Question 1182764: Problem 6 (20%)
Suppose that the prevalence of the Alzheimer’s disease is given by the following table in function of the sex and age range:
Age group; Males; Females
65-69; 0.01; 0
70-74; 0; 0.02
75-79; 0.05; 0.03
80-84; 0.09; 0.08
85+; 0.35; 0.28
Suppose an unrelated 77- year- old man, 76-year-old woman and 82-year-old woman are selected from the community. We denote by:
A: a 77 year- old man has Alzheimer; B: a 76- year-old woman has Alzheimer; C: an 82- year- old woman has Alzheimer, and we suppose that A, B and C are independent.
a- Define p(A), P(B), P(C)
b- Consider the event D: “Exactly one of the two women has Alzheimer”. Write D in terms of A, B and C in two methods.
c- What is the probability that all the three of these individuals have Alzheimer?
d- What is the probability that at least one of the women has Alzheimer?
e- Suppose we know that one women has Alzheimer but we don’t know which one. What is the probability that she is younger than 80 years of age?
f- Suppose we know two of the three people has Alzheimer but we don’t know which ones. What is the conditional probability that they are both younger than 80 years of age?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! ======M==F
65-69; 0.01; 0
70-74; 0; 0.02
75-79; 0.05; 0.03
80-84; 0.09; 0.08
85+; 0.35; 0.28
p(A) probability is 0.05 for a man between 75 and 79
p(B) probability is 0.03 for a woman
p(C) probability is 0.08
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One of the two women has Alzheimers': probability is 0.03*0.92 for one case (the second doesn't have it)=0.0276.
and 0.08* 0.97=0.0776 for the second. So the answer is the sum or 0.1052
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All three would be the product of 0.05*0.03*0.08=0.0012.
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The probability at least one of the women has it is 1- probability both don't have it, which is 1-(0.97*0.92)=0.1076. It is also the p(1) above which is 0.1052+ probability of both, which is 0.024, and that sum is 0.1076.
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The probability that one woman has Alzheimer's is 0.1052
The conditional probability given that the woman is younger than 80, which has probability 0.03, is
0.03/0.1052=0.2852
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Probability 2 of the 3 have it:
P(A,B, not C) which is 0.05*0.03*0.92=0.00138
P (A, C not B) which is 0.05*0.08*0.97=0.00388
P (B, C not A) which is 0.03*0.08*0.95=0.00228
That sum is 0.00754
the probability they are both younger than 80 is 0.00526
that quotient, 0.00526/0.00754=0.6976 is the conditional probability.
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