% useraleonel_r_kafando40Georgiasouthem eduaenecuveo wardstat223112017 (/webwork
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% useraleonel_r_kafando40Georgiasouthem eduaenecuveo wardstat223112017 (/webwork2/wardSTAT2231F2017m user-leonel-r-kafand0NA0GeorgiaSouthern edu&effectiveUser-leonel; r kafando%40GeorgiaSo homework 14 (/webwork2/wardSTAT2231F2017/Homework 14/7 userileoneLr_kafando%40GeorgiaSouthem.edu&effectiveUseraleonel; r kafando%40Georgias Homework 14: Problem 1 Previous Problem List (/webwork2/wa user.leonel r kafand0%40GeorgiaSouthern.edu&effectiveUser-leonel-r-kafando;%40Georgias Next (/webwork2/wardST kafando%40GeorgiaSouthern .edu&effectiveUser-leonel-r-kafando;%40GeorgiaSo user.leonel r (1 point) Government data assign a single cause for each death that occurs in the United States. In a certain city, the data show that the probability is 0.33 that a randomly chosen death was due to cardiovascular (mainly heart) disease, and 0.25 that it was due to cancer. (a) The probability that a death was due either to cardiovascular disease or to cancer is (b) The probability that the death was due to some other cause is Note: You can eam partial credit on this problem. Preview My Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining. Email instructorExplanation / Answer
a) Let A be the event that the death is due to cardiovascular disease, P(A)=0.33
B be the event that the death is due to cancer. P(B) =0.25
The two events are mutually exclusive
P(A or B ) = P(A) + P(B) =0.33+0.25=0.58
b) The probability that the death was due to other cause = 1-0.58=0.42
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