Of those customers using plus, \({\rm{60\% }}\)fill their tanks, whereas of those using premium, \({\rm{50\% }}\)fill their tanks.

a. What is the probability that the next customer will request plus gas and fill the tank\(\left( {{{\rm{A}}_{\rm{2}}} \cap {\rm{B}}} \right)\)?

b. What is the probability that the next customer fills the tank?

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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:

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