Probability(Bayes rule)

Question

10% of men have disability, 8% of women have disability.

First: Mother doesn't have disability => probability child will have disability is 0,05.

Second: Mother has got disability, father doesn't have disability => probability child will have disability is 0,3.

Third: Mother has got disability, father has got disability => probability child will have disability is 0,8.

Question: Let's say parents have 2 children and first child has disability, about parents I don't know anything(they haven't been tested). What is a probability that second child will have also disability?

P(H1) = 0,92 P(A|H1) = 0,05 for first.
P(H2) = 0,08*0,9= 0,072 P(A|H2) = 0,3 for second.
P(H3) = 0,08*0,1 P(A|H3) = 0,8 for third.

Answer 1

Let $M^*$ be the event of the mother having the disability, $M$ be the event of the mother not having the disability and so on. There are three main possibilities.

$M$ has probability $0.92$.

$M^*\cap F$ has probability $0.072$.

$M^*\cap F^*$ has probability $0.008$.

Probability of event $D$, a child having the disability is:

$0.92\times 0.05+0.08(0.9\times 0.3+0.1\times 0.8)=0.074$.

$p(M|D)=\frac{0.92\times0.05}{0.074}=0.6216$

$p(M^*\cap F|D)=\frac{0.072\times0.3}{0.074}=0.2919$

$p(M^*\cap F^*|D)=\frac{0.008\times0.8}{0.074}=0.0865$

So, the conditional probability of the second child having the disease is

$0.6216\times 0.05+0.2919\times 0.3+.0865\times 0.8=0.1875$.