What is the probability that Mary’s son, Bob, inherited Huntington's Disease gene?
|
|
Carrier |
Non-carrier |
|
Prior |
1/2 |
1/2 |
|
Conditional |
1/2 |
1 |
|
Joint |
1/4 |
1/2 |
|
Posterior |
1/4 / (1/4 + 1/2) = 1/3 |
1/2 / (1/4 + 1/2) = 2/3 |
The probability of Mary being a carrier of Huntington's Disease is 1/3
The probability of Mary having an unaffected son is 1/2
We multiply these probabilities to get:
1/3 x 1/2 = 1/6 (the probability that Bob inherited the Huntington's Disease gene)