Question

ons er the following information about travelers on vacation: 40% check work erna. 30% use cell phone to stay connected to work, 35% bring a prop with both check work erna" and use a cell phone to stay connected, and 54.4% neither check work email ror use a cell phone to stay out of every 100 who bring a laptop also check work email, and 70 out of every 100 who use a cell phone to stay connected also bring a laptop them, 1 m i n 9% ed nor bring a laptop. In addition, 84 (a) What is the probability that a randomly selected traveler 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 traveler checked work emalil and brought a laptop, what is the probablity that he/she uses a cell phone to stay connected? (Round your answer to four decimal places.) Need Help?ReTala to Tuter

Answer 1

Solution:

Let E denotes the event that a person checks work email, C denotes the event that the person uses a cell phone to stay connected to work, L denotes the event that the person brings a laptop with him.

We are given:

$P(E)=0.4, P(C)=0.3, P(L)=0.35$

$P(E\cap C)=0.19$

$P(E^{'}\cap C^{'}\cap L^{'} )=0.544$

$P(E|L)=0.84$

$P(L|C)=0.7$

a) We are required to find:

$P(C|E)=\frac{P(C\cap E)}{P(E)}$

  $=\frac{0.19}{0.4}$

$=0.475$

Therefore, the probability that a randomly selected traveler who checks work email also uses a cell phone to stay connected is $\boldsymbol{0.475 }$

b) We are required to find:

$P(C|L)=\frac{P(C\cap L)}{P(L)}=\frac{P(L|C)P(C)}{P(L)}=\frac{0.7 \times 0.3}{0.35}=0.6$

Therefore, the probability that someone who brings a laptop on vacation also uses a cell phone to stay connected is 0.6

c) We are required to find here:

$P(C|E\cap L)$

We are given:

$P(E^{'}\cap C^{'}\cap L^{'} )=0.544\Rightarrow P(E\cup C\cup L)^{'}=0.544$

$\therefore P(E\cup C\cup L)=1- P(E\cup C\cup L)^{'}=1-0.544=0.456$

We know that:

$\inline \small P(E\cup C\cup L)=P(E)+P(C)+P(L)-P(E|L)P(L)-P(L|C)P(C)-P(E\cap C)+P(E\cap C\cap L)$

$0.456=0.4+0.3+0.35-0.84 \times 0.35-0.7 \times 0.3-0.19+P(E\cap C\cap L)$

$0.456=0.356+P(E\cap C\cap L)$

$\inline \small P(E\cap C\cap L)=0.10$

$P(C|E\cap L)=\frac{P(E\cap C\cap L)}{P(E|L)P(L)}=\frac{0.1}{0.84 \times 0.35}=0.3401$

Therefore, If the randomly selected traveler checked work email and brought a laptop, the probability that he/she uses a cell phone to stay connected is 0.3401