Question

Consider the following information about travelers (based partly on a recent Travelocity poll): 40% check work e-mail, 30% use a cell phone to stay connected to work, 25% bring a laptop with them, 23% both check work e-mail and use a cell phone to stay connected, and 51% neither check work e-mail nor use a cell phone to stay connected nor bring a laptop. Finally, 88 out of every 100 who bring a laptop check work e-mail, 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 e-mail 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 a randomly selected traveler checked work e-mail and brought a laptop, what is the probability that s/he uses a cell phone to stay connected?

Answer 1

Step-by-step explanation:

I defined the following events:

J = travellers check email

K = they use a cell phone to stay connected

M = they have a laptop alongside

P(J) = 0.40

P(K) = 0.30

P(M) = 0.25

P(J n K) = 0.23

P(J U K U M)' = 0.51

P(J|M) = 0.88

P(M|K) = 0.70

A.

We are to find probability of K|J

P(JnK)/p(J)

= 0.23/0.40

= 0.575

B.

P(K|M)

= P(KnM)/P(m)

= 0.21/0.25

= 0.84

C.

P(J U K U M) = 1 - p(J U K U M)'

= 1-0.51

= 0.49

P(JnM) = p(J|M)p(M)

=0.88x0.25

= 0.22

P(JnKnM) = 0.49-0.40-0.30-0.25+0.23+0.22+0.21

= 0.20

P(K|JnM) = P(JnKnM)/p(JnM)

= 0.20/0.22

= 0.909