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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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