Consider the following information about passengers on a cruise ship on vacation

Question

Consider the following information about passengers on a cruise ship on vacation: 39% check work e-mail, 29% use a cell phone to stay connected to work, 26% bring a laptop with them on vacation, 21% both check work e-mail and use a cell phone to stay connected, and 52% neither check work e-mail nor use a cell phone to stay connected nor bring a laptop. In addition 89% of those who bring a laptop also check work e-mail and 70% of those who use a cell phone to stay connected also bring a laptop. With

E = event that a traveler on vacation checks work e-mail
C = event that a traveler on vacation uses a cell phone to stay connected
L = event that a traveler on vacation brought a laptop

use the given information to determine the following probabilities. A Venn diagram may help. (Round all answers to four decimal places.)

(a)    P(E) =  

(b)    P(C) =  

(c)    P(L) =  

(d)    P(E and C) =  

(e)    P(Ecand Ccand Lc ) =  

(f)    P(E or C or L) =  

(g)    P(E|L) =  

(h)    P(L|C) =  

(i)    P(E and C and L) =  

(j)    P(E and L) =  

(k)    P(C and L) =  

(l)    P(C|(E and L)) =  

Explanation / Answer

a) P(E) = 0.39

b) P(C) = 0.29

c) P(L) = 0.26

d) P(E and C) = 0.21

e) P(Ec and Cc and Lc) = 0.52

f) P(E or C or L) = 1 - 0.52 = 0.48

g) P(E|L) = 0.89

h) P(L|C) = 0.7

i) P(E and C and L) = P(E or C or L) - P(E) - P(C) - P(L) + P(E and C) + P(E and L) + P(C and L)

                                = 0.48 - 0.39 - 0.29 - 0.26 + 0.21 + 0.2314 + 0.203 = 0.1844

j) P(E and L) = P(E|L) * P(L) = 0.89 * 0.26 = 0.2314

k) P(C and L) = P(L|C) * P(C) = 0.7 * 0.29 = 0.203

l) P(C| (E and L) = P(C and E and L)/P(E and L)

                           = 0.1844/0.2314 = 0.7969

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