# Bayes' Theorem application to medicine.

Here's a disease D. Its probability of happening in general population is 3%. P(D)=0.03 If parent has D, so the chance that children will have D is 15%. P(P|C)=0.15 Parent has 2 children.One child developed D.

What is probability that parent has D as well? What is probability that another child develops D?

My solution for first question:

P(D)=0.03; P(P|C)=0.15; also P(C)=0.03 and P(P)=0.03; $P(\overline C)$=0.97 $P(P|\overline C)$=0.5 => P(C|P)=P(P|C)*P(C)/(P(P|C)*P(C) + $P(\overline C)$*$P(P|\overline C)$)=(0.15*0.03)/(0.15*0.03+0.97*0.5)=0.00919

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What are your thoughts on this problem? –  dmk Feb 23 '14 at 21:50
First question is ~1% –  user131014 Feb 23 '14 at 21:53
Second question is how to find P(C1|C2&P) or $P(C1|\overline P&C2)$ –  user131014 Feb 23 '14 at 22:10
P(C|P)=0.15 and not P(P|C). Am I correct? –  Stef Feb 23 '14 at 22:16
Stefanos, P(P|C)=0.15 if parent had D so 15% chance that children will have D –  user131014 Feb 23 '14 at 22:49