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7.2. Example 2
Load tree effet enfants normaux.arbre
It's with this kind of problem in mind that this program was developed. The demonstration is given as is . I hope not to have let some mistakes in it (let me know, my email is at the end of the help file...). Anyway, the program is kind enough to give results that are in line with the predicted ones.

graphic


Event A= n unaffected children

probability for IIe to be affected knowing that the couple I12 had n unaffected children= p (IIe affected/ A).

P( IIe affected/A) = p(IIe affected and A) / p(A) ( conditionals probabilities).

P(A/ Ile affected) = p(IIe affected and A) / p(IIe affected) (idem)

P(A/IIe affected)=(3/4)n since in this case both members of the couple I12 are necessarily heterozygous .
p(IIe affected)= p(I12 both heterozygous) x 1/4 (the only way to get an affected child)
p(IIe affected)= (a x b) /4
so p(IIe affected and A)=(3/4)n x (a x b) / 4

A event can be split in four mutually excluded events (the I12 couple is composed either of two heterozygous or one heterozygous and one unaffected homozygous or two unaffected homozygous).

p(A)= p(A et I12 both heterozygous) + p[A and (I1 hetero and I2 homo)] + p[ A and (I2 hetero and I1 homo)] +p((A and (I12 homo)

= p(A/I12 both hetero)x p(I12 both hetero)
+ p(A/(I1 hetero and I2 homo)) x p(I1 hetero and I2 homo)
+p(A/(I2 hetero et I1 homo)) x p(21 hetero et I1 homo)
+p(A/I12 both homo)*p(I12 both)
graphic

Applications
Example with a and b=1 (I12 are proven heterozygous)

p(IIe affected)=1/4

Example with a=0.5 et b =1
Number of unaffected children0123456
p(IIe/n unaffected)0.125000.107140.090000.074180.060090.04795 0.03777
Odds : one against781012162025
Example with a=0.5 et b =0.05

Number of unaffected children0123456
p(IIe/n unaffected)0.006250.004720.003550.002680.002010.001510.00114
Odds : one against159211280373497660880