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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 Winter '11
 Kumar
 Genetics, Conditional Probability, Probability, Theorem Equation

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