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: 40% check work e-mail, 31% 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 51% neither check work e-mail nor use a cell phone to stay connected nor bring a laptop. In addition 88% of those who bring a laptop also check work e-mail and 71% 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

Solution:

(a) P(E) = 0.4
(b) P(C) = 0.31
(c) P(L) = 0.26
(d) P(E and C) =0.21
(e) P(Ec and Cc and Lc ) =0.51
(f) P(E or C or L) = 0.51
(g) P(E|L) = 0.88
(h) P(L|C) = 0.71
(i) P(E and C and L) =
P(E or C or L) = P(E) +P(L)+P(C)-P(E and L)-P(E and C)-P(L and C) +P(E and C and L)
=>P(E and C and L) = P(E or C or L)-P(E) -P(L)-P(C)+P(E and L)+P(E and C)+P(L and C)
= 0.51-0.4-0.26-0.31+ 0.2288+0.21+0.2201
= 0.1989
(j) P(E and L) = P(E|L) * P(L)
= 0.88*0.26
=0.2288

(k) P(C and L) = E(L|C)*P(C)
= 0.71*0.31
= 0.2201
(l) P(C|(E and L)) = P(C and E and L)/P(E and L)
= 0.1989/0.2288
= 0.8693

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