Chapter 6, Problem 88CR

a.

To determine

Find $P\left(E\right)$.

Answer to Problem 88CR

The probability of the event that a traveler on vacation checks work e-mail is 0.4.

Explanation of Solution

Calculation:

The percentage of passengers on a cruise ship who check work e-mail, use a cell phone, bring a laptop, check work e-mail and use a cell phone and neither check work e-mail nor use a cell phone are 40%, 30%, 25%, 23%, and 51%, respectively. The percentage of passengers who bring a laptop and check work e-mail is 88% and the percentage of passengers who use a cell phone and bring a laptop is 70%.

The events that a traveler on leave checks work e-mail, uses a cell phone, and bring a laptop are denoted as E, C, and L, respectively.

The given information is the summary table of the survey.

The probability of any Event A is given below:

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Here, the percentage of passengers on a cruise ship who check work e-mail is 40%.

Thus, the probability of the event that a traveler on vacation on a cruise ship who checks work e-mail is 0.4.

b.

To determine

Obtain $P\left(C\right)$.

Answer to Problem 88CR

The probability of the event that a traveler on vacation on a cruise ship who uses a cell phone is 0.3.

Explanation of Solution

Calculation:

Here, the percentage of travelers on vacation on a cruise ship who uses a cell phone is 30%.

Thus, the probability of the event that a passenger on a cruise ship who uses a cell phone is 0.3.

c.

To determine

Compute $P\left(L\right)$.

Answer to Problem 88CR

The probability of the event that a passenger on a cruise ship who brought a laptop is 0.25.

Explanation of Solution

Calculation:

Here, the percentage of passengers on a cruise ship who brought a laptop is 25%.

Thus, the probability of the event that a traveler on vacation on a cruise ship who brought a laptop is 0.25.

d.

To determine

Calculate $P\left(E\text{ and }C\right)$.

Answer to Problem 88CR

The probability of the event that a traveler on a cruise ship who checks work e-mail and uses a cell phone is 0.23.

Explanation of Solution

Calculation:

Here, the percentage of passengers on a cruise ship who checks work e-mail and uses a cell phone is 23%.

Thus, the probability of the event that a passenger on a cruise ship who checks work e-mail and uses a cell phone is 0.23.

e.

To determine

Calculate $P\left(E^C\text{ and }{C}^C\text{and }{L}^C\right)$.

Answer to Problem 88CR

The probability of the event that a passenger on a cruise ship neither checks work e-mail nor uses a cell phone nor bring a laptop is 0.51.

Explanation of Solution

Calculation:

Here, the percentage of travelers on a cruise ship neither checks work e-mail nor uses a cell phone nor bring a laptop is 51%.

Thus, the probability of the event that a passenger on a cruise ship neither checks work e-mail nor uses a cell phone nor bring a laptop is 0.51.

f.

To determine

Compute $P\left(E\text{ and }C\text{ or }L\right)$.

Answer to Problem 88CR

$\underset{\_}{P\left(E\text{ and }C\text{ or }L\right)=0.28}$.

Explanation of Solution

Calculation:

The given information is about the percentage of passengers on a cruise ship who checks work e-mail, use a cell phone, bring a laptop, check work e-mail and use a cell phone, and neither check work e-mail nor use a cell phone nor bring a laptop are 40%, 30%, 25%, 23%, and 51%, respectively. The percentage of passengers who bring a laptop also check work e-mail is 88% and those who use a cell phone and bring a laptop is 70%.

The events that a passenger checks work e-mail, uses a cell phone, and brought a laptop are denoted as E, C, and L, respectively.

The required probability can be obtained as follows:

$\begin{array}{c}P\left(E\text{ and }C\text{ or }L\right)=P\left(\left(E\text{ and }C\right)\text{or }L\right)\\ =\left(P\left(E\cap C\right)+P\left(L\right)-P\left(E\cap C\cap L\right)\right)\end{array}$

The general formula for $P\left(E\cap C\cap L\right)$ is as follows:

$P\left(E\text{ and }C\text{ and }L\right)=\left(\begin{array}{l}P\left(E\cup C\cup L\right)-P\left(E\right)-P\left(C\right)-P\left(L\right)+P\left(E\cap C\right)+\\ P\left(E\cap L\right)+P\left(C\cap L\right)\end{array}\right)$

Here, the percentage passengers who neither check work e-mail nor use a cell phone nor bring a laptop is 51%.

Then, the probability of passengers who check work e-mail nor use a cell phone nor bring a laptop is 0.49 $\left(=1-0.51\right)$.

Therefore, $P\left(E\cup C\cup L\right)=0.49$

The given probability values are as follows:

$P\left(E\right)=0.40,P\left(C\right)=0.30,P\left(L\right)=0.25,P\left(E\cap C\right)=0.23,P\left(E|L\right)=0.88,\text{and }P\left(L|C\right)=0.70$

The probability of passengers who bring a laptop also checks work e-mail $P\left(E|L\right)$ is 0.88.

The probability of passengers who check work e-mail and brings a laptop is obtained as follows:

$\begin{array}{c}P\left(E\text{ and }L\right)=P\left(L\right)\times P\left(E|L\right)\\ =\left(0.25\right)\left(0.88\right)\\ =0.22\end{array}$

Thus, $P\left(E\text{ and }L\right)=0.22$.

The probability of passengers who uses a cell phone and brings a laptop is obtained as follows:

$\begin{array}{c}P\left(\text{C and }L\right)=P\left(C\right)\times P\left(L|C\right)\\ =\left(0.3\right)\left(0.7\right)\\ =0.21\end{array}$

Thus, $P\left(C\text{ and }L\right)=0.22$.

Then, the probability that a passenger on a cruise ship who checks work e-mail, uses a cell phone and brings a laptop is given below:

$\begin{array}{c}P\left(E\text{ and }C\text{ and }L\right)=0.49-0.4-0.3-0.25+0.23+0.22+0.21\\ =0.2\end{array}$

Thus, $P\left(E\text{ and }C\text{ and }L\right)=0.2$.

The required probability is given below:

$\begin{array}{c}P\left(E\text{ and }C\text{ or }L\right)=0.23+0.25-0.2\\ =0.28\end{array}$

Thus, $\underset{\_}{P\left(E\text{ and }C\text{ or }L\right)=0.28}$.

g.

To determine

Obtain $P\left(E\text{ |}L\right)$.

Answer to Problem 88CR

$\underset{\_}{P\left(E\text{ |}L\right)=0.88}$.

Explanation of Solution

It is known that 88% of the passengers who brought laptops also checks work e-mail. Therefore,

$\underset{\_}{P\left(E\text{ |}L\right)=0.88}$.

h.

To determine

Find $P\left(L|C\right)$.

Answer to Problem 88CR

$\underset{\_}{P\left(L|C\right)=0.70}$.

Explanation of Solution

It is known that 70% of the passengers who uses a cell phone also brought a laptop. Therefore,

$P\left(L|C\right)=0.70$.

i.

To determine

Find $P\left(E\text{ and }C\text{ and }L\right)$.

Answer to Problem 88CR

The value of $P\left(E\text{ and }C\text{ and }L\right)$ is 0.2.

Explanation of Solution

Calculation:

The percentage of passengers on a cruise ship who check work e-mail, use a cell phone, bring a laptop, check work e-mail and use a cell phone, and neither check work e-mail nor use a cell phone nor bring a laptop are 40%, 30%, 25%, 23%, and 51%, respectively. The percentage of passengers who bring a laptop also check work e-mail is 88% and use a cell phone also bring a laptop is 70%.

The events that a passenger checks work e-mail, uses a cell phone, and brought a laptop are denoted as E, C, and L, respectively.

The general formula for $P\left(E\cup C\cup L\right)$ is as follows:

$P\left(E\cup C\cup L\right)=\left(\begin{array}{l}P\left(E\right)+P\left(C\right)+P\left(L\right)-P\left(E\cap C\right)-\\ P\left(E\cap L\right)-P\left(C\cap L\right)+P\left(E\cap C\cap L\right)\end{array}\right)$

Then, the formula for the required probability is as follows:

$P\left(E\text{ and }C\text{ and }L\right)=\left(\begin{array}{l}P\left(E\cup C\cup L\right)-P\left(E\right)-P\left(C\right)-P\left(L\right)+P\left(E\cap C\right)+\\ P\left(E\cap L\right)+P\left(C\cap L\right)\end{array}\right)$

Here, the percentage passengers who neither check work e-mail nor use a cell phone nor bring a laptop is 51%.

Then, the probability of passengers who check work e-mail nor use a cell phone nor bring a laptop is $1-0.51=0.49$.

The probability of passengers, who check work e-mail is 0.40.

The probability of passengers who use a cell phone is 0.30.

The probability of passengers who bring a laptop is 0.25.

The probability of passengers who bring a laptop also check work e-mail is 0.88.

The probability of passengers who check work e-mail and bring a laptop is obtained as follows:

$\begin{array}{c}P\left(E\text{ and }L\right)=P\left(L\right)\times P\left(E|L\right)\\ =\left(0.25\right)\left(0.88\right)\\ =0.22\end{array}$

Thus, the probability of passengers who check work e-mail and bring a laptop is 0.22.

The probability of passengers who use a cell phone also bring a laptop is 0.7.

The probability of passengers who use a cell phone and bring a laptop is obtained as follows:

$\begin{array}{c}P\left(\text{C and }L\right)=P\left(C\right)\times P\left(E|L\right)\\ =\left(0.3\right)\left(0.7\right)\\ =0.21\end{array}$

Thus, the probability of passengers who use a cell phone and bring a laptop is 0.21.

Then, the probability of the event that a passenger on a cruise ship who check work e-mail, use a cell phone, and bring a laptop is given below:

$\begin{array}{c}P\left(E\text{ and }C\text{ and }L\right)=0.49-0.4-0.3-0.25+0.23+0.22+0.21\\ =0.2\end{array}$

Thus, the probability of the event that a passenger on a cruise ship who check work e-mail, use a cell phone, and bring a laptop is 0.2.

j.

To determine

Obtain $P\left(E\text{ and }L\right)$.

Answer to Problem 88CR

The probability of the event that a passenger on a cruise ship who check e-mail and bring a laptop is 0.22.

Explanation of Solution

From Part (a), the probability of passengers who check work e-mail and bring a laptop is 0.22.

k.

To determine

Obtain $P\left(C\text{ and }L\right)$.

Answer to Problem 88CR

The probability of the event that a passenger on a cruise ship who use a cell phone and bring a laptop is 0.21.

Explanation of Solution

From Part (a), the probability of passengers who use a cell phone and bring a laptop is 0.21.

l.

To determine

Calculate $P\left(C|\left(E\text{ and }L\right)\right)$.

Answer to Problem 88CR

The probability of the event that a passenger uses a cell phone, given that he or she checks work e-mail and brought a laptop is 0.909.

Explanation of Solution

Calculation:

Conditional rule:

The formula for probability of E given F is $P\left(E|F\right)=\frac{P\left(E\cap F\right)}{P\left(F\right)}$

Then, the required probability can be obtained as follows:

$P\left(C|\left(E\text{ and }L\right)\right)=\frac{P\left(C\cap E\cap L\right)}{P\left(E\text{ and }L\right)}$

From Part (a), $P\left(E\cap C\cap L\right)=0.2$ and $P\left(E\cap L\right)=0.22$.

$\begin{array}{c}P\left(C|\left(E\text{ and }L\right)\right)=\frac{0.2}{0.22}\\ =0.909\end{array}$

Thus, the probability of the event that a passenger uses a cell phone, given that he or she checks work e-mail, and brought a laptop is 0.909.