Question

Consider the following information about travelers on vacation: 40% check work email, 30% use a cell phone to stay connected to work, 35% bring a laptop with them, 16% both check work email and use a cell phone to stay connected, and 42.8% neither check work email nor use a cell phone to stay connected nor bring a laptop. In addition, 88 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.
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 email and brought a laptop, what is the probability that he/she uses a cell phone to stay connected?

Answer 1

Answer:

a. 0.4

b. 0.6

c. 0.6493

Step-by-step explanation:

p(checking work email) = p(A) = 0.40

p(staying connected with cell phone) = p(B) = 0.30

p(having laptop) = p(c) = 0.35

p(checking work mail and staying connected with cell phone) = p(AnB) = 0.16

p(neither A,B or C) = p(AuBuC)

= 1-42.8%

= 0.572

p(A|C) = 88% = 0.88

p(C|B) = 70% = 0.7

a. What is the probability that a randomly selected traveler who checks work email also uses a cell phone to stay connected?

p(B|A) = p(AnB)/p(A)

= 0.16/0.4

= 0.4

b. What is the probability that someone who brings a laptop on vacation also uses a cell phone to stay connected?

p(B|C) = P(C|B)p(B)/p(C)

= 0.7x0.3/0.35

= 0.6

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

p(A|BnC)

= P(BnAnC)/p(AnC)

= p(AnC) = p(A|C).p(C)

= 0.88x0.35

= 0.308

p(AnBnC) = p(AuBuC)-p(a)-p(b)+ p(AnB)+p(AnC)+p(BnC)

p(BnC) = 0.7x0.3

= 0.21

p(AnBnC) = 0.572-0.4-0.3-0.35+0.16+0.308+0.21

= 0.2

p(A|BnC) = 0.2/0.308

= 0.6493