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1 is not dd then ii1 must be dd refer back to

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If II.1 is not Dd, then II.1 must be dd (refer back to pedigree if confused, II.1 can only be Dd or dd; so if not Dd, then II.1 must be dd). Plug in all the values into the equation: P(A|B) = (P(B|A)P(A)) / ((P(B|A)P(A) + P(B|nA)P(nA)) = ((1/2) 3 (1/2)) / ((1/2) 3 (1/2) + (1) 3 (1/2)) = (1/16) / ((1/16) + (1/2)) = 1/9 Thus the overall probability that II.1 is Dd is 1/9. 3. Genetic Counseling Method The genetic counseling students here at UM have a really cool and systematic way of dealing with Bayes’ Theorem. They utilize the following chart: Probability Event A Event B Prior a b Condition c d Joint ac bd Posterior ac/(ac+bd) bd/(ac+bd) We need to define what Events A and B are. Let A = II.1 is Dd, and B = II.1 is dd. Event B is basically Event “ not A” (nA). Before we fill in the chart, we need to know what Prior probability and Condition probability are. The rest of the chart is straightforward. Prior probability = the “original” probability that II.1 is either Dd or dd (depending on which event you’re looking at).
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Condition probability = the additional condition that modifies the prior probability. In this problem, the condition is having 3 children who’re all dd.
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