2 is dd so 100 and the last fraction is the

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or dd (so, 1/2 for either), the second is the probability that II.2 is dd (so, 100%), and the last fraction is the probability that the children are all dd given that the II.1 is either Dd or dd and II.2 is dd.) Now add the two final fractions together: (1/16) + (1/2) = 9/16. This is the “combined” probability that II.1 is Dd or dd. But we only want the probability that II.1 is Dd, so we take the probability that II.1 is Dd divided by the “combined” probability. So: (1/16) / (9/16) = 1/9 Thus the overall probability that II.1 is Dd is 1/9. 2. Expanded Bayes’ Theorem Equation The expanded Bayes’ Theorem equation is:
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P(A|B) = (P(B|A)P(A)) / ((P(B|A)P(A) + P(B|nA)P(nA)) P(A|B) = probability of A given B P(B|A) = probability of B given A P(B|nA) = probability of B given not A P(nA) = probability of not A We need to define A and B. Let A = II.1 is Dd, let B = all 3 children are dd. What do we know for this equation? P(B|A) = probability that all 3 children are dd given II.1 is Dd = (1/2) 3 P(A) = probability that II.1 is Dd = 1/2 P(B|nA) = probability that all 3 children are dd given II.1 is not Dd (aka, is dd instead) = (1) 3 P(nA) = probability that II.1 is not Dd (aka, is dd instead) = 1/2 Note: understanding the nA may be confusing. If A is that II.1 is Dd, then nA is that II.1 is not Dd. If II.1 is not Dd, then II.1 must be dd (refer back to pedigree if confused, II.1 can only be
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