P(1st) = 0.049 P(not 1st) = 0.951 P(2nd) = 0.023 P(not 2nd) = 0.977 P(3rd) = 0.0
ID: 3048785 • Letter: P
Question
P(1st) = 0.049 P(not 1st) = 0.951
P(2nd) = 0.023 P(not 2nd) = 0.977
P(3rd) = 0.078 P(not 3rd) = 0.922
a. What is the probablility that all three subjects have Alzheimer's disease?
b. What is the probability that at least one of the randomly selected women has Alzheimer's disease?
Below is a table showing the percentage of elderly people, subdivided by age and gender, who develop Alzheimer's disease (A.D.) in the United States. An example of how to read the table is "1.6% of 65-69 year old Males have Alzheimer's disease. (adapted from Fundamentals of Biostatistics, Rosner 2011): Age Group 65-69 70-74 75-79 80-84 Males with AD 1.6% 0.0% 4.9% 8.6% Females with AD 0.0% 2.2% 2.3% 7.8% Three elderly subjects are randomly selected and screened for Alzheimer's disease Subject 1 Male, age 77 Subject 2 Female, age 76 Subject 3 -Female, age 82Explanation / Answer
a) probablility that all three subjects have Alzheimer's disease =0.049*0.023*0.078 =0.000088
b) probability that at least one of the randomly selected women has Alzheimer's disease =1-P(none have Alzheimer's disease) =1-(1-0.023)*(1-0.078) =0.099206
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