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
Answer:
which ever one you want its an opinion answer.
Step-by-step explanation:
