Answer:
Step-by-step explanation:
Let E = events that a traveler on vacation checks works email
C = events that a traveler on vacation uses a cell phone to stay connected
L = event that a traveler on vacation broughgt a laptop
from the information given ; P(E) = 0.41, P(C) = 0.31, P(L)= 0.25, P(EnC) = 0.22
50% neither check work e-mail nor use a cell phone to stay connected nor bring a laptop= P(EUCUL)' = 0.50
9% of those who bring a laptop also check work e-mail = P(E|L) = 0.89
71% of those who use a cell phone to stay connected also bring a laptop = P(L|C) = 0.71
a) P(E) = 0.41
(b) P(C) = 0.31
(c) P(L) = 0.25
(d) P(EnC) = 0.22
(e) P(E'nC'nL' ) = 0.50
(f) P(EUCUL) = 1 - P(EUCUL)' = 0.50
(g) P(E|L) = 0.89
Also P(EnL) = P(E|L)P(L) = 0.25 X 0.89 = 0.2225
(h) P(L|C) = 0.71
(I) P(EnCnL) = P(EUCUL) + P(E) + P(C) + P(L) - P(EnC) -P(EnL) -P(CnL)
But P(LnC) = P(C)P(L|C) = 0.31 X 0.71 = 0.2201 and P(EnL) = 0.2225
P(EnCnL) = 0.1926
(j) P(EnL) = 0.2225
(k) P(CnL) = 0.2201
(l) P(C|(EnL)) = P(CnEnL)/P(EnL)
= 0.8656
