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(AB) = (P(BA)P(A)) / ((P(BA)P(A) + P(BnA)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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 Winter '11
 Kumar
 Genetics, Conditional Probability, Probability, Theorem Equation

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