What is the probability that Mary’s son, Bob, inherited the HD gene?
|
Probability |
A |
B |
|
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 that Mary's son Bob inherited the HD gene = 1/3 (posterior probability that Mary has mutation) x 1/2 (Probability that Bob inherits the HD allele) = 1/6
The probability that Mary's son Bob inherited the HD gene is 1/6