Bayes Theorem

# Fill in the chart probability ii1 is dd ii1 is dd

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Fill in the chart: Probability II.1 is Dd II.1 is dd Prior 1/2 1/2 Condition (1/2) 3 (1) 3 Joint (1/2)(1/2) 3 (1/2)(1) 3 Posterior ((1/2)(1/2) 3 ) / ((1/2)(1/2) 3 + (1/2)(1) 3 ) Since we’re only interested in when II.1 is Dd, we don’t have to calculate bd/(ac+bd). Thus our final answer is the Posterior probability for Event A (II.1 is Dd). ac/(ac+bd) = ((1/2)(1/2) 3 ) / ((1/2)(1/2) 3 + (1/2)(1) 3 ) = (1/16) / ((1/16) + (1/2)) = 1/9 Thus the overall probability that II.1 is Dd is 1/9. Final Note: If you’re unclear how the probability for all 3 children being dd is calculated, you should do two Punnett squares between II.1 and II.2. In one Punnett square, you’re assuming that II.1 is Dd and in the other you’re assuming that II.1 is dd. So, if II.1 is Dd, then the cross is Dd x dd. Thus the independent probability that each child is dd is 1/2. Thus the probability that 3 children will be dd is (1/2) 3 . So, if II.1 is dd, then the cross is dd x dd. Thus the independent probability that each child is dd is 1. Thus the probability that 3 children will be dd is (1) 3 .
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