2. 1.96/5.88 points | Previous Answers DevoreStat9 2.E.066. My Notes Ask Your Te

Question

2. 1.96/5.88 points | Previous Answers DevoreStat9 2.E.066. My Notes Ask Your Teacher Consider the following information about travelers on vacation: 40% check work email, 30% use a cell phone to stay connected to work, 25% bring a laptop with them, 22% both check work email and use a cell phone to stay connected, and 59% neither check work email nor use a cell phone to stay connected nor bring a laptop. In addition, 84 out of every 100 who bring a laptop also check work email, 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 email also uses a cell phone to stay connected? 0.55 (b) What is the probability that someone who brings a laptop on vacation also uses a cell phone to stay connected? 2.8 (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? Round your answer to four decimal places.) Need Help? Read It Talk to a Tutor Submit Answer Save Progress Practice Another Version

Explanation / Answer

Ans:

a)P(cell phone/email)=P(cell phone and email)/P(email)=0.22/0.4=0.55

b)P(cell phone/laptop)=P(cell phone and laptop)/P(laptop)

=P(laptop/cell phone)*P(cell phone)/P(laptop)

=0.7*0.3/0.25=0.21/0.25=0.84

c)P(C/E and L)=P(C and E and L)/P(E and L)

Now,P(E and L)=P(E/L)*P(L)=0.84*0.25=0.21

P(C or E or L)=1-0.59=0.41

P(C or E or L)=0.41=P(C)+P(E)+P(L)-P(C and E)-P(E and L)-P(L and C)+P(C and E and L)

0.41=0.4+0.3+0.25-0.22-0.21-0.7*0.3+P(C and E and L)

P(C and E and L)=0.41-0.31=0.1

hence,

P(C/E and L)=P(C and E and L)/P(E and L)=0.1/0.21=0.4762

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