Question 1182764
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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.