Q:

Bayesian belief network

A child inherits a gene X with probability 50%. A disease will develop if child inherited gene from both parents. The disease will not develop if child got gene from just one of parents.
Jain and Max have 0.25 probability of having that gene. They are parents of two children Mark and Eva. What is probability that Eva is healthy? What probability that Eva is healthy given Mark is healthy?

A:

Let's start with the first question.
$$P(\text{Eva healthy}) = 1 - P(\text{Eva unhealthy})$$
Here we'll use law of total probability
$$P(\text{Eva unhealthy}) = P(\text{Both parents pass gene onto Eva}|\text{Both parents have the gene}).P(\text{Both parents have the gene})$$
$$P(\text{Eva healthy}) =1- 0.5^{2}0.25^{2}=\frac{63}{64}$$
Now for the second question we have more information.
The conditional will remain unchanged 
$$P(\text{Both parents pass gene onto Eva}|\text{Both parents have the gene}) = 0.5^{2}$$
However, now that we know that Mark is healthy, our belief of the parents having the gene decreases. So now,
$$P(\text{Eva unhealthy}) = P(\text{Both parents pass gene onto Eva}|\text{Both parents have the gene}).P(\text{Both parents have the gene}|\text{Mark is healthy})$$
$$P(\text{Both parents have the gene}|\text{Mark is healthy}) = \frac{3*0.25^{2}*0.5^{2}}{3*0.25^{2}*0.5^{2} + 2*0.25*0.75 + 0.75^{2}}$$
$$P(\text{Both parents have the gene}|\text{Mark is healthy}) = \frac{1}{21}$$
Consequently,
$$P(\text{Eva healthy}|\text{Mark healthy}) =1- 0.5^{2}\frac{1}{21}=\frac{83}{84}$$

