Question

66. Consider the following information about travelers on vacation (based partly on a recent Travelocity poll): 40% check work email, 30% use a cell phone to stay connected to work, 25% bring a laptop with them, 23% both check work email and use a cell phone to stay connected, and 51% neither check work email nor use a cell phone to stay connected nor bring a laptop. In addition, 88 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.

QUESTION is for part C)

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

click the link below. how is it that P(AUBUC) = 1-0.51 = 0.49 ?????
How if P(AUB) = P(A) + P(B) ??????

They do NOT add up

https://www.chegg.com/homework-help/population-particular-country-consists-three-ethnic-groups-i-chapter-2-problem-66E-solution-9781305251809-exc

Answer 1

Let A denote the event that a traveller checks work email Let B denote the event that a traveller uses a cell phone Let C denote the event that a traveller brings a laptop. From the given information, P(A)=0.40;P(B)=0.30 and P(C)=0.25 P( A B)=0.23;P(A UBUC)=1-0.513 0.49 P(A|C)=0.88 and P(C|B) = 0.70 Calculate the value of the P(A B C) [P(AUBUC)- P(A)- P(B) – P(C)+1 P(A,BC)= P(An B)+P(ANC)+P(BOC) = [0.49 – 0.40–0.30 – 0.25+0.23+0.22 +0.21] = 0.20

C) If the randomly selected traveller checked work email and brought a laptop, what is the probability that he/she uses a cell phone to stay connected. P(BA P( BAC) PA nC) 0.20 P(A|C)P(C) 0.20 0.88x0.25 0.20 0.22 = 0.909 Hence, the required probability is 0.909).